在第一个阵列模 M 中添加的最小数字以使两个阵列的频率相等

给定两个数组A[]和B[]由N 个正整数和一个整数M 组成，任务是找到X的最小值，使得操作(A[i] + X) % M对数组A 的每个元素执行[]导致形成一个数组，其元素频率与另一个给定数组B[] 中的元素频率相同。

例子：

Input: N = 4, M = 3, A[] = {0, 0, 2, 1}, B[] = {2, 0, 1, 1} 
Output: 1 
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: 0 
Explanation: 
Frequency of elements in both the arrays are already equal.

方法：这个问题可以通过使用贪心方法来解决。请按照以下步骤操作：

• X至少有一个可能的值，使得对于每个索引i ， ( A[i] + X ) % M = B[0] 。
• 找出将A[] 的每个元素转换为B[]的第一个元素的X 的所有可能值。
• 检查这些可能的X值是否满足B[]的其他剩余值。
• 如果有多个答案，取X的最小值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Utility function to find
// the answer
int moduloEquality(int A[], int B[],
                   int n, int m)
{
 
    // Stores the frequencies of
    // array elements
    map mapA, mapB;
 
    for (int i = 0; i < n; i++) {
        mapA[A[i]]++;
        mapB[B[i]]++;
    }
 
    // Stores the possible values
    // of X
    set 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 possible = 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
                possible = false;
                break;
            }
        }
 
        // Update minimum value of X
        if (possible) {
            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;
}

Java

// 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 frequencies of
    // array elements
    HashMap mapA = new HashMap();
    HashMap mapB = new HashMap();
 
    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 possibleValues = new HashSet();
 
    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 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

Python3

# 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 frequencies 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

C#

// 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 frequencies of
    // array elements
    Dictionary mapA = new Dictionary();
                    
    Dictionary mapB = new Dictionary();
  
    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 possibleValues = new HashSet();
  
    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 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

Javascript

 
// JavaScript program for the above approach
 
// Utility function to find
// the answer
function moduloEquality(A, B, n, m)
{
 
    // Stores the frequencies of
    // array elements
    var mapA = new Map(), mapB = new Map();
 
    for (var i = 0; i < n; i++) {
        if(mapA.has(A[i]))
            mapA.set(A[i], mapA.get(A[i])+1)
        else
            mapA.set(A[i], 1)
 
        if(mapB.has(B[i]))
            mapB.set(B[i], mapB.get(B[i])+1)
        else
            mapB.set(B[i], 1)
 
    }
 
    // Stores the possible values
    // of X
    var possibleValues = new Set();
 
    var FirstElement = B[0];
    for (var i = 0; i < n; i++) {
        var cur = A[i];
 
        // Generate possible positive
        // values of X
        possibleValues
            .add(
                cur > FirstElement
                    ? m - cur + FirstElement
                    : FirstElement - cur);
    }
 
    // Initialize answer
    // to MAX value
    var ans = 1000000000;
 
    possibleValues.forEach(it => {
     
 
        // Flag to check if the
        // current element of the
        // set can be considered
        var possible = true;
 
        mapA.forEach((value, key) => {
         
 
            // If the frequency of an element
            // in A[] is not equal to that
            // in B[] after the operation
            if (value
                != mapB.get((key + it) % m)) {
 
                // Current set element
                // cannot be considered
                possible = false;
            }
        });
 
        // Update minimum value of X
        if (possible) {
            ans = Math.min(ans, it);
        }
    });
    return ans;
}
 
// Driver Code
var n = 4;
var m = 3;
var A = [0, 0, 2, 1];
var B = [2, 0, 1, 1];
document.write( moduloEquality(A, B, n, m));
 
 

输出：

1

时间复杂度： O(N ² )
辅助空间： O(N)

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