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Bloomberg LP Interview Question for Financial Software Developers


Country: United States



Comment hidden because of low score. Click to expand.
13
of 13 vote

Algorithm for problem
j=1
for i = 1 to N
if (A[i] > 0)
A[j] = A[i]
j++

for i = j to N
a[i] = 0

This takes O(n) without using any buffer

- Punit Jain March 20, 2012 | Flag Reply
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0
of 0 votes

It requires only one iteration. Time complexity if O(n).

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] test = { 1, 2, 0, 4, 6, 7, 0, 9, 12, 45, 0, 49, 0, 99, 0 };

            MoveZeroAtTheEnd(test);

            Console.WriteLine(test.ToString());
        }

        static void  MoveZeroAtTheEnd(int[] array)
        {
            int nZero = 0;

            for (int i = 0; i < array.Length; ++i)
            {
                if (array[i] == 0)
                {
                    nZero++;
                    continue;
                }
                
                if( nZero > 0 )
                {
                    array[i - nZero] = array[i];

                    array[i] = 0;
                }
            }
        }
    }
}

- Anonymous April 13, 2012 | Flag
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0
of 0 votes

It requires only one iteration. Time complexity if O(n).

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] test = { 1, 2, 0, 4, 6, 7, 0, 9, 12, 45, 0, 49, 0, 99, 0 };

            MoveZeroAtTheEnd(test);

            Console.WriteLine(test.ToString());
        }

        static void  MoveZeroAtTheEnd(int[] array)
        {
            int nZero = 0;

            for (int i = 0; i < array.Length; ++i)
            {
                if (array[i] == 0)
                {
                    nZero++;
                    continue;
                }
                
                if( nZero > 0 )
                {
                    array[i - nZero] = array[i];

                    array[i] = 0;
                }
            }
        }
    }
}

- viral April 13, 2012 | Flag
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0
of 0 votes

Even shorter:

public int[] MoveZeroAtTheEnd(int[] org){
		int[] ret = new int[org.length];
		int retForwardIndex = 0;
		int retBackwardIndex = org.length-1;
		for(int i=0;i<org.length;i++){
			if(org[i] == 0){
				ret[retBackwardIndex--] = org[i];
			}else{
				ret[retForwardIndex++] = org[i];
			}
		}	
		return ret;
	}

- Ari April 12, 2013 | Flag
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3
of 3 vote

This just copying the non zero elements forward to the positions where it supposed to be. and using the zero counter, adding the zeros after the numbers into the array.

public static void main(String args[]){
		int[] arrayZeros = {1,3,0,8,12,0,4,0,7};
		
		int zeroCounter = 0;
		int indexPointer = -1;
		for(int i=0;i<arrayZeros.length;i++){
			
			if(arrayZeros[i]!=0){
				indexPointer++;
			    if(i>indexPointer)
			    	arrayZeros[indexPointer] = arrayZeros[i];
			}    
			else{
				zeroCounter++;				
			}			
		}
		int startIndex = arrayZeros.length - zeroCounter; 
		for(int i=1;i<=zeroCounter;i++){
			arrayZeros[startIndex++] = 0;
		}
		
		for(int i=0;i<arrayZeros.length;i++){
			System.out.println("["+arrayZeros[i]+"]");
		}
	}

- Ram March 15, 2012 | Flag Reply
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0
of 0 votes

Excellent! Thanks..

- Shree March 21, 2012 | Flag
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0
of 0 votes

just awesome

- pavankumartripurari November 23, 2012 | Flag
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0
of 0 vote

Code in c# :

Int32[] sortArrayPlacingZeroAtEnd(Int32[] inarray)
      {
            Int32[] outarray = new Int32[inarray.Length];
            int j = 0;
            for (int i = 0; i <= outarray.Length - 1; i++)
            {
                if (inarray[i] != 0)
                {
                  outarray[j++] = inarray[i];
                }
            }
            return outarray;
        }

- salahuddin March 15, 2012 | Flag Reply
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0
of 0 votes

Sorry, I forgot to clarify; You can't use another buffer.

