I have the following array of bytes:
01010110 01110100 00100101 01001011
These bytes are broken into two groups to encode seven integers. I know that the first group consists of 3 values 4 bits each (0101 0110 0111) that represent numbers 5,6,7. The second group consists of 4 values 5 bits each (01000 01001 01010 01011), which represent integers 8,9,10, and 11.
To extract the integers, I am currently using the following approach. Convert the array into a binary string:
public static String byteArrayToBinaryString(byte[] byteArray)
{
String[] arrayOfStrings = new String[byteArray.length];
for(int i=0; i<byteArray.length; i++)
{
arrayOfStrings[i] = byteToBinaryString(byteArray[i]);
}
String bitsetString = "";
for(String testArrayStringElement : arrayOfStrings)
{
bitsetString += testArrayStringElement;
}
return bitsetString;
}
// Taken from here: http://helpdesk.objects.com.au/java/converting-large-byte-array-to-binary-string
public static String byteToBinaryString(byte byteIn)
{
StringBuilder sb = new StringBuilder("00000000");
for (int bit = 0; bit < 8; bit++)
{
if (((byteIn >> bit) & 1) > 0)
{
sb.setCharAt(7 - bit, '1');
}
}
return sb.toString();
}
Then, I split the binary string into 2 substrings: 12 characters and 20 characters. Then I split each substring into new substrings, each of which has length that equals the number of bits. Then I convert each sub-substring into an integer.
It works but a byte array representing thousands of integers takes 30 seconds to a minute to extract.
I am a bit at a loss here. How do I do this using bitwise operators?
Thanks a lot!
I assume you have an understanding of the basic bit operations and how to express them in Java.
Use a pencil to draw a synthetic picture of the problem
byte 0 byte 1 byte 2 byte 3
01010110 01110100 00100101 01001011
\__/\__/ \__/\______/\___/\______/\___/
a b c d e f g
To extract a, b and c we need to do the following
a b c
byte 0 byte 0 byte 1
01010110 01010110 01110100
\. \. |||||||| \. \.
'\ '\ XXXX|||| '\ '\
0.. 0101 0.. 0110 0.. 0111
Shift And Shift
In Java
int a = byteArray[0] >>> 4, b = byteArray[0] & 0xf, c = byteArray[1] >>> 4;
The other values d, e, f and g are computed similarly but some of them require to read two bytes from the array (d and f actually).
d e
byte 1 byte 2 byte 2
01110100 00100101 00100101
||||\\\\ | |\\\\\
XXXX \\\\ | X \\\\\
\\\\| \\\\\
0.. 01000 01001
To compute d we need to isolate the least four bits of byte 1 with byteArray[1] & 0xf then make space for the bit from byte 2 with (byteArray[1] & 0xf) << 1, extract that bit with byteArray[1] >>> 7 and finally merge together the result.
int d = (byteArray[1] & 0xf) << 1 | byteArray[2] >>> 7;
int e = (byteArray[2] & 0x7c) >>> 2;
int f = (byteArray[2] & 0x3) << 3 | byteArray[3] >>> 5;
int g = byteArray[3] & 0x1f;
When you are comfortable with handling bits operations you may consider generalizing the function that extract the integers.
I made function int extract(byte[] bits, int[] sizes, int[] res), that given an array of bytes bits, an array of sizes sizes, where the even indices hold the size of the integers to extract in bits and the odd indices the number of integers to extract, and an output array res large enough to hold all the integers in output, extracts from bits all the integers expressed by sizes.
It returns the number of integers extracted.
For example the original problem can be solved as
int res[] = new int[8];
byte bits[] = new byte[]{0x56, 0x74, 0x25, 0x4b};
//Extract 3 integers of 4 bits and 4 integers of 5 bits
int ints = BitsExtractor.extract(bits, new int[]{4, 3, 5, 4}, res);
public class BitsExtractor
{
public static int extract(byte[] bits, int[] sizes, int[] res)
{
int currentByte = 0; //Index into the bits array
int intProduced = 0; //Number of ints produced so far
int bitsLeftInByte = 8; //How many bits left in the current byte
int howManyInts = 0; //Number of integers to extract
//Scan the sizes array two items at a time
for (int currentSize = 0; currentSize < sizes.length - 1; currentSize += 2)
{
//Size, in bits, of the integers to extract
int intSize = sizes[currentSize];
howManyInts += sizes[currentSize+1];
int temp = 0; //Temporary value of an integer
int sizeLeft = intSize; //How many bits left to extract
//Do until we have enough integer or we exhaust the bits array
while (intProduced < howManyInts && currentByte <= bits.length)
{
//How many bit we can extract from the current byte
int bitSize = Math.min(sizeLeft, bitsLeftInByte); //sizeLeft <= bitsLeftInByte ? sizeLeft : bitsLeftInByte;
//The value to mask out the number of bit extracted from
//The current byte (e.g. for 3 it is 7)
int byteMask = (1 << bitSize) - 1;
//Extract the new bits (Note that we extract starting from the
//RIGHT so we need to consider the bits left in the byte)
int newBits = (bits[currentByte] >>> (bitsLeftInByte - bitSize)) & byteMask;
//Create the new temporary value of the current integer by
//inserting the bits in the lowest positions
temp = temp << bitSize | newBits;
//"Remove" the bits processed from the byte
bitsLeftInByte -= bitSize;
//Is the byte has been exhausted, move to the next
if (bitsLeftInByte == 0)
{
bitsLeftInByte = 8;
currentByte++;
}
//"Remove" the bits processed from the size
sizeLeft -= bitSize;
//If we have extracted all the bits, save the integer
if (sizeLeft == 0)
{
res[intProduced++] = temp;
temp = 0;
sizeLeft = intSize;
}
}
}
return intProduced;
}
}
Well I did the first group , the second can be done in similar fashion
public static void main(String args[]) {
//an example 32 bits like your example
byte[] bytes = new byte[4];
bytes[0] = 31;//0001 1111
bytes[1] = 54;//0011 0110
bytes[2] = 67;
bytes[3] = 19;
//System.out.println(bytes[0]);
int x = 0;
int j = -1; // the byte number
int k = 0; // the bit number in that byte
int n = 0; // the place of the bit in the integer we are trying to read
for (int i = 0; i < 32; i++) {
if (i < 12) { //first group
if (i % 8 == 0) {
j++;
k = 0;
}
if (i % 4 == 0) {
x = 0;
n = 0;
}
byte bit = (byte) ((bytes[j] & (1 << (7 - k))) >> (7 - k));
System.out.println("j is :" + j + " k is :" + k + " " + bit);
x = x | bit << (3 - n);
if ((i + 1) % 4 == 0) {
System.out.println(x);
}
k++;
n++;
} else {
}
}
}
It's a bit tricky because you are trying to encode an integer on less than what java allocates (8 bits). So I had to take each bit and "construct" the int from them
To get each bit
byte bit = (byte) ((bytes[j] & (1 << (7 - k))) >> (7 - k));
this takes the byte we are at and does And operation. For example I want the 3rd bit of the 1st byte, I do
bytes[0] & 1 << (7 - 3)
but this gives me an integer encoded over 8 bits, so I still have to shift it to get that single bit with >> (7 - 3)
Then I just Or it with x (the int we are trying to decode). All while putting it at the right position with << (3 - n) . 3 because your integer is encoded over 4 bits
Try running the code and reading the output.
