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Given an array of numbers, except for one number all the
others occur twice. Give an algorithm to find that number
which occurs only once in the array.
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Answer Posted / ruchi
#include<stdio.h>
#include<conio.h>
int main()
{
int a[15],i,j,n,temp,p,k;
printf("\nHow many elements are there in a array ");
scanf("%d",&n);
printf("\nEnter the elements ");
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[i]>a[j])
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
printf("\nThe element which occur once is ");
for(i=0;i<n;i++)
{
k=2*i;
p=2*i+1;
if(a[k]!=a[p])
{
printf("%d\n",a[k]);
break;
}
}
getch();
}
| Is This Answer Correct ? | 0 Yes | 3 No |
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