Simple Program to same Element in an Array code example

Example 1: hopw to check how many duplicates in an array c

/**
 * C program to count total number of duplicate elements in an array
 */

#include <stdio.h>

#define MAX_SIZE 100  // Maximum array size

int main()
{
    int arr[MAX_SIZE];
    int i, j, size, count = 0;

    /* Input size of array */
    printf("Enter size of the array : ");
    scanf("%d", &size);

    /* Input elements in array */
    printf("Enter elements in array : ");
    for(i=0; i<size; i++)
    {
        scanf("%d", &arr[i]);
    }

    /*
     * Find all duplicate elements in array
     */
    for(i=0; i<size; i++)
    {
        for(j=i+1; j<size; j++)
        {
            /* If duplicate found then increment count by 1 */
            if(arr[i] == arr[j])
            {
                count++;
                break;
            }
        }
    }

    printf("\nTotal number of duplicate elements found in array = %d", count);

    return 0;
}

Example 2: Find the duplicate in an array of N integers.

// 287. Find the Duplicate Number
// Medium

// Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

// Example 1:

// Input: [1,3,4,2,2]
// Output: 2
// Example 2:

// Input: [3,1,3,4,2]
// Output: 3
// Note:

// You must not modify the array (assume the array is read only).
// You must use only constant, O(1) extra space.
// Your runtime complexity should be less than O(n2).
// There is only one duplicate number in the array, but it could be repeated more than once.

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int n=nums.size();
        int s=nums[0];
        int f=nums[nums[0]];
        while(s!=f) {
            s=nums[s];
            f=nums[nums[f]];
        }
        f=0;
        while(s!=f) {
            s=nums[s];
            f=nums[f];
        }
        return s;
        
    }
};