Smallest number to be added in first Array modulo M to make frequencies of both Arrays equal

Given two arrays A[] and B[] consisting of N positive integers and a integer M, the task is to find the minimum value of X such that operation (A[i] + X) % M performed on every element of array A[] results in the formation of an array with frequency of elements same as that in another given array B[].

Examples: 

Input: N = 4, M = 3, A[] = {0, 0, 2, 1}, B[] = {2, 0, 1, 1} 
Output:
Explanation: 
Modifying the given array A[] to { (0+1)%3, (0+1)%3, (2+1)%3, (1+1)%3 } 
= { 1%3, 1%3, 3%3, 2%3 }, 
= { 1, 1, 0, 2 }, which is equivalent to B[] in terms of frequency of distinct elements.

Input: N = 5, M = 10, A[] = {0, 0, 0, 1, 2}, B[] = {2, 1, 0, 0, 0} 
Output:
Explanation: 
Frequency of elements in both the arrays are already equal.

Approach: This problem can be solved by using Greedy Approach. Follow the steps below: 



Below is the implementation of the above approach: 

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to find
// the answer
int moduloEquality(int A[], int B[],
                   int n, int m)
{
 
    // Stores the frequncies of
    // array elements
    map<int, int> mapA, mapB;
 
    for (int i = 0; i < n; i++) {
        mapA[A[i]]++;
        mapB[B[i]]++;
    }
 
    // Stores the possible values
    // of X
    set<int> possibleValues;
 
    int FirstElement = B[0];
    for (int i = 0; i < n; i++) {
        int cur = A[i];
 
        // Generate possible positive
        // values of X
        possibleValues
            .insert(
                cur > FirstElement
                    ? m - cur + FirstElement
                    : FirstElement - cur);
    }
 
    // Initialize answer
    // to MAX value
    int ans = INT_MAX;
 
    for (auto it :
         possibleValues) {
 
        // Flag to check if the
        // current element of the
        // set can be considered
        bool posible = true;
 
        for (auto it2 : mapA) {
 
            // If the frequency of an element
            // in A[] is not equal to that
            // in B[] after the operation
            if (it2.second
                != mapB[(it2.first + it) % m]) {
 
                // Current set element
                // cannot be considered
                posible = false;
                break;
            }
        }
 
        // Update minimum value of X
        if (posible) {
            ans = min(ans, it);
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    int n = 4;
    int m = 3;
 
    int A[] = { 0, 0, 2, 1 };
    int B[] = { 2, 0, 1, 1 };
 
    cout << moduloEquality(A, B, n, m)
         << endl;
 
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Utility function to find
// the answer
static int moduloEquality(int A[], int B[],
                          int n, int m)
{
     
    // Stores the frequncies of
    // array elements
    HashMap<Integer,
            Integer> mapA = new HashMap<Integer,
                                        Integer>();
    HashMap<Integer,
            Integer> mapB = new HashMap<Integer,
                                        Integer>();
 
    for(int i = 0; i < n; i++)
    {
        if (mapA.containsKey(A[i]))
        {
            mapA.put(A[i], mapA.get(A[i]) + 1);
        }
        else
        {
            mapA.put(A[i], 1);
        }
        if (mapB.containsKey(B[i]))
        {
            mapB.put(B[i], mapB.get(B[i]) + 1);
        }
        else
        {
            mapB.put(B[i], 1);
        }
    }
 
    // Stores the possible values
    // of X
    HashSet<Integer> possibleValues = new HashSet<Integer>();
 
    int FirstElement = B[0];
    for(int i = 0; i < n; i++)
    {
        int cur = A[i];
 
        // Generate possible positive
        // values of X
        possibleValues.add(cur > FirstElement ?
                       m - cur + FirstElement :
                  FirstElement - cur);
    }
 
    // Initialize answer
    // to MAX value
    int ans = Integer.MAX_VALUE;
 
    for(int it : possibleValues)
    {
         
        // Flag to check if the
        // current element of the
        // set can be considered
        boolean posible = true;
 
        for(Map.Entry<Integer,
                      Integer> it2 : mapA.entrySet())
        {
             
            // If the frequency of an element
            // in A[] is not equal to that
            // in B[] after the operation
            if (it2.getValue() !=
                mapB.get((it2.getKey() + it) % m))
            {
                 