I am honestly not sure if this is the best way, but I believe it's at least faster than dealing with Strings
I'm trying to write a program which takes as arguments a number of digits and a base, and counts upward through the numbers that have their nonzero digits in ascending order. For instance, in base 4 with 3 digits, it should print:
000 001 002 003 010 011 012 013 020 022 023 030 033 100 101 102 103
110 111 112 113 120 122 123 130 133 200 202 203 220 222 223 230 233
300 303 330 333
and in base 3 with 4 digits it should print:
0000 0001 0002 0010 0011 0012 0020 0022 0100 0101 0102 0110 0111 0112
0120 0122 0200 0202 0220 0222 1000 1001 1002 1010 1011 1012 1020 1022
1100 1101 1102 1110 1111 1112 1120 1122 1200 1202 1220 1222 2000 2002
2020 2022 2200 2202 2220 2222
I have done this successfully, but my algorithm seems unnecessarily complicated and time-consuming (time is very important for my application). Is there any way of either making it faster, or simplifying it if the speed cannot be improved?
Here is the program:
public static void count(int base, int size)
{
int[] array = new int[size];
print(array); // private print method prints out the array
int index = 0;
while (index < array.length)
{
if (array[index] < base - 1)
{
// check whether we need to increase array[index] by extra to maintain the order
if (array[index] == 0)
{
int i;
// search for the next nonzero digit
// this search seems to take unnecessary time; is there a faster alternative?
for (i = index + 1; i < array.length && array[i] == 0; i++);
// check whether there was, in fact, some later nonzero digit
if (i < array.length) array[index] = array[i];
else array[index] = 1;
}
else array[index]++;
print(array);
index = 0;
}
// carry over to the next digit
else array[index++] = 0;
}
}
I would go for a recursive solution:
public static void count(int base, int size) {
int[] configuration = new int[size];
placeDigits(configuration, base, 0, 1);
}
public static void placeDigits(int[] configuration, int base, int pos, int minNonZero) {
if (pos >= configuration.length) {
print(configuration);
} else {
// 0 is a possible candidate
configuration[pos] = 0;
placeDigits(configuration, base, pos + 1, minNonZero);
// digits between minNonZero and base
for (int d = minNonZero; d < base; d++) {
configuration[pos] = d;
placeDigits(configuration, base, pos + 1, d);
}
}
}
It places digits one after the other into the array and observes the constraint that the non-zero digits must be non decreasing.
Okay, this is a bit of a cheat, but here's a solution expressed in pseudocode:
results : list
for i in 1..max
if '0' not in str(i)
append i to results
fi
rof
print results
On the other hand, is it a cheat? "numbers with nonzero digits" is inherently a question about the decimal representation of the numbers, not in some sense the numbers themselves.
Time complexity is O(n) of course -- at least counting str(i) as a single step, which is where it is a little bit of a cheat.
Just for fun, here's the same solution in Python:
print [i for i in xrange(max) if '0' not in str(i)]
And a sketch of a recursive solution:
Let dig be a list of the nonzero digits, i.e., ['1','2','3','4','5','6','7','8','9']. Enumerate all strings on that list of length ceil(log10(max)) (quiz question, why that limit?).
Print those strings in order, stopping when max is exceeded.
If you don't mind keeping the numbers in memory, you could code the following algorithm:
Start with the numbers 0,1...base-1
For each added digit, d, first add zero, then all previous numbers that begin with digits d or higher (indexing those by starting digit and number of digits, you could access them directly).
Or, as some like to phrase, dp style: Let dp[i][j] represent the sequence of numbers with i digits and left-most digit j. Then dp[i][j] = [d] ++ map (d +) dp[l][k], for all l < i and k >= j, where d = j * 10 ^ (i - 1)
(I borrowed the ++ from Haskell, where it often means to concat lists).
