Given an integer A representing the square blocks. The height of each square block is 1. The task is to create a staircase of max height using these blocks. The first stair would require only one block, the second stair would require two blocks and so on. Find and return the maximum height of the staircase.
Your submission failed for the following input: A : 92761
Your function returned the following : 65536
The expected returned value : 430
Approach:
We are interested in the number of steps and we know that each step Si uses exactly Bi number of bricks. We can represent this problem as an equation:
n * (n + 1) / 2 = T (For Natural number series starting from 1, 2, 3, 4, 5 …)
n * (n + 1) = 2 * T
n-1 will represent our final solution because our series in problem starts from 2, 3, 4, 5…
Now, we just have to solve this equation and for that we can exploit binary search to find the solution to this equation. Lower and Higher bounds of binary search are 1 and T.
CODE
public int solve(int A) {
int l=1,h=A,T=2*A;
while(l<=h)
{
int mid=l+(h-l)/2;
if((mid*(mid+1))==T)
return mid;
if((mid*(mid+1))>T && (mid!=0 && (mid*(mid-1))<=T) )
return mid-1;
if((mid*(mid+1))>T)
h=mid-1;
else
l=mid+1;
}
return 0;
}
To expand on the comment by Matt Timmermans:
You know that for n steps, you need (n * (n + 1))/2 blocks. You want know, if given B blocks, how many steps you can create.
So you have:
(n * (n + 1))/2 = B
(n^2 + n)/2 = B
n^2 + n = 2B
n^2 + n - 2B = 0
That looks suspiciously like something for which you'd use the quadratic formula.
In this case, a=1, b=1, and c=(-2B). Plugging the numbers into the formula:
n = ((-b) + sqrt(b^2 - 4*a*c))/(2*a)
= (-1 + sqrt(1 - 4*1*(-2B)))/(2*a)
= (-1 + sqrt(1 + 8B))/2
= (sqrt(1 + 8B) - 1)/2
So if you have 5050 blocks, you get:
n = (sqrt(1 + 40400) - 1)/2
= (sqrt(40401) - 1)/2
= (201 - 1)/2
= 100
Try it with the quadratic formula calculator. Use 1 for the value of a and b, and replace c with negative two times the number of blocks you're given. So in the example above, c would be -10100.
In your program, since you can't have a partial step, you'd want to truncate the result.
Why are you using all these formulas? A simple while() loop should do the trick, eventually, it's just a simple Gaussian Sum ..
public static int calculateStairs(int blocks) {
int lastHeight = 0;
int sum = 0;
int currentHeight = 0; //number of bricks / level
while (sum <= blocks) {
lastHeight = currentHeight;
currentHeight++;
sum += currentHeight;
}
return lastHeight;
}
So this should do the job as it also returns the expected value. Correct me if im wrong.
public int solve(int blocks) {
int current; //Create Variables
for (int x = 0; x < Integer.MAX_VALUE; x++) { //Increment until return
current = 0; //Set current to 0
//Implementation of the Gauss sum
for (int i = 1; i <= x; i++) { //Sum up [1,*current height*]
current += i;
} //Now we have the amount of blocks required for the current height
//Now we check if the amount of blocks is bigger than
// the wanted amount, and if so we return the last one
if (current > blocks) {
return x - 1;
}
}
return current;
}
I implemented Median of medians selection algorithm based on algs4 quickselect using the Wikipedia article, but my code doesn't work well:
1) it is said that median of medians finds kth largest element. However, my code finds kth smallest element.
2) my implementation runs 1-20 times slower than quickselect, but the median of medians algorithm should be asymptotically faster.
I've checked everything several times, but I cannot find the issue.