                // Current set element
                // cannot be considered
                posible = false;
                break;
            }
        }
 
        // Update minimum value of X
        if (posible)
        {
            ans = Math.min(ans, it);
        }
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 4;
    int m = 3;
 
    int A[] = { 0, 0, 2, 1 };
    int B[] = { 2, 0, 1, 1 };
 
    System.out.print(moduloEquality(A, B, n, m) + "\n");
}
}
 
// This code is contributed by Amit Katiyar
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
import sys
from collections import defaultdict
 
# Utility function to find
# the answer
def moduloEquality(A, B, n, m):
 
    # Stores the frequncies of
    # array elements
    mapA = defaultdict(int)
    mapB = defaultdict(int)
 
    for i in range(n):
        mapA[A[i]] += 1
        mapB[B[i]] += 1
 
    # Stores the possible values
    # of X
    possibleValues = set()
 
    FirstElement = B[0]
    for i in range(n):
        cur = A[i]
 
        # Generate possible positive
        # values of X
        if cur > FirstElement:
            possibleValues.add(m - cur + FirstElement)
        else:
            possibleValues.add(FirstElement - cur)
 
    # Initialize answer
    # to MAX value
    ans = sys.maxsize
 
    for it in possibleValues:
 
        # Flag to check if the
        # current element of the
        # set can be considered
        posible = True
 
        for it2 in mapA:
 
            # If the frequency of an element
            # in A[] is not equal to that
            # in B[] after the operation
            if (mapA[it2] !=
                mapB[(it2 + it) % m]):
 
                # Current set element
                # cannot be considered
                posible = False
                break
 
        # Update minimum value of X
        if (posible):
            ans = min(ans, it)
             
    return ans
 
# Driver Code
if __name__ == "__main__":
     
    n = 4
    m = 3
 
    A = [ 0, 0, 2, 1 ]
    B = [ 2, 0, 1, 1 ]
 
    print(moduloEquality(A, B, n, m))
 
# This code is contributed by chitranayal
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Utility function to find
// the answer
static int moduloEquality(int[] A, int[] B,
                          int n, int m)
{
     
    // Stores the frequncies of
    // array elements
    Dictionary<int,
               int> mapA = new Dictionary<int,
                                          int>();
                    
    Dictionary<int,
               int> mapB = new Dictionary<int,
                                          int>();
  
    for(int i = 0; i < n; i++)
    {
        if (mapA.ContainsKey(A[i]))
        {
            mapA[A[i]] = mapA[A[i]] + 1;
        }
        else
        {
            mapA.Add(A[i], 1);
        }
        if (mapB.ContainsKey(B[i]))
        {
            mapB[B[i]] = mapB[B[i]] + 1;
        }
        else
        {
            mapB.Add(B[i], 1);
        }
    }
  
    // Stores the possible values
    // of X
    HashSet<int> possibleValues = new HashSet<int>();
  
    int FirstElement = B[0];
    for(int i = 0; i < n; i++)
    {
        int cur = A[i];
  
        // Generate possible positive
        // values of X
        possibleValues.Add(cur > FirstElement ?
                       m - cur + FirstElement :
                  FirstElement - cur);
    }
  
    // Initialize answer
    // to MAX value
    int ans = Int32.MaxValue;
    
    foreach(int it in possibleValues)
    {
          
        // Flag to check if the
        // current element of the
        // set can be considered
        bool posible = true;
         
        foreach(KeyValuePair<int, int> it2 in mapA)
        {
              
            // If the frequency of an element
            // in A[] is not equal to that
            // in B[] after the operation
            if (it2.Value != mapB[(it2.Key + it) % m])
            {
                  
                // Current set element
                // cannot be considered
                posible = false;
                break;
            }
        }
  
        // Update minimum value of X
        if (posible)
        {
            ans = Math.Min(ans, it);
        }
    }
    return ans;
}
 
// Driver code
static void Main()
{
    int n = 4;
    int m = 3;
  
    int[] A = { 0, 0, 2, 1 };
    int[] B = { 2, 0, 1, 1 };
   
    Console.WriteLine(moduloEquality(A, B, n, m));
}
}
 
// This code is contributed by divyeshrabadiya07
chevron_right

Output: 
1




 

Time Complexity: O(N2
Auxiliary Space: O(N)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :