Minimum operations to make counts of remainders same in an array

Given an array arr[] of N integers and an integer M where N % M = 0. The task is to find the minimum number of operations that need to be performed on the array to make c0 = c1 = ….. = cM – 1 = N / M where cr is the number of elements in the given array having remainder r when divided by M. In each operation, any array element can be incremented by 1.

Examples:

Input: arr[] = {1, 2, 3}, M = 3
Output: 0
After performing the modulus operation on the given array, the array becomes {0, 1, 2}
And count of c0 = c1 = c2 = n / m = 1.
So, no any additional operations are required.

Input: arr[] = {3, 2, 0, 6, 10, 12}, M = 3
Output: 3
After performing the modulus operation on the given array, the array becomes {0, 2, 0, 0, 1, 0}
And count of c0 = 4, c1 = 1 and c2 = 1. To make c0 = c1 = c2 = n / m = 2.
Add 1 to 6 and 2 to 12 then the array becomes {3, 2, 0, 7, 10, 14} and c0 = c1 = c2 = n / m = 2.

Approach: For each i from 0 to m – 1, find all the elements of the array that are congruent to i modulo m and store their indices in a list. Also, create a vector called extra, and let k = n / m.

We have to cycle from 0 to m – 1 twice. For each i from 0 to m – 1, if there are more elements than k in the list, remove the extra elements from this list and add them to extra. If instead there are lesser elements than k then remove the last few elements from the vector extra. For every removed index idx, increase arr[idx] by (i – arr[idx]) % m.

It is obvious that after the first m iterations, every list will have size at most k and after m more iterations all lists will have the same sizes i.e. k.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum
// number of operations required
int minOperations(int n, int a[], int m)
{
    int k = n / m;
  
    // To store modulos values
    vector<vector<int> > val(m);
    for (int i = 0; i < n; ++i) {
        val[a[i] % m].push_back(i);
    }
  
    long long ans = 0;
    vector<pair<int, int> > extra;
  
    for (int i = 0; i < 2 * m; ++i) {
        int cur = i % m;
  
        // If it's size greater than k
        // it needed to be decreased
        while (int(val[cur].size()) > k) {
            int elem = val[cur].back();
            val[cur].pop_back();
            extra.push_back(make_pair(elem, i));
        }
  
        // If it's size is less than k
        // it needed to be increased
        while (int(val[cur].size()) < k && !extra.empty()) {
            int elem = extra.back().first;
            int mmod = extra.back().second;
            extra.pop_back();
            val[cur].push_back(elem);
            ans += i - mmod;
        }
    }
  
    return ans;
}
  
// Drive code
int main()
{
    int m = 3;
  
    int a[] = { 3, 2, 0, 6, 10, 12 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minOperations(n, a, m);
  
    return 0;
}

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Python3

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# Python3 implementation of the approach 
  
# Function to return the minimum 
# number of operations required 
def minOperations(n, a, m): 
  
    k = n //
  
    # To store modulos values 
    val = [[] for i in range(m)] 
    for i in range(0, n): 
        val[a[i] % m].append(i) 
      
    ans = 0
    extra = [] 
  
    for i in range(0, 2 * m): 
        cur = i %
  
        # If it's size greater than k 
        # it needed to be decreased 
        while len(val[cur]) > k: 
            elem = val[cur].pop() 
            extra.append((elem, i)) 
  
        # If it's size is less than k 
        # it needed to be increased 
        while (len(val[cur]) < k and
               len(extra) > 0): 
            elem = extra[-1][0
            mmod = extra[-1][1
            extra.pop() 
            val[cur].append(elem) 
            ans += i - mmod 
  
    return ans 
  
# Drive code 
if __name__ == "__main__"
  
    m = 3
  
    a = [3, 2, 0, 6, 10, 12
    n = len(a) 
    print(minOperations(n, a, m))
      
# This code is contributed by Rituraj Jain

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Output:

3


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Improved By : rituraj_jain