Minimum operations required to make all the elements distinct in an array

Given an array of N integers. If a number occurs more than once, choose any number y from the array and replace the x in the array to x+y such that x+y is not in the array. The task is to find the minimum number of operations to make the array a distinct one.


Input: a[] = {2, 1, 2}
Output: 1
Let x = 2, y = 1 then replace 2 by 3.
Performing the above step makes all the elements in the array distinct.

Input: a[] = {1, 2, 3}
Output: 0

Approach: If a number appears more than once, then the summation of (occurrences-1) for all duplicate elements will be the answer. The main logic behind this is if x is replaced by x+y where y is the largest element in the array, then x is replaced by x+y which is the largest element in the array. Use a map to store the frequency of the numbers of array. Traverse in the map, and if the frequency of an element is more than 1, add it to the count by subtracting one.

Below is the implementation of the above approach:





// C++ program to find Minimum number
// of  changes to make array distinct
#include <bits/stdc++.h>
using namespace std;
// Fucntion that returns minimum number of changes
int minimumOperations(int a[], int n)
    // Hash-table to store frequency
    unordered_map<int, int> mp;
    // Increase the frequency of elements
    for (int i = 0; i < n; i++)
        mp[a[i]] += 1;
    int count = 0;
    // Traverse in the map to sum up the (occurences-1)
    // of duplicate elements
    for (auto it = mp.begin(); it != mp.end(); it++) {
        if ((*it).second > 1)
            count += (*it).second-1;
    return count;
// Driver Code
int main()
    int a[] = { 2, 1, 2, 3, 3, 4, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimumOperations(a, n);
    return 0;




Time Complexity: O(N)

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