- sunghw March 15, 2012 | Flag
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0
of 0 vote

#include<stdio.h>
#include<stdlib.h>

void print(int *a, int n) {
	int i;
	for(i=0;i<n;i++)
		printf("%d, ", a[i]);
	printf("\n");
}

void swap(int *a, int *b) {
	int tmp;
	tmp=*a;
	*a=*b;
	*b=tmp;
}

int main(void) {
	int *a, n, i, end;
	scanf("%d", &n);
	a=(int *)calloc(n,sizeof(int));
	for(i=0;i<n;i++)
		scanf("%d", &a[i]);
	print(a,n);
	end=i=n-1;
	while(i>=0){
		if(a[i]) {
			if(i<end)
				swap(&a[i],&a[end]);
			end--;
		}
		i--;
	}	
	print(a,n);
	free(a);
	return 0;
}

- puri March 15, 2012 | Flag Reply
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0
of 0 votes

Wait, if it's 1 2 3 0 7,
when i = 4 and end = 4, a[4] !=0, but i==end, so both i and end decrement.
i = 3, end = 3, a[3]==0, so only i decrements.
i=2, end=3, a[2] = 3, and i<end, so they swap.
12037.
I do not think it is working... There is no way you will swap the end elements 3, 7 with the zero element in the middle at this point... isn't it?

- sunghw March 15, 2012 | Flag
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0
of 0 votes

You are right "sunghw", the code above would push non-zero integers toward right of the array and zeros toward left ( for input 1 2 3 0 7, out put would be 0 1 2 3 7). *Didn't notice the output formated asked in the question*. Below is the corrected code :

#include<stdio.h>
#include<stdlib.h>



void print(int *a, int n) {
	int i;
	for(i=0;i<n;i++)
		printf("%d, ", a[i]);
	printf("\n");
}


void swap(int *a, int *b) {
	int tmp;
	tmp=*a;
	*a=*b;
	*b=tmp;
}


int main(void) {
	int *a, n, i=0, start=0;
	scanf("%d", &n);
	a=(int *)calloc(n,sizeof(int));
	for(i=0;i<n;i++)
		scanf("%d", &a[i]);
	print(a,n);
	i=0;
	while(i<n){
		if(a[i]) {
			if(i>start)
				swap(&a[i],&a[start]);
			start++;
		}
		i++;
	}
	print(a,n);
	free(a);
	return 0;
}

- puri March 17, 2012 | Flag
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0
of 0 vote

is there any restriction on the usage of extra space. because, we can iterate through the array and add non-zero's to a queue simulataneously zeroing each element.

Once complete, we can deque elements in queue and add it back to array.
time: O(n)
size: O(n)

- Somisetty March 15, 2012 | Flag Reply
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0
of 0 votes

no, that would be easy. you can't use any additional buffer.

- sunghw March 15, 2012 | Flag
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0
of 0 votes

too bad. i had another solution similar to merge sort ready.
in this one, we will divide and merge the sets recursively. except that while merging new set will have non zero's first and zero's last.

- Somisetty March 15, 2012 | Flag
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0
of 0 vote

public static void moveZero(int[] arr) {
    if (arr == null || arr.length == 0) return;
    for (int i = 0; i < arr.length - 1; i++) {
      if (arr[i] == 0) {
        for (int j = i + 1; j < arr.length; j++) {
          if (arr[j] != 0) {
            swap(arr, i, j);
            break;
          }
        }
        if (arr[i] == 0) return;
      }
    }
  }
  
  private static void swap(int[] arr, int i, int j) {
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
  }

- Zhengyang.Feng2011 March 15, 2012 | Flag Reply
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0
of 0 vote

public static void moveZero(int[] arr) {
    if (arr == null || arr.length == 0) return;
    for (int i = 0; i < arr.length - 1; i++) {
      if (arr[i] == 0) {
        for (int j = i + 1; j < arr.length; j++) {
          if (arr[j] != 0) {
            swap(arr, i, j);
            break;
          }
        }
        if (arr[i] == 0) return;
      }
    }
  }
  
  private static void swap(int[] arr, int i, int j) {
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
  }

- Zhengyang.Feng2011 March 15, 2012 | Flag Reply
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0
of 0 vote

/*
A program to add all zero at the end of array

*/
#include<stdio.h>
void zero(int a[],int n)
{

int i,j,tmp;
i=0;
while(i<n) //find first zero
{
if(a[i]==0)
break;
else
i++;
}

if(i==n-1||i==n) // if last element is zero or no zero element
return ;

j=i+1;

while(j<n&&i<n)
{
if(a[j]!=0)
{
tmp=a[i];
a[i]=a[j];
a[j]=tmp;
i++;
j++;
}
else
j++;
}
}
int main()
{

int a[]= {0,1 ,2 ,0 , 3,0,4,0,0,0,0,5,6,7,8,9};
int n=sizeof(a)/sizeof(int);
zero(a,n);

int i;

for(i=0;i<n;i++)
printf("%d",a[i]);

return 0;
}

- NaiveCoder March 15, 2012 | Flag Reply
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0
of 0 vote

