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Discover Ok such that repeated subtraction of Ok from Array components make the Array equal


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Given an array arr[] of size N, the task is to find the value of an integer K such that its repeated subtraction from array elements will make all array elements in minimum operations.


Input: arr[] = {5, 3, 3, 7}
Output: 2
Explanation: Minimum 2 operations must be performed:
1st operation: subtract 2 from elements at indices {0, 3}, arr[] = {3, 3, 3, 5}
2nd operation: subtract 2 from element at index 3, arr[] = {3, 3, 3, 3}
So, the value of K = 2

Input: arr[] = {-1, 0, -1, 0, -1}
Output: 1


Approach: To make all elements in the array arr equal, all elements need to be changed to the smallest value in the array. So, the gcd of the difference of all array elements to the smallest element will be the value of K. Now, to solve the following problem, follow the below steps:

  1. Create a variable mn to store the minimum element in the array arr.
  2. Now, traverse the array arr and find the gcd of the differences between all array elements and mn. Store this value in a variable K.
  3. Return K, as the answer to this question.

Below is the implementation of the above approach:


#include <bits/stdc++.h>

using namespace std;


int findtheValue(int arr[], int N)



    int mn = *min_element(arr, arr + N);


    int K = arr[0] - mn;





    for (int i = 1; i < N; ++i) {

        K = __gcd(K, arr[i] - mn);



    return K;



int main()


    int arr[] = { 5, 3, 3, 7 };

    int N = sizeof(arr) / sizeof(arr[0]);

    cout << findtheValue(arr, N);


    return 0;



import math


def findtheValue(arr, N):



    mn = min(arr)


    K = arr[0] - mn





    for i in range(1, N):

        K = math.gcd(K, arr[i] - mn)


    return K


if __name__ == "__main__":


    arr = [5, 3, 3, 7]

    N = len(arr)

    print(findtheValue(arr, N))




Time Complexity: O(Nlog(mn))
Auxiliary Space: O(1)


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