For example, base 4, 3 digits:
Start with one digit:
0,1,2,3
Add to the second digit from the first sequence:
10,11,12,13
20,22,23
30,33
Third digit, add from all previous sequences:
100,101,102,103
110,111,112,113
120,122,123
130,133
200,202,203
220,222,223
230,233
300,303
330,333
JavaScript code:
var base = 4;
var dp = [,[]];
for (var j=0; j<base; j++){
dp[1][j] = [j];
}
for (var i=2; i<4; i++){
dp[i] = [];
for (var j=1; j<base; j++){
var d = j * Math.pow(10,i - 1);
dp[i][j] = [d];
for (var l=1; l<i; l++){
for (var k=j; k<base; k++){
dp[i][j] = dp[i][j].concat(
dp[l][k].map(function(x){
return d + x;
}));
}
}
}
}
console.log(JSON.stringify(dp))
/*
[null,[[0],[1],[2],[3]]
,[null,[10,11,12,13]
,[20,22,23]
,[30,33]]
,[null,[100,101,102,103,110,111,112,113,120,122,123,130,133]
,[200,202,203,220,222,223,230,233]
,[300,303,330,333]]]
*/
Late to the party for this faster answer:
Base 8
Size 20 digits
Current solution: 79 seconds (76~82)
Solution below: 23 seconds (22~24)
Possible numbers: 12245598208
without prints. Principle:
The rule "a digit may be followed by a 0 or a digit >= preceding ones" is also valid for (valid) groups of digits: "a group may be followed by a group of zeroes, or a group which smaller digit is >= any of the preceding ones among the preceding groups". Processing is done at the group level, rather than at the digit level.
Given T total size, and N smaller number of digits in each group (T % N == 0), by calculating all possible groups of N digits they can then be assembled together (T / N groups per solution).
pre-calculate all possible digits on a smaller size, eg 5 (2668 numbers), in an array (takes less than half a second)
keep the maximum digit for each of the "parts" in another array
set in another "atleast" array the indexes of groups based on their smaller digit
build the large numbers by sticking all possible chunks (eg 4x5), provided that the lower digit of a group has to be >= highest of the preceding groups.
Sample code to precalculate the small chunks (parts)
static ArrayList<int[]> parts = new ArrayList<int[]>();
static ArrayList<ArrayList<Integer>> atleast = new ArrayList<ArrayList<Integer>>();
static ArrayList<Integer> maxi = new ArrayList<Integer>();
static int stick[];
static int base;
static long num = 0;
public static void makeParts(int min, int ptr)
{
int me = 0;
do {
array[ptr] = me;
if (ptr > 0) makeParts(Math.max(me,min), ptr-1);
else {
// add part
int[] newa = new int [array.length];
int i,mi,ma,last=array.length-1;
for (i=0 ; i<array.length ; i++) newa[i] = array[i];
parts.add(newa);
// maxi
for (i=0 ; i<=last && newa[i]==0 ; i++) /* */;
maxi.add(ma = i<=last ? newa[i] : 0);
// mini
for (i=last ; i>=0 && newa[i]==0 ; i--) /* */;
mi = i>=0 ? newa[i] : 0;
// add to atleast lists
int pi = parts.size() - 1;
ArrayList<Integer> l;
int imi = mi == 0 ? base-1 : mi;
for (i=0 ; i<=imi ; i++) {
if (i < atleast.size()) l = atleast.get(i);
else {
l = new ArrayList<Integer>();
atleast.add(i, l);
}
l.add(pi);
}
}
me = me == 0 ? (min > 0 ? min : 1) : me+1;
} while (me < base);
}
Sticking the "parts"
public static void stickParts(int minv, int ptr)
{
// "atleast" gives indexes in "parts" of groups which min digit
// is at least "minv" (or only zeroes)
for (int pi: atleast.get(minv)) {
stick[ptr] = pi;
if (ptr > 0) {
stickParts(Math.max(minv,maxi.get(pi)), ptr-1);
}
else {
// count solutions
// the number is made of "parts" from indexes
// stored in "stick"
num++;
}
}
}
Calling this in "main"
base = 8;
int leng = 20;
int pleng = 4;
array = new int [pleng];
makeParts(0,array.length-1);
num = 0;
stick = new int [leng / pleng];
stickParts(0, (leng/pleng) - 1);
out.print(String.format("Got %d numbers\n", num));
If T (total size) is prime, for instance, another specific group has to be calculated, eg for size 17, we could have 3 groups (of 5 digits) + one group of two digits.
Quite an interesting program you have written.
I've tried to increase the performance of the nested search, but so far I haven't found a way to make the worst-case scenario of searching for the next nonzero digit less than O(n).
In the worst-case scenario, the subarray A[i..array.length-1] is not sorted, and array[i] = 0,therefore to find the next non-zero digit, you have to do a linear search.
Aditionally, if there is no next non-zero digit, you have to search the whole array to "find it".
(For example: we have that i = 1 for the sequence '0040'. The subarray [0, 4, 0] is not sorted, so you have to do a linear search to find the next largest/smallest nonzero digit, which would be located in array[2])
The complexity for the worst case will be O(n).
Can you improve running time? I guess you can if you do some parallel programming, but I have no knowledge of that field to help you, unfortunately.
This recursive function tries to avoid any unnecessary loop
public static void count0(int min, int ptr)
{
int me = 0; // joker
do {
array[ptr] = me;
if (ptr > 0) count0(Math.max(me,min), ptr-1);
else print(array);
me = me == 0 ? (min > 0 ? min : 1) : me+1;
} while (me < base);
}
Called like this (base 8 for length of 17) to carry less arguments:
static int array[];
static int base;
int leng = 17;
base = 8;
array = new int [leng];
count0 (0, array.length-1);
Recursivity has its price, though.
I didn't measure performance, but think my code is better readable.
The idea is, to produce every number of base b and length l by Integer-iteration from 0 to the known number in decimal, using the Java-build-in conversion decimal to base b, then removing the zeros in that number (which is of type String) and testing for ascending order.
The output has to be padded with zeros, so therefore the complicated printf in the end.
public static boolean isAscending (String digits) {
for (int i = 1; i < digits.length (); ++i)
if (digits.charAt (i-1) > digits.charAt (i))
return false;
return true;
}
public static void count (int base, int size)
{
/**
Build numbers,i.e. from 000 to 333, for base 4 at length 3
or 4^3 = 4*4*4 = 64 combinations
*/
double max = Math.pow (base, size);
for (int i = 0; i < max; ++i)
{
String res = Integer.toString (i, base);
if (isAscending (res.replaceAll ("0", "")))
System.out.printf ("%0"+size+"d ", Long.parseLong (res));
}
}
I am calculating the int equivalent of a given set of bits and storing that in memory. From there, I would like to determine all 1 value bits from the original bitmask. Example:
33 --> [1,6]
97 --> [1,6,7]
Ideas for an implementation in Java?
