find duplicate elements in array in c++ code example

Example 1: find duplicate in an array using xor

int DuplicateNumber(int arr[], int size){
    int ans=0;
    for(int i=0;i<size;i++){
        ans= ans ^ arr[i] ; 
    }
    for(int i=0;i<=size-2;i++){
        ans= ans ^ i;
    }
    return ans;
   }

Example 2: find duplicate in an array using xor

int DuplicateNumber(int arr[], int size){
    /* Don't write main().
     * Don't read input, it is passed as function argument.
     * Return output and don't print it.
     * Taking input and printing output is handled automatically.
     */
    
    int ans=0;
    for(int i=0;i<size;i++){
        ans= ans ^ arr[i] ;
       
    }
    for(int i=0;i<=size-2;i++){
        ans= ans ^ i;
       
    }
    return ans;
    

}

Example 3: 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;
        
    }
};