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Minimum operations required to make all Array elements divisible by K

  • Difficulty Level : Medium
  • Last Updated : 13 May, 2021

Given an array a[], integer K and an integer X (which is initially initialized to 0). Our task is to find the minimum number of moves required to update the array such that each of its elements is divisible by K by performing the following operations:

  • Choose one index i from 1 to N and increase ai by X and then increase X by 1. This operation cannot be applied more than once to each element of the array
  • Only increase the value of X by 1.

Examples:

Input: K = 3, a = [1, 2, 2, 18] 
Output:
Explanation: 
Initially X = 0 hence update X to 1. 
For X = 1 add X to the second element of array to make the array [1, 3, 2, 18] and increase X by 1. 
For X = 2 add X to the first element of array [3, 3, 2, 18] and increase X by 1. 
For X = 3 just increase X by 1. 
For X = 4 add X to the third element of array to make the array [3, 3, 6, 18] and increase X by 1. 
At last, the array becomes [3, 3, 6, 18] where all the elements are divisible by K = 3.
Input: K = 5, a[] = [15, 25, 5, 10, 20, 1005, 70, 80, 90, 100] 
Output:
Explanation: 
Here all elements are already divisible by 5.

Approach: The main idea is to find the maximum value of X that is needed to update the elements of the array to make it divisible by K.



  • To do this we need to find the maximum value of (K – (ai mod K)) to add to the array elements to make it divisible by K.
  • However, there can be equal elements so keep track of the number of such elements, using map data structures.
  • When found another such element in the array then update the answer with (K – (ai mod K)) + (K * number of Equal Elements) because with every move we increase X by 1.

Below is the implementation of the above approach:

C++




// C++ implementation to find the
// Minimum number of moves required to
// update the array such that each of
// its element is divisible by K
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// Minimum number of moves required to
// update the array such that each of
// its element is divisible by K
void compute(int a[], int N, int K)
{
    // Initialize Map data structure
    map<long, long> eqVal;
 
    long maxX = 0;
 
    // Iterate for all the elements
    // of the array
    for (int i = 0; i < N; i++) {
 
        // Calculate the
        // value to be added
        long val = a[i] % K;
 
        val = (val == 0 ? 0 : K - val);
 
        // Check if the value equals
        // to 0 then simply continue
        if (val == 0)
            continue;
 
        // Check if the value to be
        // added is present in the map
        if (eqVal.find(val) != eqVal.end()) {
 
            long numVal = eqVal[val];
            // Update the answer
            maxX = max(maxX,
                    val + (K * numVal));
 
            eqVal[val]++;
        }
 
        else {
            eqVal[val]++;
            maxX = max(maxX, val);
        }
    }
 
    // Print the required result
    // We need to add 1 to maxX
    // because we cant ignore the
    // first move where initially X=0
    // and we need to increase it by 1
    // to make some changes in array
    cout << (maxX == 0 ? 0 : maxX + 1)
        << endl;
}
 
// Driver code
int main()
{
    int K = 3;
    int a[] = { 1, 2, 2, 18 };
    int N = sizeof(a) / sizeof(a[0]);
    compute(a, N, K);
    return 0;
}

Java




// Java implementation to find the
// minimum number of moves required to
// update the array such that each of
// its element is divisible by K
import java.util.*;
 
class GFG{
     
// Function to find the minimum
// number of moves required to
// update the array such that each of
// its element is divisible by K
static void compute(int a[], int N, int K)
{
     
    // Initialize Map data structure
    Map<Long,
        Long> eqVal = new HashMap<Long,
                                Long>();
 
    long maxX = 0;
 
    // Iterate for all the elements
    // of the array
    for(int i = 0; i < N; i++)
    {
         
        // Calculate the
        // value to be added
        long val = a[i] % K;
 
        val = (val == 0 ? 0 : K - val);
 
        // Check if the value equals
        // to 0 then simply continue
        if (val == 0)
            continue;
 
        // Check if the value to be
        // added is present in the map
        if (eqVal.containsKey(val))
        {
            long numVal = eqVal.get(val);
             
            // Update the answer
            maxX = Math.max(maxX,
                            val + (K * numVal));
 
            eqVal.put(val,
            eqVal.getOrDefault(val, 0l) + 1l);
        }
        else
        {
            eqVal.put(val, 1l);
            maxX = Math.max(maxX, val);
        }
    }
 
    // Print the required result
    // We need to add 1 to maxX
    // because we cant ignore the
    // first move where initially X=0
    // and we need to increase it by 1
    // to make some changes in array
    System.out.println(maxX == 0 ? 0 : maxX + 1);
}
 