On BitSet
Use java.util.BitSet to store, well, a set of bits.
Here's how you can convert from an int to a BitSet, based on which bits in the int is set:
static BitSet fromInt(int num) {
BitSet bs = new BitSet();
for (int k = 0; k < Integer.SIZE; k++) {
if (((num >> k) & 1) == 1) {
bs.set(k);
}
}
return bs;
}
So now you can do the following:
System.out.println(fromInt(33)); // prints "{0, 5}"
System.out.println(fromInt(97)); // prints "{0, 5, 6}"
And just for completeness, here's the reverse transformation:
static int toInt(BitSet bs) {
int num = 0;
for (int k = -1; (k = bs.nextSetBit(k + 1)) != -1; ) {
num |= (1 << k);
}
return num;
}
So composing both together, we always get back the original number:
System.out.println(toInt(fromInt(33))); // prints "33"
System.out.println(toInt(fromInt(97))); // prints "97"
On 0-based indexing
Note that this uses 0-based indexing, which is the more commonly used indexing for bits (and most everything else in Java). This is also more correct. In the following, ^ denotes exponentiation:
33 = 2^0 + 2^5 = 1 + 32 97 = 2^0 + 2^5 + 2^6 = 1 + 32 + 64
33 -> {0, 5} 97 -> {0, 5, 6}
If you insist on using 1-based indexing, however, you can use bs.set(k+1); and (1 << (k-1)) in the above snippets. I would advise strongly against this recommendation, however.
Related questions
What does the ^ operator do in Java? -- it's actually not exponentiation
For bit fiddling, java.lang.Integer has some very helpful static methods. Try this code as a starting base for your problem:
public int[] extractBitNumbers(int value) {
// determine how many ones are in value
int bitCount = Integer.bitCount(value);
// allocate storage
int[] oneBits = new int[bitCount];
int putIndex = 0;
// loop until no more bits are set
while (value != 0) {
// find the number of the lowest set bit
int bitNo = Integer.numberOfTrailingZeros(value);
// store the bit number in array
oneBits[putIndex++] = bitNo+1;
// clear the bit we just processed from the value
value &= ~(1 << bitNo);
}
return oneBits;
}
I can show you C# implementation, Java should be very similar.
int value = 33;
int index = 1;
while (value > 0)
{
if ((value % 2) == 1)
Console.WriteLine(index);
index++;
value /= 2;
}
If you want to get an array like that you'll likely need to loop the number of bits you want to check & the integer with a bit shifted 1 for each step.
Something like (pseudo):
Init array
mask = 1
for (0 to BitCount):
if Integer & mask
array[] = pos
mask << 1
A bit-crunching variation would be something like:
int[] getBits(int value) {
int bitValue = 1;
int index = 1;
int[] bits = new int[33];
while (value >= bitValue)
{
bits[index++] = (value & bitValue);
bitValue << 1; // or: bitValue *= 2;
}
return bits;
}
Note that since the bits are indexed from 1 as you requested, bits[0] is left unused.
This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
I have an array of numbers from 1 to 100 (both inclusive). The size of the array is 100. The numbers are randomly added to the array, but there is one random empty slot in the array.
What is the quickest way to find that slot as well as the number that should be put in the slot? A Java solution is preferable.
You can do this in O(n). Iterate through the array and compute the sum of all numbers. Now, sum of natural numbers from 1 to N, can be expressed as Nx(N+1)/2. In your case N=100.
Subtract the sum of the array from Nx(N+1)/2, where N=100.
That is the missing number. The empty slot can be detected during the iteration in which the sum is computed.
// will be the sum of the numbers in the array.
int sum = 0;
int idx = -1;
for (int i = 0; i < arr.length; i++)
{
if (arr[i] == 0)
{
idx = i;
}
else
{
sum += arr[i];
}
}
// the total sum of numbers between 1 and arr.length.
int total = (arr.length + 1) * arr.length / 2;
System.out.println("missing number is: " + (total - sum) + " at index " + idx);
We can use XOR operation which is safer than summation because in programming languages if the given input is large it may overflow and may give wrong answer.
Before going to the solution, know that A xor A = 0. So if we XOR two identical numbers the value is 0.
Now, XORing [1..n] with the elements present in the array cancels the identical numbers. So at the end we will get the missing number.
// Assuming that the array contains 99 distinct integers between 1..99
// and empty slot value is zero
int XOR = 0;
for(int i=0; i<100; i++) {
if (ARRAY[i] != 0) // remove this condition keeping the body if no zero slot
XOR ^= ARRAY[i];
XOR ^= (i + 1);
}
return XOR;
//return XOR ^ ARRAY.length + 1; if your array doesn't have empty zero slot.
Let the given array be A with length N. Lets assume in the given array, the single empty slot is filled with 0.
We can find the solution for this problem using many methods including algorithm used in Counting sort. But, in terms of efficient time and space usage, we have two algorithms. One uses mainly summation, subtraction and multiplication. Another uses XOR. Mathematically both methods work fine. But programatically, we need to assess all the algorithms with main measures like
Limitations(like input values are large(A[1...N]) and/or number of
input values is large(N))
Number of condition checks involved
Number and type of mathematical operations involved
etc. This is because of the limitations in time and/or hardware(Hardware resource limitation) and/or software(Operating System limitation, Programming language limitation, etc), etc. Lets list and assess the pros and cons of each one of them.
Algorithm 1 :
In algorithm 1, we have 3 implementations.
Calculate the total sum of all the numbers(this includes the unknown missing number) by using the mathematical formula(1+2+3+...+N=(N(N+1))/2). Here, N=100. Calculate the total sum of all the given numbers. Subtract the second result from the first result will give the missing number.
Missing Number = (N(N+1))/2) - (A[1]+A[2]+...+A[100])
Calculate the total sum of all the numbers(this includes the unknown missing number) by using the mathematical formula(1+2+3+...+N=(N(N+1))/2). Here, N=100. From that result, subtract each given number gives the missing number.
Missing Number = (N(N+1))/2)-A[1]-A[2]-...-A[100]
(Note:Even though the second implementation's formula is derived from first, from the mathematical point of view both are same. But from programming point of view both are different because the first formula is more prone to bit overflow than the second one(if the given numbers are large enough). Even though addition is faster than subtraction, the second implementation reduces the chance of bit overflow caused by addition of large values(Its not completely eliminated, because there is still very small chance since (N+1) is there in the formula). But both are equally prone to bit overflow by multiplication. The limitation is both implementations give correct result only if N(N+1)<=MAXIMUM_NUMBER_VALUE. For the first implementation, the additional limitation is it give correct result only if Sum of all given numbers<=MAXIMUM_NUMBER_VALUE.)
Calculate the total sum of all the numbers(this includes the unknown missing number) and subtract each given number in the same loop in parallel. This eliminates the risk of bit overflow by multiplication but prone to bit overflow by addition and subtraction.
//ALGORITHM
missingNumber = 0;
foreach(index from 1 to N)
{
missingNumber = missingNumber + index;
//Since, the empty slot is filled with 0,
//this extra condition which is executed for N times is not required.
//But for the sake of understanding of algorithm purpose lets put it.
if (inputArray[index] != 0)
missingNumber = missingNumber - inputArray[index];
}
In a programming language(like C, C++, Java, etc), if the number of bits representing a integer data type is limited, then all the above implementations are prone to bit overflow because of summation, subtraction and multiplication, resulting in wrong result in case of large input values(A[1...N]) and/or large number of input values(N).
Algorithm 2 :
We can use the property of XOR to get solution for this problem without worrying about the problem of bit overflow. And also XOR is both safer and faster than summation. We know the property of XOR that XOR of two same numbers is equal to 0(A XOR A = 0). If we calculate the XOR of all the numbers from 1 to N(this includes the unknown missing number) and then with that result, XOR all the given numbers, the common numbers get canceled out(since A XOR A=0) and in the end we get the missing number. If we don't have bit overflow problem, we can use both summation and XOR based algorithms to get the solution. But, the algorithm which uses XOR is both safer and faster than the algorithm which uses summation, subtraction and multiplication. And we can avoid the additional worries caused by summation, subtraction and multiplication.
In all the implementations of algorithm 1, we can use XOR instead of addition and subtraction.
Lets assume, XOR(1...N) = XOR of all numbers from 1 to N
Implementation 1 => Missing Number = XOR(1...N) XOR (A[1] XOR A[2] XOR...XOR A[100])
Implementation 2 => Missing Number = XOR(1...N) XOR A[1] XOR A[2] XOR...XOR A[100]
Implementation 3 =>
//ALGORITHM
missingNumber = 0;
foreach(index from 1 to N)
{
missingNumber = missingNumber XOR index;
//Since, the empty slot is filled with 0,
//this extra condition which is executed for N times is not required.
//But for the sake of understanding of algorithm purpose lets put it.
if (inputArray[index] != 0)
missingNumber = missingNumber XOR inputArray[index];
}
All three implementations of algorithm 2 will work fine(from programatical point of view also). One optimization is, similar to
1+2+....+N = (N(N+1))/2
We have,
1 XOR 2 XOR .... XOR N = {N if REMAINDER(N/4)=0, 1 if REMAINDER(N/4)=1, N+1 if REMAINDER(N/4)=2, 0 if REMAINDER(N/4)=3}
We can prove this by mathematical induction. So, instead of calculating the value of XOR(1...N) by XOR all the numbers from 1 to N, we can use this formula to reduce the number of XOR operations.
Also, calculating XOR(1...N) using above formula has two implementations. Implementation wise, calculating
// Thanks to https://a3nm.net/blog/xor.html for this implementation
xor = (n>>1)&1 ^ (((n&1)>0)?1:n)
is faster than calculating
xor = (n % 4 == 0) ? n : (n % 4 == 1) ? 1 : (n % 4 == 2) ? n + 1 : 0;
So, the optimized Java code is,
long n = 100;
long a[] = new long[n];
//XOR of all numbers from 1 to n
// n%4 == 0 ---> n
// n%4 == 1 ---> 1
// n%4 == 2 ---> n + 1
// n%4 == 3 ---> 0
//Slower way of implementing the formula
// long xor = (n % 4 == 0) ? n : (n % 4 == 1) ? 1 : (n % 4 == 2) ? n + 1 : 0;
//Faster way of implementing the formula
// long xor = (n>>1)&1 ^ (((n&1)>0)?1:n);
long xor = (n>>1)&1 ^ (((n&1)>0)?1:n);
for (long i = 0; i < n; i++)
{
xor = xor ^ a[i];
}
//Missing number
System.out.println(xor);
This was an Amazon interview question and was originally answered here: We have numbers from 1 to 52 that are put into a 51 number array, what's the best way to find out which number is missing?
It was answered, as below:
1) Calculate the sum of all numbers stored in the array of size 51.
2) Subtract the sum from (52 * 53)/2 ---- Formula : n * (n + 1) / 2.
It was also blogged here: Software Job - Interview Question
Here is a simple program to find the missing numbers in an integer array
ArrayList<Integer> arr = new ArrayList<Integer>();
int a[] = { 1,3,4,5,6,7,10 };
int j = a[0];
for (int i=0;i<a.length;i++)
{
if (j==a[i])
{
j++;
continue;
}
else
{
arr.add(j);
i--;
j++;
}
}
System.out.println("missing numbers are ");
for(int r : arr)
{
System.out.println(" " + r);
}
Recently I had a similar (not exactly the same) question in a job interview and also I heard from a friend that was asked the exactly same question in an interview.