public class MedianOfMedians {
public static Comparable medianOfMedians(Comparable[] nums, int k) {
return nums[select(nums, 0, nums.length - 1, k)];
}
private static int select(Comparable[] nums, int lo, int hi, int k) {
while (lo < hi) {
int pivotIndex = pivot(nums, lo, hi);
int j = partition(nums, lo, hi, pivotIndex);
if (j < k) {
lo = j + 1;
} else if (j > k) {
hi = j - 1;
} else {
return j;
}
}
return lo;
}
private static int pivot(Comparable[] list, int left, int right) {
// for 5 or less elements just get median
if (right - left < 5) {
return partition5(list, left, right);
}
// otherwise move the medians of five-element subgroups to the first n/5 positions
for (int i = left; i <= right; i += 5) {
// get the median of the i'th five-element subgroup
int subRight = i + 4;
if (subRight > right) {
subRight = right;
}
int median5 = partition5(list, i, subRight);
exch(list, median5, (int) (left + Math.floor((i - left) / 5d)));
}
// compute the median of the n/5 medians-of-five
return select(list,
left,
(int) (left + Math.ceil((right - left) / 5d) - 1),
(int) (left + (right - left) / 10d));
}
private static int partition5(Comparable[] list, int lo, int hi) {
for (int i = lo; i <= hi; i++) {
for (int j = i; j > lo; j--) {
if (less(list[j - 1], list[j])) {
exch(list, j, j - 1);
}
}
}
return (hi + lo) / 2;
}
private static int partition(Comparable[] a, int lo, int hi, int pivotIndex) {
exch(a, lo, pivotIndex);
int i = lo;
int j = hi + 1;
Comparable v = a[lo];
while (true) {
while (less(a[++i], v) && i != hi) { }
while (less(v, a[--j]) && j != lo) { }
if (j <= i) break;
exch(a, i, j);
}
exch(a, j, lo);
return j;
}
private static void exch(Comparable[] nums, int i, int j) { }
private static boolean less(Comparable v, Comparable w) { }
}
JUnit test:
public class MedianOfMediansTest {
private final static int TESTS_COUNT = 100;
#org.junit.Test
public void test() {
// generate TESTS_COUNT arrays of 10000 entries from 0..Integer.MAX_VALUE
Integer[][] tests = generateTestComparables(TESTS_COUNT, 10000, 10000, 0, Integer.MAX_VALUE);
for (int i = 0; i < tests.length; i++) {
Integer[] array1 = Arrays.copyOf(tests[i], tests[i].length);
Integer[] array2 = Arrays.copyOf(tests[i], tests[i].length);
Integer[] array3 = Arrays.copyOf(tests[i], tests[i].length);
long time = System.nanoTime();
final int a = (Integer) MedianOfMedians.medianOfMedians(array1, 0);
long nanos_a = System.nanoTime() - time;
time = System.nanoTime();
final int b = (Integer) Quick.select(array2, 0);
long nanos_b = System.nanoTime() - time;
time = System.nanoTime();
Arrays.sort(array3);
final int c = array3[0];
long nanos_c = System.nanoTime() - time;
System.out.println("MedianOfMedians: " + a + " (" + nanos_a + ") " +
"QuickSelect: " + b + " (" + nanos_b + ") " +
"Arrays.sort: " + c + " (" + nanos_c + ")");
System.out.println(((double) nanos_a) / ((double) nanos_b));
Assert.assertEquals(c, a);
Assert.assertEquals(b, a);
}
}
public static Integer[][] generateTestComparables(int numberOfTests,
int arraySizeMin, int arraySizeMax,
int valueMin, int valueMax) {
Random rand = new Random(System.currentTimeMillis());
Integer[][] ans = new Integer[numberOfTests][];
for (int i = 0; i < ans.length; i++) {
ans[i] = new Integer[randInt(rand, arraySizeMin, arraySizeMax)];
for (int j = 0; j < ans[i].length; j++) {
ans[i][j] = randInt(rand, valueMin, valueMax);
}
}
return ans;
}
public static int randInt(Random rand, int min, int max) {
return (int) (min + (rand.nextDouble() * ((long) max - (long) min)));
}
}
1) it is said that median of medians finds kth largest element.
However, my code finds kth smallest element.
This is not strictly true. Any selection algorithm can find either smallest or largest element because that's essentially the same task. It depends on how you compare elements and how you partition them (and you can always do something like length - 1 - result later). Your code indeed seems to find the kth smallest element, which is by the way the most typical and intuitive way of implementing a selection algorithm.
2) my implementation runs 1-20 times slower than quickselect, but the
median of medians algorithm should be asymptotically faster.
Not just asymptotically faster. Asymptotically faster in the worst case. In the average case, both are linear, but MoM has higher constant factors. Since you generate your tests randomly, you are very unlikely to hit the worst case. If you used randomized quickselect, then for any input it's unlikely to hit the worst case, otherwise the probability will depend on the pivot selection algorithm used.
With that in mind, and the fact that median of medians has high constant factors, you should not expect it to perform better than quickselect! It might outperform sorting, though, but even then—those logarithmic factors in sorting aren't that large for small inputs (lg 10000 is about 13-14).
Take my MoM solution for a LeetCode problem, for example. Arrays.sort sometimes outperforms MoM for arrays with 500 million elements. In the best case it runs about twice faster, though.
Therefore, MoM is mostly of theoretical interest. I could imagine a practical use case when you need 100% guarantee of not exceeding some time limit. Say, some real-time system on an aircraft, or spacecraft, or nuclear reactor. The time limit is not very tight, but exceeding it even by one nanosecond is catastrophic. But it's an extremely contrived example, and I doubt that it's actually the way it works.