Is this okay ?

static void Main(string[] args)
        {
            int[] a = {1,2,3,4,4,5,5,6,6,6};

            int i=0, j=0;

            while (j < a.Length)
            {
                if (a[j] != 0)
                {
                    a[i++] = a[j];
                }
                j++;
            }

            while (i < a.Length)
            {
                a[i++] = 0;
            }

            for (int n = 0; n < a.Length; n++)
            {
                Console.Write("{0} ", a[n]);
            }

            Console.Read();
        }

- Srivathsan March 15, 2012 | Flag Reply
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0
of 0 votes

I think it pretty good

- Jenny March 17, 2012 | Flag
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0
of 0 vote

Have 2 indexes i & j, both initialized to 0. as long as a[j] is not zero, copy it into a[i].
increment i when u copy from a[j]; The moment u reach the end of the array, you would have copied all the non zero elements to the left.
from this point on, fill the elements at a[i] with zeroes.

I guess it should work fine.

- Srivathsan March 15, 2012 | Flag Reply
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0
of 0 vote

hope this is what you are looking for

#include<iostream>
using namespace std;

int main()
{	
	int a[]={1,0,2,0,0,3,0,4,0};
	int count = 0;
	int temp;
	
	for(int i=0;i<8;i++)
	{
		for(int j=0;j<8-i;j++)
		{
			if(a[j] == 0)
			{
				while(a[j+1] !=0)
				{temp=a[j];
				a[j]=a[j+1];
				a[j+1]=temp;j++;}	
			}
		}	
		
		
	}
	
	for(int i=0;i<8;i++)
	{
		cout << a[i];
		}
	
	return 0;
}

- abhinav March 16, 2012 | Flag Reply
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0
of 0 vote

If we have extra space, we can allocate a new array buffer.

if the current element in the original array is non-zero, copy it to the new array and increase new array's index.
Otherwise just continue without updating new array's index.
When we reach the end of the original array, fill the new array with 0s.

This algorithm takes extra n space and takes liner time.

- Jason March 16, 2012 | Flag Reply
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0
of 0 vote

I wrote a code using vector hope this helps

for(int i=0;i<n;){
    if(b[i]==0){
      b.erase(b.begin()+i);
      b.push_back(0);
      n=n-1;
    }
    else
      i++;
  }

here I used erase the number 0 from a list and pushed a 0 at the back of the vector

- harish_venkat March 16, 2012 | Flag Reply
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0
of 0 vote

int a[] ={1,3,0,0,4,10,0,18,3,0,6,0,1,4};   
        int current_zero_index = 0;
		int count =0 ,flag=0;
		
		for (int i=0; i<a.length;i++) {
			
			if(a[i]==0 && flag==0) {
				current_zero_index=i-count;
				flag=1;
			//	count++;
			}
			
			if(a[i]>0 && current_zero_index>0) {
			    a[current_zero_index++] = a[i];
			    count++;
			    a[i]=0;
			    flag=0;
			}
			
		}

- Anonymous March 17, 2012 | Flag Reply
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0
of 0 vote

public class fixArray {
public static void main(String[] args) {
int[] arr = {5,0,3,7,8,0,1,0};
int j = 0;
for(int i = 0; i < arr.length; i++) {
if (arr[i] != 0) {
arr[j] = arr[i];
if(i != j) arr[i] = 0;
++j;
}
}
for(int i = 0; i < arr.length; i++) {
System.out.println(arr[i]);
}
}
}

- SolutionYODA March 18, 2012 | Flag Reply
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0
of 0 votes

Just a minor comment:

if (arr[i] != 0) {
if(i != j) {
arr[j] = arr[i]; // this can be done only if i not equal to j
arr[i] = 0;
}
++j;
}

of course your thing works

- venkat January 08, 2013 | Flag
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0
of 0 vote

This works!!

