C++ Exercises: Find a number which occurs odd number of times of a given array of positive integers

Last update on December 20 2024 12:26:44 (UTC/GMT +8 hours)

Write a C++ program to find a number that occurs an odd number of times in a given array of positive integers. In the array, all numbers occur an even number of times.

Sample Solution:

C++ Code :

#include <iostream> // Header file for input/output stream
using namespace std; // Using the standard namespace

// Function to find the element occurring an odd number of times in an array
int getOddOccurrence(int nums[], int n)
{
    for (int i = 0; i < n; i++) { // Loop through each element in the array

        int ctr = 0; // Counter to count occurrences of nums[i]

        for (int j = 0; j < n; j++) // Loop to count occurrences of nums[i] in the array
        {
            if (nums[i] == nums[j]) // If the current element matches nums[i]
                ctr++; // Increment the counter
        }
        if (ctr % 2 != 0) // If the counter is odd
            return nums[i]; // Return the element occurring odd times
    }
    return -1; // Return -1 if no element occurs an odd number of times
}

int main()
{
    int nums[] = {5, 7, 8, 8, 5, 8, 7, 7}; // Declaration and initialization of an integer array
    int n = sizeof(nums)/sizeof(nums[0]); // Calculate the number of elements in the array
    cout << "Original array: ";
    for (int i=0; i < n; i++) 
    cout << nums[i] <<" "; // Output each element of the original array
    cout << "\nNumber which occurs odd number of times: " << getOddOccurrence(nums, n); // Call function to find the element occurring odd times
    return 0; // Return 0 to indicate successful execution
}

Sample Output:

Original array: 5 7 8 8 5 8 7 7 
Number which occurs odd number of times: 7

Flowchart:

C++ Code Editor:

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