So here is an answer to the OP question and a few more variations that can be potentially asked.
The answers example are given in Java because, it's stated that:
A Java solution is preferable.
Variation 1:
Array of numbers from 1 to 100 (both inclusive) ... The numbers are randomly added to the array, but there is one random empty slot in the array
public static int findMissing1(int [] arr){
int sum = 0;
for(int n : arr){
sum += n;
}
return (100*(100+1)/2) - sum;
}
Explanation:
This solution (as many other solutions posted here) is based on the formula of Triangular number, which gives us the sum of all natural numbers from 1 to n (in this case n is 100). Now that we know the sum that should be from 1 to 100 - we just need to subtract the actual sum of existing numbers in given array.
Variation 2:
Array of numbers from 1 to n (meaning that the max number is unknown)
public static int findMissing2(int [] arr){
int sum = 0, max = 0;
for(int n : arr){
sum += n;
if(n > max) max = n;
}
return (max*(max+1)/2) - sum;
}
Explanation:
In this solution, since the max number isn't given - we need to find it. After finding the max number - the logic is the same.
Variation 3:
Array of numbers from 1 to n (max number is unknown), there is two random empty slots in the array
public static int [] findMissing3(int [] arr){
int sum = 0, max = 0, misSum;
int [] misNums = {};//empty by default
for(int n : arr){
sum += n;
if(n > max) max = n;
}
misSum = (max*(max+1)/2) - sum;//Sum of two missing numbers
for(int n = Math.min(misSum, max-1); n > 1; n--){
if(!contains(n, arr)){
misNums = new int[]{n, misSum-n};
break;
}
}
return misNums;
}
private static boolean contains(int num, int [] arr){
for(int n : arr){
if(n == num)return true;
}
return false;
}
Explanation:
In this solution, the max number isn't given (as in the previous), but it can also be missing of two numbers and not one. So at first we find the sum of missing numbers - with the same logic as before. Second finding the smaller number between missing sum and the last (possibly) missing number - to reduce unnecessary search. Third since Javas Array (not a Collection) doesn't have methods as indexOf or contains, I added a small reusable method for that logic. Fourth when first missing number is found, the second is the subtract from missing sum.
If only one number is missing, then the second number in array will be zero.
Variation 4:
Array of numbers from 1 to n (max number is unknown), with X missing (amount of missing numbers are unknown)
public static ArrayList<Integer> findMissing4(ArrayList<Integer> arr){
int max = 0;
ArrayList<Integer> misNums = new ArrayList();
int [] neededNums;
for(int n : arr){
if(n > max) max = n;
}
neededNums = new int[max];//zero for any needed num
for(int n : arr){//iterate again
neededNums[n == max ? 0 : n]++;//add one - used as index in second array (convert max to zero)
}
for(int i=neededNums.length-1; i>0; i--){
if(neededNums[i] < 1)misNums.add(i);//if value is zero, than index is a missing number
}
return misNums;
}
Explanation:
In this solution, as in the previous, the max number is unknown and there can be missing more than one number, but in this variation, we don't know how many numbers are potentially missing (if any). The beginning of the logic is the same - find the max number. Then I initialise another array with zeros, in this array index indicates the potentially missing number and zero indicates that the number is missing. So every existing number from original array is used as an index and its value is incremented by one (max converted to zero).
Note
If you want examples in other languages or another interesting variations of this question, you are welcome to check my Github repository for Interview questions & answers.
(sum of 1 to n) - (sum of all values in the array) = missing number
int sum = 0;
int idx = -1;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == 0) idx = i; else sum += arr[i];
}
System.out.println("missing number is: " + (5050 - sum) + " at index " + idx);
On a similar scenario, where the array is already sorted, it does not include duplicates and only one number is missing, it is possible to find this missing number in log(n) time, using binary search.
public static int getMissingInt(int[] intArray, int left, int right) {
if (right == left + 1) return intArray[right] - 1;
int pivot = left + (right - left) / 2;
if (intArray[pivot] == intArray[left] + (intArray[right] - intArray[left]) / 2 - (right - left) % 2)
return getMissingInt(intArray, pivot, right);
else
return getMissingInt(intArray, left, pivot);
}
public static void main(String args[]) {
int[] array = new int[]{3, 4, 5, 6, 7, 8, 10};
int missingInt = getMissingInt(array, 0, array.length-1);
System.out.println(missingInt); //it prints 9
}
Well, use a bloom filter.
int findmissing(int arr[], int n)
{
long bloom=0;
int i;
for(i=0; i<;n; i++)bloom+=1>>arr[i];
for(i=1; i<=n, (bloom<<i & 1); i++);
return i;
}
This is c# but it should be pretty close to what you need:
int sumNumbers = 0;
int emptySlotIndex = -1;
for (int i = 0; i < arr.length; i++)
{
if (arr[i] == 0)
emptySlotIndex = i;
sumNumbers += arr[i];
}
int missingNumber = 5050 - sumNumbers;
The solution that doesn't involve repetitive additions or maybe the n(n+1)/2 formula doesn't get to you at an interview time for instance.
You have to use an array of 4 ints (32 bits) or 2 ints (64 bits). Initialize the last int with (-1 & ~(1 << 31)) >> 3. (the bits that are above 100 are set to 1) Or you may set the bits above 100 using a for loop.
Go through the array of numbers and set 1 for the bit position corresponding to the number (e.g. 71 would be set on the 3rd int on the 7th bit from left to right)
Go through the array of 4 ints (32 bit version) or 2 ints(64 bit version)
public int MissingNumber(int a[])
{
int bits = sizeof(int) * 8;
int i = 0;
int no = 0;
while(a[i] == -1)//this means a[i]'s bits are all set to 1, the numbers is not inside this 32 numbers section
{
no += bits;
i++;
}
return no + bits - Math.Log(~a[i], 2);//apply NOT (~) operator to a[i] to invert all bits, and get a number with only one bit set (2 at the power of something)
}
Example: (32 bit version) lets say that the missing number is 58. That means that the 26th bit (left to right) of the second integer is set to 0.