#include <iostream>
using namespace std;

int main()
{
	int n = 15;
	int a[15] = {1,0,0,8,12,0,4,0,7,0,0,0,0,1,1};
	int i = 0, j= 0;
	for (int k = 0; k<n; k++) cout <<a[k]<<'\t';
	cout <<'\n'<<'\n';
	while(j<n)
	{

		if(a[j]==0 && a[i]!=0)i=j++;

		else if(a[i]==0 && a[j]!=0)
		{
			int temp = a[i];
			a[i] = a[j];
			a[j] = temp;
			i++;j++;
		}
		else if (a[j]==0 && a[i] ==0) j++;
		else j++;
	}
	cout <<'\n'<<'\n';
	for (int k = 0; k<n; k++) cout <<a[k]<<'\t';
}

- Luc March 19, 2012 | Flag Reply
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0
of 0 votes

This works in O(n). crawling the array just ones

- Luc March 19, 2012 | Flag
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0
of 0 vote

for(i=0; i<5; i++)
{
if(arr[i]== 0)
{
shift_left++;
}
else
{
arr[i-shift_left] = arr[i];
if(shift_left!=0)
{
arr[i] = 0;
}
}
}

- tosha shah March 22, 2012 | Flag Reply
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0
of 0 vote

void PutZroatEnd(int src[],int size)
{
int distancetoCursor = 1;
for(int i =0;i<size-distancetoCursor;++i)
{
if(src[i] == 0)
{
if(src[i+distancetoCursor] == 0)
++distancetoCursor;
int temp = src[i+distancetoCursor];
src[i+distancetoCursor] = src[i];
src[i] = temp;
}
}
}

- Anonymous March 28, 2012 | Flag Reply
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0
of 0 vote

This one works for the data set provided in the code. I guess I can improve this a lot. There is a bug though. The code is getting into infinite loop if I have zeros in the beginning.

The code is in C# (Windows Forms)

private void button1_Click(object sender, EventArgs e)
        {
            int[] a = new int[] { 13, 8, 0, 12, 0, 4, 7, 0, 8, 9, 0, 5, 6, 0, 17, 12, 34, 56, 78, 0, 98, 87, 0, 65, 0, 54, 0, 0, 0, 101, 0, 1, 0, 2, 0, 3, 0, 4 };
            int nextcurr = -1;
            int noofzeros = 0;
            int count = 0;

            // 1 iteration to get the number of zeros
            for (int i = 0; i < a.Length; i++)
            {
                if (a[i] == 0)
                    noofzeros++;
            }

            int curr = 0;

            while (nextcurr != (a.Length - noofzeros))
            {
                // Check to swap
                if (curr < a.Length - 1)
                {
                    if (a[curr] == 0 && a[curr + 1] != 0)
                        Swap(a, curr, curr + 1);
                    else if (a[curr] == 0 && a[curr + 1] == 0 && nextcurr < 0)
                        nextcurr = curr;
                    
                    curr++;
                    Display(a);
                }
                else if (curr >= a.Length - 1 && nextcurr > 0)
                {
                    curr = nextcurr;
                    nextcurr = -1;
                }

                count++;
            }

            richTextBox1.AppendText("\nTotal iterations: " + count.ToString());
        }

- Nitin April 10, 2012 | Flag Reply
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0
of 0 vote

Here is the output for the code above:

13 ,8 ,0 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,0 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,0 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,0 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,0 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,0 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,0 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,0 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,0 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,0 ,98 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,0 ,87 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,0 ,65 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
13 ,8 ,12 ,0 ,4 ,7 ,0 ,8 ,9 ,0 ,5 ,6 ,0 ,17 ,12 ,34 ,56 ,78 ,0 ,98 ,87 ,0 ,65 ,0 ,0 ,54 ,0 ,0 ,0 ,101 ,0 ,1 ,0 ,2 ,0 ,3 ,0 ,4
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13 ,8 ,12 ,4 ,7 ,8 ,9 ,5 ,6 ,17 ,12 ,34 ,56 ,78 ,98 ,87 ,65 ,54 ,101 ,1 ,2 ,3 ,0 ,4 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0
13 ,8 ,12 ,4 ,7 ,8 ,9 ,5 ,6 ,17 ,12 ,34 ,56 ,78 ,98 ,87 ,65 ,54 ,101 ,1 ,2 ,3 ,0 ,4 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0
13 ,8 ,12 ,4 ,7 ,8 ,9 ,5 ,6 ,17 ,12 ,34 ,56 ,78 ,98 ,87 ,65 ,54 ,101 ,1 ,2 ,3 ,4 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0
13 ,8 ,12 ,4 ,7 ,8 ,9 ,5 ,6 ,17 ,12 ,34 ,56 ,78 ,98 ,87 ,65 ,54 ,101 ,1 ,2 ,3 ,4 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0

