In this C program, we are going to learn **how to find out integers which come odd number of times in an integer array**?

Submitted by Shubham Singh Rajawat, on January 21, 2018

An array of integers is given in which one number (integer) must repeat odd number of times and we have to find that one integer.

**Example:**

Input: Array elements are: [1, 4, 6, 1, 9, 6, 4] Output: In this array output will be 9

**There are two ways to find it:**

In this method we will have to traverse the array multiple number of times and count the occurrence of every integer in that array. This will gives us O(n^{2}) time complexity as for every number we have to traverse all n elements.

This method is the best as it has O(n) time complexity because traverse the array only once to find the solution. This method uses bit manipulation.

A B Y0 0 0 0 1 1 1 0 1 1 1 0

As it gives zero as output when two same numbers are given as input and when a number (say x) and zero will be given as input it will give that same number (x) as output.

Now let’s take one example to make it much easier to understand.

[1, 4, 6, 1, 9, 6, 4]At first result=0 Then, 0^1 = 1 result = 1 1^4 = 5 result = 5 5^6 = 3 result = 3 3^1 = 2 result = 2 2^9 = 11 result = 11 11^6 = 13 result = 13 13^4 = 9 result = 9

**C++ code:**

#include <iostream> #include <vector> using namespace std; //function to find odd integer int oddInteger(vector <int> a) { int result=0; for(unsigned int i=0;i<a.size();i++) { result=result^a[i]; } return result; } //main function to test code int main() { int n; //input total number of elements cin >> n; vector<int> a(n); //read n numbers for(int i=0;i<n;i++) { cin>>a[i]; } //find and print result int result = oddInteger(a); cout << result << endl; return 0; }

**Output**

9

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