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How to sum consecutive equal digits in a number in Java

The following question was asked in my last interview (yesterday), and I'm trying to solve it since then (couldn't solve it in the interview).
Sorry for any grammar mistakes or any logical mistakes, I don't have the question, it was written by memory:
You are given a number in a string format, for example: "14438832066".
You got to sum up the consecutive equal digits in that number. If no
consecutive equal digit was found, just add the digit to the result.
for example: solution(19938832066) => 11831632012
Explanation: first digit is 1.
The second and third digits are both 9 which means they will turn into 18 in the result string.
So on
with the rest of the digits (as you can see, the last 2 digits are both 6 which means they will turn into 12 in the result string).
You are required to do that for the result string as well, if needed, until no equal consecutive digits are found in the number.
Example:: number: 14438832066 solution( "19938832066") ->"11831632012" -> "2831632012"
Explanation: first result is 11831632012, but then you can see that there are still equal consecutive digits : the first and the
second digits are both 1. So process that number as well.
You are given a string and must return a string.
My solution:
I couldn't write the solution, I don't know why. It's a pretty simple question, I thought going recursive at first but didn't want to complex things.
I wrote 2 helper methods:
one that returns a boolean whether the number consists of equal consecutive digits.
one that actually makes the business logic:
turn the string into a char array
create a counter that will count instances of the same digit - (int counter = 1).
loop on the array from the first to the one before the last element :
inside the loop:
//equal digit was found - increment counter and continue to next digit
if char[i] == char[i+1] then counter++
//calculation in case we are done counting the same digit
else if counter > 0 then result.append(counter*digit[i])
// if no consecutive equal digit was found
else result.append(digit[i])
end loop: return result
Problems I had:
I created the counter inside the loop, so each iteration it got rested. took me few minutes to realize.
I had troubles realizing that 'int(digit[i])' doesn't give me the numeric value of the char, it gives the ASCII value. I had to use "Character.getNumericValue" (don't remember the exact name of the method).
Because of these problems, it took me 45 minutes to write the solution which in the end didn't even work.
I'll be glad to get a working solution, and even better - to get any feedback and tips on my solution and what, in your opinion, were my mistakes.
Thank you.

Your pseudo-code seems alright, as far as it goes. What's missing is that you don't repeatedly check the result string to see if another pass is required. I also show how you don't need to remember the API to convert a character to a digit; if you know the digits are decimal, you can interpret them yourself. As an interviewer, I would accept that there is an API that you can't precisely remember or your home-brew solution as equally valid.
String transform(String number) {
while (true) {
String result = collapse(number);
if (result.equals(number)) return result;
number = result;
}
}
private static String collapse(String number) {
StringBuilder result = new StringBuilder();
for (idx = 0; idx < number.length(); ) {
int mark = idx;
int digit = digitAt(number, idx++);
while (idx < number.length() && digitAt(number, idx) == digit) ++idx;
result.append((idx - mark) * digit);
}
return result.toString();
}
private static int digitAt(String num, int index) {
char ch = number.charAt(index);
if (ch < '0' || ch > '9') throw new IllegalArgumentException();
return ch - '0';
}
The preceding is a naïve approach that transforms the string until there are no changes. I suspect there might be a more "elegant" approach that works from left to right through the input in a single pass, but it would take some thought, and I probably couldn't come up with that in an interview.

Here's an algorithm that uses recursion and a for-loop to add consecutive equal digits. I think the code is pretty self-explanatory but please ask if you have any queries.
public static String addConsecutiveDigits(String number) {
char[] arr = number.toCharArray();
StringBuilder result = new StringBuilder();
boolean foundConsecutive = false; // boolean flag for checking if the number contained consecutive equal digits
for (int i = 0; i < arr.length; i++) {
int digit = arr[i] - '0'; //Subtracting ascii values to get integer values
int newNumber = digit;
if (i != arr.length - 1) {
int nextDigit = arr[i + 1] - '0';
if (digit == nextDigit) { // check if the digits are consecutive digits
newNumber = digit + nextDigit;
i++; // increment i as we have already added the i+1 digit
foundConsecutive = true;
}
}
result.append(newNumber);
}
if (!foundConsecutive) // if no consecutive equal digits were found then return the result;
return result.toString();
else // recurse to check for more consecutive equal digits
return addConsecutiveDigits(result.toString());
}

I'm not a Java guy, so this code might not be ideal but I would do something like this:
public String solve(String input)
{
String result = "";
int i = 0;
while (i < input.length())
{
var first = input.charAt(i);
if (i == input.length() - 1){
result += first;
break;
}
var second = input.charAt(i + 1);
if (first == second){
result += (Character.getNumericValue(first) + Character.getNumericValue(second));
i += 2;
} else {
result += first;
i += 1;
}
}
return result;
}
For the second part, I would just run the function in a loop until the result matches the input.