The first int is -1 (all bits are set) so, we go ahead for the second one and add to "no" the number 32. The second int is different from -1 (a bit is not set) so, by applying the NOT (~) operator to the number we get 64. The possible numbers are 2 at the power x and we may compute x by using log on base 2; in this case we get log2(64) = 6 => 32 + 32 - 6 = 58.
Hope this helps.
I think the easiest and possibly the most efficient solution would be to loop over all entries and use a bitset to remember which numbers are set, and then test for 0 bit. The entry with the 0 bit is the missing number.
This is not a search problem. The employer is wondering if you have a grasp of a checksum. You might need a binary or for loop or whatever if you were looking for multiple unique integers, but the question stipulates "one random empty slot." In this case we can use the stream sum. The condition: "The numbers are randomly added to the array" is meaningless without more detail. The question does not assume the array must start with the integer 1 and so tolerate with the offset start integer.
int[] test = {2,3,4,5,6,7,8,9,10, 12,13,14 };
/*get the missing integer*/
int max = test[test.length - 1];
int min = test[0];
int sum = Arrays.stream(test).sum();
int actual = (((max*(max+1))/2)-min+1);
//Find:
//the missing value
System.out.println(actual - sum);
//the slot
System.out.println(actual - sum - min);
Success time: 0.18 memory: 320576 signal:0
I found this beautiful solution here:
http://javaconceptoftheday.com/java-puzzle-interview-program-find-missing-number-in-an-array/
public class MissingNumberInArray
{
//Method to calculate sum of 'n' numbers
static int sumOfNnumbers(int n)
{
int sum = (n * (n+1))/ 2;
return sum;
}
//Method to calculate sum of all elements of array
static int sumOfElements(int[] array)
{
int sum = 0;
for (int i = 0; i < array.length; i++)
{
sum = sum + array[i];
}
return sum;
}
public static void main(String[] args)
{
int n = 8;
int[] a = {1, 4, 5, 3, 7, 8, 6};
//Step 1
int sumOfNnumbers = sumOfNnumbers(n);
//Step 2
int sumOfElements = sumOfElements(a);
//Step 3
int missingNumber = sumOfNnumbers - sumOfElements;
System.out.println("Missing Number is = "+missingNumber);
}
}
function solution($A) {
// code in PHP5.5
$n=count($A);
for($i=1;$i<=$n;$i++) {
if(!in_array($i,$A)) {
return (int)$i;
}
}
}
Finding the missing number from a series of numbers. IMP points to remember.
the array should be sorted..
the Function do not work on multiple missings.
the sequence must be an AP.
public int execute2(int[] array) {
int diff = Math.min(array[1]-array[0], array[2]-array[1]);
int min = 0, max = arr.length-1;
boolean missingNum = true;
while(min<max) {
int mid = (min + max) >>> 1;
int leftDiff = array[mid] - array[min];
if(leftDiff > diff * (mid - min)) {
if(mid-min == 1)
return (array[mid] + array[min])/2;
max = mid;
missingNum = false;
continue;
}
int rightDiff = array[max] - array[mid];
if(rightDiff > diff * (max - mid)) {
if(max-mid == 1)
return (array[max] + array[mid])/2;
min = mid;
missingNum = false;
continue;
}
if(missingNum)
break;
}
return -1;
}
One thing you could do is sort the numbers using quick sort for instance. Then use a for loop to iterate through the sorted array from 1 to 100. In each iteration, you compare the number in the array with your for loop increment, if you find that the index increment is not the same as the array value, you have found your missing number as well as the missing index.
Below is the solution for finding all the missing numbers from a given array:
public class FindMissingNumbers {
/**
* The function prints all the missing numbers from "n" consecutive numbers.
* The number of missing numbers is not given and all the numbers in the
* given array are assumed to be unique.
*
* A similar approach can be used to find all no-unique/ unique numbers from
* the given array
*
* #param n
* total count of numbers in the sequence
* #param numbers
* is an unsorted array of all the numbers from 1 - n with some
* numbers missing.
*
*/
public static void findMissingNumbers(int n, int[] numbers) {
if (n < 1) {
return;
}
byte[] bytes = new byte[n / 8];
int countOfMissingNumbers = n - numbers.length;
if (countOfMissingNumbers == 0) {
return;
}
for (int currentNumber : numbers) {
int byteIndex = (currentNumber - 1) / 8;
int bit = (currentNumber - byteIndex * 8) - 1;
// Update the "bit" in bytes[byteIndex]
int mask = 1 << bit;
bytes[byteIndex] |= mask;
}
for (int index = 0; index < bytes.length - 2; index++) {
if (bytes[index] != -128) {
for (int i = 0; i < 8; i++) {
if ((bytes[index] >> i & 1) == 0) {
System.out.println("Missing number: " + ((index * 8) + i + 1));
}
}
}
}
// Last byte
int loopTill = n % 8 == 0 ? 8 : n % 8;
for (int index = 0; index < loopTill; index++) {
if ((bytes[bytes.length - 1] >> index & 1) == 0) {
System.out.println("Missing number: " + (((bytes.length - 1) * 8) + index + 1));
}
}
}
public static void main(String[] args) {
List<Integer> arrayList = new ArrayList<Integer>();
int n = 128;
int m = 5;
for (int i = 1; i <= n; i++) {
arrayList.add(i);
}
Collections.shuffle(arrayList);
for (int i = 1; i <= 5; i++) {
System.out.println("Removing:" + arrayList.remove(i));
}
int[] array = new int[n - m];
for (int i = 0; i < (n - m); i++) {
array[i] = arrayList.get(i);
}
System.out.println("Array is: " + Arrays.toString(array));
findMissingNumbers(n, array);
}
}
Lets say you have n as 8, and our numbers range from 0-8 for this example
we can represent the binary representation of all 9 numbers as follows
0000
0001
0010
0011
0100
0101
0110
0111
1000
in the above sequence there is no missing numbers and in each column the number of zeros and ones match, however as soon as you remove 1 value lets say 3 we get a in balance in the number of 0's and 1's across the columns. If the number of 0's in a column is <= the number of 1's our missing number will have a 0 at this bit position, otherwise if the number of 0's > the number of 1's at this bit position then this bit position will be a 1. We test the bits left to right and at each iteration we throw away half of the array for the testing of the next bit, either the odd array values or the even array values are thrown away at each iteration depending on which bit we are deficient on.