Even if you can find a practical use case for MoM, you can probably use something like Introselect instead. It essentially starts with quickselect, and then switches to MoM if things don't look good. But testing it would be a nightmare—how would you come up with a test that actually forces the algorithm to switch (and therefore test the MoM part), especially if it's randomized?
Your code looks fine overall, but I'd make some helper methods package-private or even moved them to another class to test separately because such things are very hard to get right. And you may not notice the effect if the result is right. I'm not sure that your groups-of-five code is 100% correct, for example. Sometimes you use right - left where I'd expect to see element count, which should be right - left + 1.
Also, I would replace those ceil/floor calls with pure integer arithmetic equivalents. That is, Math.floor((i - left) / 5d)) => (i - left) / 5, Math.ceil((right - left) / 5d) => (right - left + 4) / 5 (this is the part where I don't like the right - left thing, by the way, but I'm not sure if it's wrong).
I am trying to solve this question as the preparation for a programming interview:
A frog only moves forward, but it can move in steps 1 inch long or in jumps 2 inches long. A frog can cover the same distance using different combinations of steps and jumps.
Write a function that calculates the number of different combinations a frog can use to cover a given distance.
For example, a distance of 3 inches can be covered in three ways: step-step-step, step-jump, and jump-step.
I think there is a quite simple solution to this, but I just can't seem to find it. I would like to use recursion, but I can't see how. Here is what I have so far:
public class Frog {
static int combinations = 0;
static int step = 1;
static int jump = 2;
static int[] arr = {step, jump};
public static int numberOfWays(int n) {
for (int i = 0; i < arr.length; i++) {
int sum = 0;
sum += arr[i];
System.out.println("SUM outer loop: " + sum + " : " + arr[i]);
while (sum != 3) {
for (int j = 0; j < arr.length; j++) {
if (sum + arr[j] <= 3) {
sum += arr[j];
System.out.println("SUM inner loop: " + sum + " : " + arr[j]);
if (sum == 3) {
combinations++;
System.out.println("Combinations " + combinations);
}
}
}
}
}
return combinations;
}
public static void main(String[] args) {
System.out.println(numberOfWays(3));
}
}
It doesn't find all combinations, and I think the code is quite bad. Anyone have a good solution to this question?
Think you have an oracle that knows how to solve the problem for "smaller problems", you just need to feed it with smaller problems. This is the recursive method.
In your case, you solve foo(n), by splitting the possible moves the frog can do in the last step, and summing them):
foo(n) = foo(n-1) + foo(n-2)
^ ^
1 step 2 steps
In addition, you need a stop clause of foo(0) = 1, foo(1)=1 (one way to move 0 or 1 inches).
Is this recursive formula looks familiar? Can you solve it better than the naive recursive solution?
Spoiler:
Fibonacci Sequence
Here's a simple pseudo-code implementation that should work:
var results = []
function plan(previous, n){
if (n==0) {
results.push(previous)
} else if (n > 0){
plan(previous + ' step', n-1)
plan(previous + ' hop', n-2)
}
}
plan('', 5)
If you want to improve the efficiency of an algorithm like this you could look into using memoization
Here's a combinatoric way: think of n as 1 + 1 + 1 ... = n. Now bunch the 1's in pairs, gradually increasing the number of bunched 1's, summing the possibilities to arrange them.
For example, consider 5 as 1 1 1 1 1:
one bunch => (1) (1) (1) (11) => 4 choose 1 possibilities to arrange one 2 with three 1's
two bunches => (1) (11) (11) => 3 choose 2 possibilities to arrange two 2's with one 1
etc.
This seems directly related to Wikipedia's description of Fibonacci numbers' "Use in Mathematics," for example, in counting "the number of compositions of 1s and 2s that sum to a given total n" (http://en.wikipedia.org/wiki/Fibonacci_number).
This logic is working fine. (Recursion)
public static int numberOfWays(int n) {
if (n== 1) {
return 1; // step
} else if (n== 2) {
return 2; // (step + step) or jump
} else {
return numberOfWays(n- 1)
+ numberOfWays(n- 2);
}
}
The accepted answer fails performance test for larger sets. Here is a version with for loop that satisfies performance tests at testdome.
using System;
public class Frog
{
public static int NumberOfWays (int n)
{
int first = 0, second = 1;
for ( int i = 0; i<n; i++ )
{
int at = first;
first = second;
second = at + second;
}
return second;
}
public static void Main (String[] args)
{
Console.WriteLine (NumberOfWays (3));
}
}
C++ code works fine.
static int numberOfWays(int n)
{
if (n == 1) return 1;
else if (n == 2) return 2;
else
{
static std::unordered_map<int,int> m;
auto i = m.find(n);
if (i != m.end())
return i->second;
int x = numberOfWays(n - 1) + numberOfWays(n - 2);
m[n] = x;
return x;
}
}
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