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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 / deepak
/*
you can also do some pre tests like array length should be odd.
just take an XOR of all the numbers it will give u the
number that occured single time.
it assumes that data is in correct form i.e. there is one
and only one number that occurs once.
*/
public static int whoOccursSingleTime(int a[]){
int s=0;
for(int i=0;i<a.length;i++){
s=s^a[i];
}
return s;
}
| Is This Answer Correct ? | 19 Yes | 3 No |
Answer / priyanshu
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
long int find(long int a[],long int n,long int f,long int l)
{
long int mid;
mid=(f+l)/2;
if(a[mid]==a[mid+1])
{
return find(a,mid,mid+1,n-1);
}
else if(a[mid]==a[mid-1])
{
return find(a,mid,0,mid-1);
}
else
return a[mid];
}
int main()
{
long int n,i;
scanf("%ld",&n);
long int a[n];
for(i=0;i<n;i++)
scanf("%ld",&a[i]);
sort(a,a+n);
printf("%ld",find(a,n,0,n-1));
return 0;
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / pushpa
int num[3];
int i,j,p=0,count_1=0,count=0,n;
n= 3;
for(i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(num[i] == num[j])
count_1=0;
else
count++;
}
if(count == n-1)
{
printf("The number which occurs once is %d\n",num[i]);
}
count = 0;
}
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / don151
#include<stdio.h>
#include<conio.h>
void main()
{
int a[10],i,j,n,b[10],m,k,temp;
clrscr();
printf("eneett the v vallrue of n");
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
printf("before sorting array");
for(i=0;i<n;i++)
{
printf("%d\n",a[i]);
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[j]>=a[i])
{
temp=a[j];
a[j]=a[i];
a[i]=temp;
}
}
}
printf("after sorting");
for(i=0;i<n;i++)
{
printf("\n%d",a[i]);
}
for(i=0;i<n;i+=2)
{
if(a[i]!=a[i+1])
{
printf("resutl for %d,%d\n",i,a[i]);
i++;
}
}
getch();
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / 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 |
Answer / guest
int arr[n]; //n' numbers
for (int i=0;i<n;i++)
for (int j=0;j<n;j++)
{ if ( (a[i]==a[j])&&i!=j ) break; //ITS NOT THE NUM
if ( (a[i]!=a[j])&&j==n-1) return i; //IT IS!!
return -1; //IF THERE IS NO SUCH NUMBER...
| Is This Answer Correct ? | 3 Yes | 7 No |
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