Total iterations: 349

- Nitin April 10, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
using namespace std;

// version 1: using a separate buffer
//void separateZeros(int *input, int *output, int size){
//     for(int i=0, j=0, k=size-1; i<size; ++i){   
//          if(input[i] != 0){
//               output[j++] = input[i]; 
//          }
//          else{
//               output[k--] = input[i];    
//          }
//     }
//}

// version 2: no extra buffer
void separateZeros(int *arr, int size){
     int i = 0;
     while(i < size && arr[i] != 0)
          ++i;     //find first 0

     if(i < size-1){     
          int j = i+1;  
          while(j < size){
               if(arr[j] == 0){
                    ++j;
               }
               else{
                    arr[i] = arr[j];
                    arr[j] = 0; 
                    ++i;
                    ++j;
               }
          }
     }
}

int main(){
     int a[] = {1, 3, 0, 8, 12, 0, 4, 0, 7};     
     int n = sizeof(a)/sizeof(a[0]); 

     separateZeros(a, n);
     cout << "after separating 0's and non-0's: " << endl;
     for(int i=0; i<n; ++i){
          cout << a[i] << " ";
     }
     cout << endl;

     return 0;
}

- phe April 10, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Use stable_sort to sort and preserve ordering of certain elements.

bool q1_comp(int i_first,int i_second){return (i_first==0)<(i_second==0);}
void main(){
   int q1_ints[]={1,3,0,8,12,0,4,0,7};
   vector<int> q1_array(q1_ints,q1_ints+sizeof(q1_ints)/sizeof(q1_ints[0]));
   stable_sort(q1_array.begin(),q1_array.end(),q1_comp);
}

- stevenamolzen@hotmail.com April 24, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

for(int j=0; j<n; j++)
{
for(i=0;i<n&&a[i]=0;i++)
{
int temp=a[i];
a[i]=a[i+1];
}
a[i]=temp;
}

- Anonymous May 10, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

for(int j=0; j<n; j++)
{
for(i=0;i<n&&a[i]=0;i++)
{
int temp=a[i];
a[i]=a[i+1];
}
a[i]=temp;
}

- Anonymous May 10, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<iostream>
using namespace std;

void swap(int a[],int length)
{
     int zeroPointer = 0;
     int nonZeroPointer = 0;
     while((zeroPointer < length) &&(nonZeroPointer < length))
     {
                        while(a[zeroPointer] != 0)
                        {
                                             zeroPointer++;
                                             if(zeroPointer == length)
                                             {
                                                            break;
                                             }
                        }
                        while((nonZeroPointer < zeroPointer) || (a[nonZeroPointer] == 0))
                        {
                                                 nonZeroPointer++;
                                                 if(nonZeroPointer == length)
                                                 {
                                                                   break;
                                                 }
                        }
                        if((zeroPointer < length) && (nonZeroPointer < length))
                        {
                                        int temp = a[nonZeroPointer];
                                        a[nonZeroPointer] = a[zeroPointer];
                                        a[zeroPointer]= temp;
                                        if(a[zeroPointer + 1] == 0)
                                        {
                                             zeroPointer++;
                                        }
                        }
     }
}                        
int main()
{
    int a[]={0,1,3,0,0,8,12,4,7};
    swap(a,9); 
    for(int i=0;i<9;i++)
    {
            cout << a[i] << " ";
    }
    cout << endl
}

- Anonymous June 29, 2012 | Flag Reply
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0
of 0 vote

import java.util.Arrays;

public class CareerUp_ArrayIndex {

public static void main(String[] args) {
process();
}

public static void process() {

int[] price = {1,3,0,8,12, 0, 4, 0,7};
int[] _price = new int[price.length];
int j = price.length-1;
int k = 0;

for (int i=0;i<price.length;i++){
if (price[i] != 0){
_price[k] = price[i];
k++;
}
if (price[i] == 0){
_price[j] = price[i];
j--;
}
}
System.out.println(Arrays.toString(_price));
}
}

- Xiaonb July 25, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

// Move all zeros to the end
void MoveZerosToEnd(int a[], size_t n)
{
	size_t nCurIndex = 0;
	for (size_t i = 0; i < n; ++i)
	{
		if (0 != a[i])
		{
			a[nCurIndex] = a[i];
			++nCurIndex;
		}
	}
	for (int i = nCurIndex; i < n ; ++i)
	{
		a[i] = 0;
	}
}