Getting the second digit from a long variable [duplicate]

This question already has answers here:
Java: parse int value from a char
(9 answers)
Closed 5 years ago.
I am trying to fetch second digit from a long variable.
long mi = 110000000;
int firstDigit = 0;
String numStr = Long.toString(mi);
for (int i = 0; i < numStr.length(); i++) {
System.out.println("" + i + " " + numStr.charAt(i));
firstDigit = numStr.charAt(1);
}
When I am printing firstDigit = numStr.charAt(1) on console. I am getting 1 which is expected but when the loop finishes firstDigit has 49.
Little confused why.

Because 49 is the ASCII value of char '1'.
So you should not assign a char to int directly.
And you don't need a loop here which keeps ovveriding the current value with charAt(1) anyway.
int number = numStr.charAt(1) - '0'; // substracting ASCII start value
The above statement internally works like 49 -48 and gives you 1.
If you feel like that is confusious, as others stated use Character.getNumericValue();
Or, although I don't like ""+ hack, below should work
int secondDigit = Integer.parseInt("" + String.valueOf(mi).charAt(1));

You got confused because 49 is ASCII value of integer 1. So you may parse character to integer then you can see integer value.
Integer.parseInt(String.valueOf(mi).charAt(1)):

You're probably looking for Character.getNumericValue(...) i.e.
firstDigit = Character.getNumericValue(numStr.charAt(1));
Otherwise, as the variable firstDigit is of type int that means you're assigning the ASCII representation of the character '1' which is 49 rather than the integer at the specified index.
Also, note that since you're interested in only a particular digit there is no need to put the statement firstDigit = numStr.charAt(1); inside the loop.
rather, just do the following outside the loop.
int number = Character.getNumericValue(numStr.charAt(1));

you only need define firstDigit as a char type variable, so will print as character.
since you define as int variable, it's value is the ASCII value of char '1': 49. this is why you get 49 instead of 1.

the answer Integer.parseInt(String.valueOf(mi).charAt(1)+""); is correct.
However, if we want to consider performace in our program, we need some improvements.
We have to time consuming methods, Integer.parseInt() and String.valueOf(). And always a custom methods is much faster than Integer.parseInt() and String.valueOf(). see simple benchmarks.
So, high performance solution can be like below:
int y=0;
while (mi>10)
{
y=(int) (mi%10);
mi=mi/10;
}
System.out.println("Answer is: " + y);
to test it:
long mi=4642345432634278834L;
int y=0;
long start = System.nanoTime();
//first solution
//y=Integer.parseInt(String.valueOf(mi).charAt(1)+"");
//seconf solution
while (mi>10)
{
y=(int) (mi%10);
mi=mi/10;
}
long finish = System.nanoTime();
long d = finish - start;
System.out.println("Answer is: " + y + " , Used time: " + d);
//about 821 to 1232 for while in 10 runs
//about 61225 to 76687 for parseInt in 10 runs

Doing string manipulation to work with numbers is almost always the wrong approach.
To get the second digit use the following;
int digitnum = 2;
int length = (int)Math.log10(mi));
int digit = (int)((mi/Math.pow(base,length-digitnum+1))%base);
If you want a different digit than the second change digitnum.
To avoid uncertainty with regards to floating point numbers you can use a integer math library like guavas IntMath

Let's take a look
System.out.println(numStr.charAt(1));
firstDigit = numStr.charAt(1);
System.out.println(firstDigit);
The result wouldn't be the same you will get
1
49
This happens because your firstDigit is int. Change it to char and you will get expected result

You can also do like below,
firstDigit = Integer.parseInt( numStr.charAt(1)+"");
So it will print second digit from long number.

