使数组中的剩余数相同的最少操作

给定N个整数的数组arr []和一个整数M ，其中N％M = 0 。任务是找到使数组c ₀ = c ₁ =….. = c _{M – 1} = N / M所需在数组上执行的最小操作数，其中c _r是给定数组中的元素数当除以M时，余数为r 。在每个操作中，任何数组元素都可以增加1 。

例子：

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. 

方法：对于从0到m – 1的每个i ，找到数组中所有与i取模m一致的元素，并将它们的索引存储在列表中。另外，创建一个称为extra的向量，并使k = n / m 。

我们必须从0循环到m – 1两次。对于从0到m – 1的每个i ，如果列表中的元素多于k ，请从此列表中删除多余的元素，并将其添加到多余的元素中。相反，如果元素少于k，则从矢量多余的元素中删除最后几个元素。对于每个删除的索引idx ，将arr [idx]增加(i – arr [idx])％m 。

显然，在前m次迭代之后，每个列表的大小最多为k，而在m次迭代之后，所有列表的大小都相同，即k 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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 > val(m);
    for (int i = 0; i < n; ++i) {
        val[a[i] % m].push_back(i);
    }
 
    long long ans = 0;
    vector > 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;
}
 
// Driver 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;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG{
 
static class pair
{
    int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
 
// Function to return the minimum
// number of operations required
static int minOperations(int n, int a[], int m)
{
    int k = n / m;
 
    // To store modulos values
    @SuppressWarnings("unchecked")
    Vector []val = new Vector[m];
    for(int i = 0; i < val.length; i++)
        val[i] = new Vector();
         
    for(int i = 0; i < n; ++i)
    {
        val[a[i] % m].add(i);
    }
 
    long ans = 0;
    Vector extra = new Vector<>();
 
    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 ((val[cur].size()) > k)
        {
            int elem = val[cur].lastElement();
            val[cur].removeElementAt(val[cur].size() - 1);
            extra.add(new pair(elem, i));
        }
 
        // If it's size is less than k
        // it needed to be increased
        while (val[cur].size() < k && !extra.isEmpty())
        {
            int elem = extra.get(extra.size() - 1).first;
            int mmod = extra.get(extra.size() - 1).second;
             
            extra.remove(extra.size() - 1);
            val[cur].add(elem);
            ans += i - mmod;
        }
    }
    return (int)ans;
}
 
// Driver code
public static void main(String[] args)
{
    int m = 3;
    int a[] = { 3, 2, 0, 6, 10, 12 };
    int n = a.length;
     
    System.out.print(minOperations(n, a, m));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
 
# Function to return the minimum
# number of operations required
def minOperations(n, a, m):
 
    k = n // m
 
    # 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 % m
 
        # 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
 
# Driver 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

C#

// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
class GFG{
 
public class pair
{
  public int first,
             second;
 
  public pair(int first,
              int second) 
  {
    this.first = first;
    this.second = second;
  }   
}
 
// Function to return the minimum
// number of operations required
static int minOperations(int n,
                         int []a,
                         int m)
{
  int k = n / m;
 
  // To store modulos values
  List []val =
       new List[m];
   
  for(int i = 0;
          i < val.Length; i++)
    val[i] = new List();
 
  for(int i = 0; i < n; ++i)
  {
    val[a[i] % m].Add(i);
  }
 
  long ans = 0;
  List extra =
       new List();
 
  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 ((val[cur].Count) > k)
    {
      int elem = val[cur][val[cur].Count - 1];
      val[cur].RemoveAt(val[cur].Count - 1);
      extra.Add(new pair(elem, i));
    }
 
    // If it's size is less than k
    // it needed to be increased
    while (val[cur].Count < k &&
           extra.Count != 0)
    {
      int elem = extra[extra.Count - 1].first;
      int mmod = extra[extra.Count - 1].second;
      extra.RemoveAt(extra.Count - 1);
      val[cur].Add(elem);
      ans += i - mmod;
    }
  }
  return (int)ans;
}
 
// Driver code
public static void Main(String[] args)
{
  int m = 3;
  int []a = {3, 2, 0, 6, 10, 12};
  int n = a.Length;
  Console.Write(minOperations(n, a, m));
}
}
 
// This code is contributed by Princi Singh

输出：

3

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