- Saranya July 30, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

#include <stdlib.h>
#include <stdio.h>

void printarray(int *a, int n);

void main()
{
int A[]={0, 3, 0, 5, 7, 0, 0, 8};
int i, j, k;
int n = sizeof(A)/sizeof(int);

printarray(A, n);
for (i=0; i<n; i++)
{
if (A[i]==0)
{
for (j=i+1; j<n;j++)
{
if (A[j]!=0)
{
k=A[i];
A[i] = A[j];
A[j] = k;
break;
}
}
}
}
printarray(A, n);

}

void printarray(int *a, int n)
{
int i;
for (i=0;i<n;i++)
{
printf("%3d", a[i]);
}
printf("\n");

}

- jing September 23, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void moveBackZero(int[] a){
int i, j;

for(i=a.length-1; i>=0; i--){
if(a[i]==0){
j=i;
while(j<a.length-1){
a[j] = a[j+1];
j++;
}
a[j] = 0;
}
}
}

- cli22 October 09, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Using the STL

int nums[] = { 1, 3, 0, 8, 12, 0, 4, 0, 7 };

std::stable_partition(std::begin(nums), std::end(nums), std::bind(std::greater<int>(), std::placeholders::_1, 0));

- Anonymous October 18, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

FWIW,

void zero_sort(int* arr, int length){
	int* end=arr+length;
	int* write=arr--;
	while (++arr!=end) if (*arr!=0) *(write++)=*arr;
	while (write!=end) *(write++)=0;
}

- Anonymous December 03, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

In place and linear time:

public static void fun(int[] data) {
		int offset=0;
		for (int i=0; i<data.length; i++) {
			if (data[i]==0)
				offset++;
			else
				data[i-offset]=data[i];
			if (i+offset>=data.length-1)
				data[i]=0;
		}
	}

- Anonymous March 29, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void zerosToEnd(int[] ints)
{
    if(ints != null)
    {
        int insertpos = 0;
        while (insertpos < ints.length && ints[insertpos] != 0)
            insertpos++;
        
        for (int i = insertpos + 1; i < ints.length; i++)
        {
            if (ints[i] != 0)
            {
                ints[insertpos] = ints[i];
                insertpos++;
            }
        }
        
        for (; insertpos < ints.length; insertpos++)
            ints[insertpos] = 0;
    }
}

- Anonymous December 14, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int arr[] = { 0, 0, 1, 8, 0, 1, 0, 1, 0 };
	int n = 9;

	int k = 0;
	for (int i = 0; i < n; i++)
	{
		if (arr[i] != 0)
			arr[k++] = arr[i];
	}

	for (; k < n; k++)
		arr[k] = 0;

- Anonymous October 02, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is a Hybrid ( Iterative-Recursive ) version:

...
void ShuffleArray( int a[], int n, int v )
{
	if( --n == 0 )
		return;
	ShuffleArray( &a[1], n, v );
	int t;
	if( a[0] == v && a[1] != v )
	{
		t = a[0];
		a[0] = a[1];
		a[1] = t;
	}
}
...
	int a[] = { 1, 3, 0, 8, 12, 0, 4, 0, 7 };
	int n = 9;
	int v = 0;
	for( int i = 0; i < n; i++ )
		ShuffleArray( &a[i], n-i, v );
...

- Sergey Kostrov November 04, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void movezerotoend1(vector<int> & vec){
    auto i=vec.begin();
    auto j=vec.begin();
    while (i!=vec.end()&&j!=vec.end()) {
        while (*i!=0) ++i;
        j=i;
        while (*j==0) ++j;
        swap(*(i++), *(j++));
    }
    
}

or one line:

for_each(std::remove_if(vec.begin(), vec.end(), zeroval()),vec.end(),setzero());

with zeroval() and setzero():

class zeroval:public unary_function<int, bool> {
    
public:
    bool operator()(int x);
};
class setzero:public unary_function<int, void> {
    
public:
    void operator()(int &x);
};

bool zeroval::operator()(int x){
    return x==0;
}

void setzero::operator()(int& x){
    x=0;
}

- andrey November 24, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void movezerotoend(int[]arr){
		int startindex,endindex;
		startindex=endindex=0;
		while(endindex<arr.length){
			if(arr[endindex]!=0){
				arr[startindex++]=arr[endindex];
			}
			endindex++;
		}
		while(startindex<arr.length){
			arr[startindex++]=0;
		}

}

- Avik Das March 04, 2015 | Flag Reply


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