int main() {
int a[] = {1,2,2,3,3,3,4,4};
int size = sizeof(a)/sizeof(a[0]);
int xorr=0;
for(int i=0;i<size;i++) {
xorr^=a[i];
}
int x=0,y=0;
int xor1=xorr & ~(xorr-1);
for(int i=0;i<size;i++) {
if(xor1 & a[i]) x^=a[i];
else y^=a[i];
}
cout<< x <<" " << y;
}
in this 1 and 3 would be output. as it would be print the number repeated
odd times. so ur answer would be in y ie 3 .
If there issue with the code let me knw.
On Fri, Apr 19, 2013 at 10:52 AM, Krishnan <aariyankrish...@gmail.com>wrote:
> In an array, some numbers occur only once, some numbers occur twice, only
> one number occur thrice. Find the number occuring thrice ? Space complexity
> O(1) Time Complexity O(n). We should not use Hash Maps.
>
> Please someone help..
>
>
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