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the code can replace 0 with 5 in a integer, suppose no pre-zeros in the integer.

int replace0to5(int n){
        if (n == 0) return 5;
        int res = n;
        int added = 5;
       while (n > 0) {
           if (n % 10 == 0) res += added;
           added *= 10;
           n /= 10;
       }
      return res;
   }

means we need to convert 60 to 65 or.. only the whole number '0' to '5'

You means like the below ex:
i/p: 1 0 3 0
o/p: 1 5 3 5

Correct me if I am wrong

1)if 60 need to be replaced as 65

for(int i=0;i<array.length;i++)
      if(array[i]%10==0) array[i]+=5;

2) if just 0 need to be replaced with 5

for(int i=0;i<array.length;i++)
    if(array[i]==0) array[i]+=5;

3) if 0 need to be replaced with 5 in a sorted array

public static void main(String[] args) {
		int arr[]= {-2,-2,-1,0,0,0,0,0,0,1,2,3,4};
		int index = binarysearch(arr, 0, arr.length-1,0);//get the index of 0
		System.out.println(index);
		int index2=index-1;
		while(true){//replace all the 0's starting from index to upwards
			if(arr[index]!=0)
				break;
			arr[index++] =5;
		}
		while(true){//replace all the 0's downwards starting from index
			if(arr[index2]!=0)
				break;
			arr[index2--] =5;
		}
		for(int i:arr)
		 System.out.print(i+ " ");
	}
	
	static int binarysearch(int[] arr, int minSize, int maxSize, int key){
		if(minSize ==maxSize) return minSize;
		int mid = (maxSize-minSize)/2;
		if(arr[mid]==key) return mid;
		if(arr[mid]>key)
			return binarysearch(arr, minSize, mid-1, key);
		else 
			return binarysearch(arr, mid+1,maxSize, key);
	}

#include <stdio.h>

int main(void)
{
int arr[]={1,0,50,55,60,78,100,450,87,0,};
int i;
for(i=0;i<10;i++)
{
if((arr[i] % 10) == 0)
{
arr[i]=arr[i]+5;
}
}
for (i=0;i<10;i++)
printf("%d ",arr[i]);
printf("\n");
return 0;
}

for(int i=0;i<array.length;i++)
if(array[i]%10==0) array[i]+=5*pow(10,i);

testing if a[i]%10==0 and adding 5 is insufficient. What happens if a[i] is 100? Then you would replace it with 105, and you would still have a 0.

class intToString{
public static void main(String[] args){
int i;
int array[]={1,0,50,55,60,78,100,450,87,0,};
String s;
for(i=0;i<10;i++)
{
s=""+array[i];
String newStr =s.replaceAll("0","5");
array[i]=Integer.parseInt(newStr);
//System.out.println(array[i]);
}
}
}

#include<stdio.h>

main()
{
  int n,arr[10],i,fact,k;
  printf("Enter the limit:\n");
  scanf("%d",&n);

  for(i=0;i<n;i++)
    scanf("%d",&arr[i]);

  for(i=0;i<n;i++)
    {
      for(k=arr[i],fact = 1;k>0;k/=10,fact*=10)
	{
	  if(k%10==0)
	    arr[i]+=5*fact;
	}
    }

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


}

void replace (int array1[])
{
int i = 0,j=10,k=5,temp;
for(;i<SIZE;i++)
{
if(array1[i]==0)
{
array1[i] = 5;
}
else
{
j=10;
temp=array1[i];
while(temp>9)
{
if((temp%10)==0)
{
array1[i]= array1[i]+j/2;

}
temp =temp/10;
j=j*10;
}
}
}
}

int n;
printf("The original list is \n");
for(i=0; i<10; i++)
{ printf("%d, ",a[i]);
}
printf("After conversion the list is \n");
for(i=0; i<10; i++)
{ j=0;
sum=0;
while(a[i]>0)
{ k=a[i]%10;
a[i]=a[i]/10;
if(k==0)
k=k+5;
sum=sum+pow(10,j)*k;
j++;
}
a[i]=sum;
}
for(i=0; i<10; i++)
printf("%d, ",a[i]);
printf("\n");
}
27,1 Bot

Could you change the int to a string and throw each char into an array.
Put 1004 into an array like: [1][0][0][4] and then go through the array and change the 0's to 5's and then re-create the number from a string. [1][5][5][4] = 1554.

If regular expression can be used, following is the code in java..
Please comment if there are any errors.

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class Test {
	
	public static void main(String args[]){
		
		int a[] = {3, 89, 4, 45, 233, 900004, 30, 606, 670, 8097};
		
		Pattern pat = Pattern.compile("0");
		
		for(int i=0; i<a.length; i++){
		Matcher mat = pat.matcher(""+a[i]);		
		a[i] = Integer.parseInt(mat.replaceAll("5"));
		System.out.println(a[i]);
		}
		
		
	}
}

Here is a simple Python code.

def replace0to5 (num):
m = 10
n = 1
d = len(str(num))

for i in xrange(d):
if ((num % m) / n == 0):
num = num + n * 5
m = m * 10
n = n * 10

return num

Ankush's code worked for me

simple O(n) solution in ruby

# O(n)
def replace_zeroes zero_arr = []
  0.upto(zero_arr.size) {|i| zero_arr[i] += 5 if zero_arr[i] == 0}
  zero_arr
end

public class ArrEx{
        public static void main(String[] arg)
        {
                int arr[] = {1,4,0,2,5,0,6};
                for(int i=0; i<arr.length; i++)
                {
                        if(arr[i]==0)
                        {
                                arr[i]=5;
                        }
                }

                for(int x:arr)
                {
                        System.out.println(x);
                }
        }
}

public class ArrEx{
        public static void main(String[] arg)
        {
                int arr[] = {1,4,0,2,5,0,6};
                for(int i=0; i<arr.length; i++)
                {
                        if(arr[i]==0)
                        {
                                arr[i]=5;
                        }
                }

                for(int x:arr)
                {
                        System.out.println(x);
                }
        }
}

if you find zero anywhere , do 0! + 0! +0 ! +0! + 0! for five you will get value 5 as 0! = 1.