Some things which have not been mentioned yet:
The second digit for integer datatypes is undefined if the long number is 0-9 (No, it is not zero. Integers do not have decimal places, this is only correct for floating-point numbers. Even then you must return undefined for NaN or an infinity value). In this case you should return a sentinel like e.g. -1 to indicate that there is no second digit.
Using log10 to get specific digits looks elegant, but they are 1. one of the numerically most expensive functions and 2. do often give incorrect results in edge cases. I will give some counterexamples later.
Performance could be improved further:
public static int getSecondDigit(long value) {
long tmp = value >= 0 ? value : -value;
if (tmp < 10) {
return -1;
}
long bigNumber = 1000000000000000000L;
boolean isBig = value >= bigNumber;
long decrement = isBig ? 100000000000000000L : 1;
long firstDigit = isBig ? bigNumber : 10;
int result = 0;
if (!isBig) {
long test = 100;
while (true) {
if (test > value) {
break;
}
decrement = firstDigit;
firstDigit = test;
test *= 10;
}
}
// Remove first
while (tmp >= firstDigit) {
tmp -= firstDigit;
}
// Count second
while (tmp >= decrement) {
tmp -= decrement;
result++;
}
return result;
}
Comparison:
1 000 000 random longs
String.valueOf()/Character.getNumericValue(): 106 ms
Log/Pow by Taemyr: 151 ms
Div10 by #Gholamali-Irani: 45 ms
Routine above: 30 ms
This is not the end, it can be even faster by lookup tables
decrementing 1/2/4/8, 10/20/40/80 and avoid the use of multiplication.

try this to get second char of your long
mi.toString().charAt(1);

How to get ASCII code
int ascii = 'A';
int ascii = 'a';
So if you assign a character to an integer, the integer will be holding the ASCII value of that character. Here I explicitly gave the values, in your code you are calling a method that returns a character, that's why you are getting ASCII instead of digit.

Converting a decimal number to binary in java not showing leading zeroes

This is my function in Java:
public static String convertFromDecimal(int number, int base)
{
String result = "";
/*
* This while loop will keep running until 'number' is not 0
*/
while(number != 0)
{
result = (number%base) + result; // Appending the remainder
number = number / base; // Dividing the number by the base so we can get the next remainder
}
// If the number is already 0, then the while loop will ignore it, so we will return "0"
if(result == "")
{
return "0";
}
return result;
}
It works fine for numbers that convert to numbers not beginning with 0, if the number is supposed to have a zero at the start, it will not record it, could anyone tell me why?
For example, if I print out
convertFromDecimal(13,2) it returns
1101
Which is correct, but if I print out
convertFromDecimal(461,2), I get
111001101
Where the actual answer is
0000000111001101
So it's the same as my answer without the leading zeroes, if anyone knows why I would appreciate the help, thank you.
EDIT My question is different because I don't want 16 digits, I want the binary number of the given decimal, a calculator like this can explain what I want.

I assume you are looking to format all your answers as shorts (16 bits).
In this case, simply check the length of your current string, and add on zeroes as needed.
int zeroesRemaining = 16 - result.length();
for (int i = 0; i < zeroesRemaining; i++) {
result = "0" + result;
}
Alternatively, if you want to do it faster, use a StringBuilder.
int zeroesRemaining = 16 - result.length();
StringBuilder tempBuilder = new StringBuilder(result);
for (int i = 0; i < zeroesRemaining; i++) {
tempBuilder.insert(0, 0); //inserts the integer 0 at position 0 of the stringbuilder
}
return tempBuilder.toString(); //converts to string format
There is also probably a formatter that could do this, but I don't know of such.
If you want to change the number of zeroes to be the closest integer primitive, just set zeroesRemaining to be the (least power of 2 that is greater than the number of bits) minus (the number of bits).

Since you want fixed lengths for your result, in groups of 8 bits, the easiest way is to append 0 to the front of your result until its length is a multiple of 8.
That is as simple as
wile (result.length() % 8 > 0)
{
result = "0" + result;
}
return result;

Converting int to char in same integer format [duplicate]

Below is a code snippet,
int a = 1;
char b = (char) a;
System.out.println(b);
But what I get is empty output.
int a = '1';
char b = (char) a;
System.out.println(b);
I will get 1 as my output.
Can somebody explain this? And if I want to convert an int to a char as in the first snippet, what should I do?

int a = 1;
char b = (char) a;
System.out.println(b);
will print out the char with Unicode code point 1 (start-of-heading char, which isn't printable; see this table: C0 Controls and Basic Latin, same as ASCII)
int a = '1';
char b = (char) a;
System.out.println(b);
will print out the char with Unicode code point 49 (one corresponding to '1')
If you want to convert a digit (0-9), you can add 48 to it and cast, or something like Character.forDigit(a, 10);.
If you want to convert an int seen as a Unicode code point, you can use Character.toChars(48) for example.

