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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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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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