The below solution is in C++
int getMissingNumber(vector<int>* input, int bitPos, const int startRange)
{
vector<int> zeros;
vector<int> ones;
int missingNumber=0;
//base case, assume empty array indicating start value of range is missing
if(input->size() == 0)
return startRange;
//if the bit position being tested is 0 add to the zero's vector
//otherwise to the ones vector
for(unsigned int i = 0; i<input->size(); i++)
{
int value = input->at(i);
if(getBit(value, bitPos) == 0)
zeros.push_back(value);
else
ones.push_back(value);
}
//throw away either the odd or even numbers and test
//the next bit position, build the missing number
//from right to left
if(zeros.size() <= ones.size())
{
//missing number is even
missingNumber = getMissingNumber(&zeros, bitPos+1, startRange);
missingNumber = (missingNumber << 1) | 0;
}
else
{
//missing number is odd
missingNumber = getMissingNumber(&ones, bitPos+1, startRange);
missingNumber = (missingNumber << 1) | 1;
}
return missingNumber;
}
At each iteration we reduce our input space by 2, i.e N, N/2,N/4 ... = O(log N), with space O(N)
//Test cases
[1] when missing number is range start
[2] when missing number is range end
[3] when missing number is odd
[4] when missing number is even
Solution With PHP $n = 100;
$n*($n+1)/2 - array_sum($array) = $missing_number
and array_search($missing_number) will give the index of missing number
Here program take time complexity is O(logn) and space complexity O(logn)
public class helper1 {
public static void main(String[] args) {
int a[] = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12};
int k = missing(a, 0, a.length);
System.out.println(k);
}
public static int missing(int[] a, int f, int l) {
int mid = (l + f) / 2;
//if first index reached last then no element found
if (a.length - 1 == f) {
System.out.println("missing not find ");
return 0;
}
//if mid with first found
if (mid == f) {
System.out.println(a[mid] + 1);
return a[mid] + 1;
}
if ((mid + 1) == a[mid])
return missing(a, mid, l);
else
return missing(a, f, mid);
}
}
public class MissingNumber {
public static void main(String[] args) {
int array[] = {1,2,3,4,6};
int x1 = getMissingNumber(array,6);
System.out.println("The Missing number is: "+x1);
}
private static int getMissingNumber(int[] array, int i) {
int acctualnumber =0;
int expectednumber = (i*(i+1)/2);
for (int j : array) {
acctualnumber = acctualnumber+j;
}
System.out.println(acctualnumber);
System.out.println(expectednumber);
return expectednumber-acctualnumber;
}
}
Use sum formula,
class Main {
// Function to ind missing number
static int getMissingNo (int a[], int n) {
int i, total;
total = (n+1)*(n+2)/2;
for ( i = 0; i< n; i++)
total -= a[i];
return total;
}
/* program to test above function */
public static void main(String args[]) {
int a[] = {1,2,4,5,6};
int miss = getMissingNo(a,5);
System.out.println(miss);
}
}
Reference http://www.geeksforgeeks.org/find-the-missing-number/
simple solution with test data :
class A{
public static void main(String[] args){
int[] array = new int[200];
for(int i=0;i<100;i++){
if(i != 51){
array[i] = i;
}
}
for(int i=100;i<200;i++){
array[i] = i;
}
int temp = 0;
for(int i=0;i<200;i++){
temp ^= array[i];
}
System.out.println(temp);
}
}
//Array is shorted and if writing in C/C++ think of XOR implementations in java as follows.
int num=-1;
for (int i=1; i<=100; i++){
num =2*i;
if(arr[num]==0){
System.out.println("index: "+i+" Array position: "+ num);
break;
}
else if(arr[num-1]==0){
System.out.println("index: "+i+ " Array position: "+ (num-1));
break;
}
}// use Rabbit and tortoise race, move the dangling index faster,
//learnt from Alogithimica, Ameerpet, hyderbad**
If the array is randomly filled, then at the best you can do a linear search in O(n) complexity. However, we could have improved the complexity to O(log n) by divide and conquer approach similar to quick sort as pointed by giri given that the numbers were in ascending/descending order.
This Program finds missing numbers
<?php
$arr_num=array("1","2","3","5","6");
$n=count($arr_num);
for($i=1;$i<=$n;$i++)
{
if(!in_array($i,$arr_num))
{
array_push($arr_num,$i);print_r($arr_num);exit;
}
}
?>
Now I'm now too sharp with the Big O notations but couldn't you also do something like (in Java)
for (int i = 0; i < numbers.length; i++) {
if(numbers[i] != i+1){
System.out.println(i+1);
}
}
where numbers is the array with your numbers from 1-100.
From my reading of the question it did not say when to write out the missing number.
Alternatively if you COULD throw the value of i+1 into another array and print that out after the iteration.
Of course it might not abide by the time and space rules. As I said. I have to strongly brush up on Big O.
========Simplest Solution for sorted Array===========
public int getMissingNumber(int[] sortedArray)
{
int missingNumber = 0;
int missingNumberIndex=0;
for (int i = 0; i < sortedArray.length; i++)
{
if (sortedArray[i] == 0)
{
missingNumber = (sortedArray[i + 1]) - 1;
missingNumberIndex=i;
System.out.println("missingNumberIndex: "+missingNumberIndex);
break;
}
}
return missingNumber;
}
Another homework question. A sequential search is the best that you can do. As for a Java solution, consider that an exercise for the reader. :P