My answer is similar to jh314's answer but I'll explain a little deeper.
What you should do in this case is:
int a = 1;
char b = (char)(a + '0');
System.out.println(b);
Here, we used '0' because chars are actually represented by ASCII values. '0' is a char and represented by the value of 48.
We typed (a + '0') and in order to add these up, Java converted '0' to its ASCII value which is 48 and a is 1 so the sum is 49. Then what we did is:
(char)(49)
We casted int to char. ASCII equivalent of 49 is '1'. You can convert any digit to char this way and is smarter and better way than using .toString() method and then subtracting the digit by .charAt() method.

It seems like you are looking for the Character.forDigit method:
final int RADIX = 10;
int i = 4;
char ch = Character.forDigit(i, RADIX);
System.out.println(ch); // Prints '4'
There is also a method that can convert from a char back to an int:
int i2 = Character.digit(ch, RADIX);
System.out.println(i2); // Prints '4'
Note that by changing the RADIX you can also support hexadecimal (radix 16) and any radix up to 36 (or Character.MAX_RADIX as it is also known as).

int a = 1;
char b = (char) a;
System.out.println(b);
hola, well i went through the same problem but what i did was the following code.
int a = 1
char b = Integer.toString(a).charAt(0);
System.out.println(b);
With this you get the decimal value as a char type. I used charAt() with index 0 because the only value into that String is 'a' and as you know, the position of 'a' into that String start at 0.
Sorry if my english isn't well explained, hope it helps you.

you may want it to be printed as '1' or as 'a'.
In case you want '1' as input then :
int a = 1;
char b = (char)(a + '0');
System.out.println(b);
In case you want 'a' as input then :
int a = 1;
char b = (char)(a-1 + 'a');
System.out.println(b);
java turns the ascii value to char :)

int a = 1;
char b = (char) (a + 48);
In ASCII, every char have their own number. And char '0' is 48 for decimal, '1' is 49, and so on. So if
char b = '2';
int a = b = 50;

Nobody has answered the real "question" here: you ARE converting int to char correctly; in the ASCII table a decimal value of 01 is "start of heading", a non-printing character. Try looking up an ASCII table and converting an int value between 33 and 7E; that will give you characters to look at.

Whenever you type cast integer to char it will return the ascii value of that int (once go through the ascii table for better understanding)
int a=68;
char b=(char)a;
System.out.println(b);//it will return ascii value of 68
//output- D

If we are talking about class types - not primitives, the following trick has to be done:
Integer someInt;
Character someChar;
someChar = (char)Integer.parseInt(String.valueOf(someInt));

First, convert the int (or another type) to String,
int a = 1;
String value = String.valueOf(a);
Then, convert that String to char.
char newValue = value.charAt(0);
You can avoid empty output in this way...
System.out.println(newValue);

In java a char is an int. Your first snippet prints out the character corresponding to the value of 1 in the default character encoding scheme (which is probably Unicode). The Unicode character U+0001 is a non-printing character, which is why you don't see any output.
If you want to print out the character '1', you can look up the value of '1' in the encoding scheme you are using. In Unicode this is 49 (the same as ASCII). But this will only work for digits 0-9.
You might be better off using a String rather than a char, and using Java's built-in toString() method:
int a = 1;
String b = toString(a);
System.out.println(b);
This will work whatever your system encoding is, and will work for multi-digit numbers.

if you want to print ascii characters based on their ascii code and do not want to go beyond that (like unicode characters), you can define your variable as a byte, and then use the (char) convert. i.e.:
public static void main(String[] args) {
byte b = 65;
for (byte i=b; i<=b+25; i++) {
System.out.print((char)i + ", ");
}
BTW, the ascii code for the letter 'A' is 65

Make sure the integer value is ASCII value of an alphabet/character.
If not then make it.
for e.g. if int i=1
then add 64 to it so that it becomes 65 = ASCII value of 'A'
Then use
char x = (char)i;
print x
// 'A' will be printed

There is one method by which int can be converted to char and even without using ASCII values.
Example:
int i = 2;
char ch = Integer.toString(i).charAt(0);
System.out.println(ch);
Explanation :
First the integer is converted to string and then by using String function charAt(), character is extracted from the string. As the integer only has one single digit, the index 0 is given to charAt() function.