// Driver code
public static void main(String[] args)
{
    int K = 3;
    int a[] = { 1, 2, 2, 18 };
    int N = a.length;
     
    compute(a, N, K);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 implementation to find the
# Minimum number of moves required to
# update the array such that each of
# its element is divisible by K
from collections import defaultdict
 
# Function to find the Minimum number
# of moves required to update the
# array such that each of its
# element is divisible by K
def compute(a, N, K):
 
    # Initialize Map data structure
    eqVal = defaultdict(int)
 
    maxX = 0
 
    # Iterate for all the elements
    # of the array
    for i in range(N):
 
        # Calculate the
        # value to be added
        val = a[i] % K
 
        if (val != 0):
            val = K - val
 
        # Check if the value equals
        # to 0 then simply continue
        if (val == 0):
            continue
 
        # Check if the value to be
        # added is present in the map
        if (val in eqVal):
            numVal = eqVal[val]
             
            # Update the answer
            maxX = max(maxX,
                    val + (K * numVal))
 
            eqVal[val] += 1
        else:
            eqVal[val] += 1
            maxX = max(maxX, val)
 
    # Print the required result
    # We need to add 1 to maxX
    # because we cant ignore the
    # first move where initially X=0
    # and we need to increase it by 1
    # to make some changes in array
    if maxX == 0:
        print(0)
    else:
        print(maxX + 1)
     
# Driver code
if __name__ == "__main__":
 
    K = 3
    a = [ 1, 2, 2, 18 ]
    N = len(a)
     
    compute(a, N, K)
 
# This code is contributed by chitranayal

C#




// C# implementation to find the
// minimum number of moves required to
// update the array such that each of
// its element is divisible by K
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the minimum
// number of moves required to
// update the array such that each of
// its element is divisible by K
static void compute(int []a, int N, int K)
{
     
    // Initialize Map data structure
    Dictionary<long,
               long> eqVal = new Dictionary<long,
                                            long>();
 
    long maxX = 0;
 
    // Iterate for all the elements
    // of the array
    for(int i = 0; i < N; i++)
    {
         
        // Calculate the
        // value to be added
        long val = a[i] % K;
 
        val = (val == 0 ? 0 : K - val);
 
        // Check if the value equals
        // to 0 then simply continue
        if (val == 0)
            continue;
 
        // Check if the value to be
        // added is present in the map
        if (eqVal.ContainsKey(val))
        {
            long numVal = eqVal[val];
             
            // Update the answer
            maxX = Math.Max(maxX,
                            val + (K * numVal));
                     
            eqVal[val] = 1 +
            eqVal.GetValueOrDefault(val, 0);
        }
        else
        {
            eqVal.Add(val, 1);
            maxX = Math.Max(maxX, val);
        }
    }
 
    // Print the required result
    // We need to add 1 to maxX
    // because we cant ignore the
    // first move where initially X=0
    // and we need to increase it by 1
    // to make some changes in array
    Console.Write(maxX == 0 ? 0 : maxX + 1);
}
 
// Driver code
public static void Main(string[] args)
{
    int K = 3;
    int []a = { 1, 2, 2, 18 };
    int N = a.Length;
     
    compute(a, N, K);
}
}
 
// This code is contributed by rutvik_56

Javascript




<script>
 
// Javascript implementation to find the
// minimum number of moves required to
// update the array such that each of
// its element is divisible by K
     
     
 // Function to find the minimum
// number of moves required to
// update the array such that each of
// its element is divisible by K
    function compute(a,N,K)
    {
        // Initialize Map data structure
    let eqVal = new Map();
   
    let maxX = 0;
   
    // Iterate for all the elements
    // of the array
    for(let i = 0; i < N; i++)
    {
           
        // Calculate the
        // value to be added
        let val = a[i] % K;
   
        val = (val == 0 ? 0 : K - val);
   
        // Check if the value equals
        // to 0 then simply continue
        if (val == 0)
            continue;
   
        // Check if the value to be
        // added is present in the map
        if (eqVal.has(val))
        {
            let numVal = eqVal.get(val);
               
            // Update the answer
            maxX = Math.max(maxX,
                            val + (K * numVal));
   
            eqVal.set(val,eqVal.get(val) + 1);
        }
        else
        {
            eqVal.set(val, 1);
            maxX = Math.max(maxX, val);
        }
    }
   
    // Print the required result
    // We need to add 1 to maxX
    // because we cant ignore the
    // first move where initially X=0
    // and we need to increase it by 1
    // to make some changes in array
    document.write(maxX == 0 ? 0 : maxX + 1);
    }
     
    // Driver code
    let K = 3;
    let a=[1, 2, 2, 18];
    let N = a.length;
    compute(a, N, K);
     
     
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
5

Time Complexity: O(N)
 




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