There is an array of numbers, where each number occurs even number of times except for one number which occurs odd number of times. Find out the number.

When you XOR a number even number of times, it becomes zero. So XOR all the numbers in the array, the result will be the number which occurred odd times.

Output:-

When you XOR a number even number of times, it becomes zero. So XOR all the numbers in the array, the result will be the number which occurred odd times.

#### C++ program to find the number occurring odd number of times

#include <iostream> using namespace std; const int MAX_INPUT = 7; int main() { int input[MAX_INPUT] = { 123, 456, 123, 678, 985, 985, 678 }; int output = 0; for(int i=0;i<MAX_INPUT;i++) output ^= input[i]; cout << "Odd number out is: " << output << endl; return 0; }

Output:-

Odd number out is: 456

ReplyDeleteuggs outletcoach outlet store onlineray ban glassesoakley vaultthunder jerseysjordan femme pas chercoach outletugg australiaray ban wayfarercoach outlet store online clearances2016.12.29xukaimin