My solution is for converting lower case alphabets (a-z) to (0-25) and vice versa.
My answer is for a specific use-case it is not generic solution my solution will help you if you want to store the frequency of character into an integer array of size 26 instead of using Hashmap<Character,Integer>
----> for converting 0 to 25 into a-z
char ch=(char)(0+'a'); // output 'a' // input 0(as integer)
char ch=(char)(25+'a'); // output 'z' // input 25(as integer)
---->for converting a to z into 0-25
int freq='a'-'a' // output 0 // input 'a'
int freq='b'-'a' // output 1 // input 'b'
int freq='c'-'a' // output 2 // input 'c'
int freq='z'-'a' // output 25 // input 'z'
Again this approach will help you to get the frequency of characters as well as characters
public class Main
{
public static void main(String[] args) {
String s="rajatfddfdf";
int freq[]= new int[26];
for(int i=0;i<s.length();i++){
char characterAtIndex=s.charAt(i);
freq[characterAtIndex-'a']+=1;
}
for(int i=0;i<26;i++){
System.out.println((char)('a'+i)+" frequency="+freq[i]);
}
}
}
by using the above code we can get the frequency as well as character using integer array of size 26 . We can write if-else logic if you don't want to include the character with frequency 0.

look at the following program for complete conversion concept
class typetest{
public static void main(String args[]){
byte a=1,b=2;
char c=1,d='b';
short e=3,f=4;
int g=5,h=6;
float i;
double k=10.34,l=12.45;
System.out.println("value of char variable c="+c);
// if we assign an integer value in char cariable it's possible as above
// but it's not possible to assign int value from an int variable in char variable
// (d=g assignment gives error as incompatible type conversion)
g=b;
System.out.println("char to int conversion is possible");
k=g;
System.out.println("int to double conversion is possible");
i=h;
System.out.println("int to float is possible and value of i = "+i);
l=i;
System.out.println("float to double is possible");
}
}
hope ,it will help at least something

If you want to convert a character to its corresponding integer, you can do something like this:
int a = (int) 'a';
char b = (char) a;
System.out.println(b);
This happens because in ASCII there are some items that can not be printed normally.
For example, numbers 97 to 122 are integers corresponding to the lowercase letters a to z.

public class String_Store_In_Array
{
public static void main(String[] args)
{
System.out.println(" Q.37 Can you store string in array of integers. Try it.");
String str="I am Akash";
int arr[]=new int[str.length()];
char chArr[]=str.toCharArray();
char ch;
for(int i=0;i<str.length();i++)
{
arr[i]=chArr[i];
}
System.out.println("\nI have stored it in array by using ASCII value");
for(int i=0;i<arr.length;i++)
{
System.out.print(" "+arr[i]);
}
System.out.println("\nI have stored it in array by using ASCII value to original content");
for(int i=0;i<arr.length;i++)
{
ch=(char)arr[i];
System.out.print(" "+ch);
}
}
}

how does Float.toString() and Integer.toString() works?

How can i implement an algorithm to convert float or int to string?
I found one link
http://geeksforgeeks.org/forum/topic/amazon-interview-question-for-software-engineerdeveloper-0-2-years-about-algorithms-13
but i cant understand the algorithm given there

the numbers 0-9 are sequential in most character encoding so twiddling with the integral value of it will help here:
int val;
String str="";
while(val>0){
str = ('0'+(val%10)) + str;
val /= 10;
}

Here's a sample of how to do the integer to string, from it I hope you'll be able to figure out how to do the float to string.
public String intToString(int value) {
StringBuffer buffer = new StringBuffer();
if (value < 0) {
buffer.append("-");
}
// MAX_INT is just over 2 billion, so start by finding the number of billions.
int divisor = 1000000000;
while (divisor > 0) {
int digit = value / divisor; // integer division, so no remainder.
if (digit > 0) {
buffer.append('0'+digit);
value = value - digit * divisor; // subtract off the value to zero out that digit.
}
divisor = divisor / 10; // the next loop iteration should be in the 10's place to the right
}
}
This is of course, very unoptimized, but it gives you a feel for how the most basic formatting is accomplished.
Note that the technique of "" + x is actually rewritten to be something like
StringBuffer buffer = new StringBuffer();
buffer.append("");
buffer.append(String.valueOf(x));
buffer.toString();
So don't think that what is written is 100% exactly HOW it is done, look at is as what must happen in a larger view of things.

The general idea is to pick off the least significant digit by taking the number remainder ten. Then divide the number by 10 and repeat ... until you are left with zero.
Of course, it is a bit more complicated than that, especially in the float case.
if i have a single digit in int fomrat then i need to insert it into char , how to convert int to char?
Easy:
int digit = ... /* 0 to 9 */
char ch = (char)('0' + digit);

Well, you can read the code yourself.