计数递减到使所有数组元素相等所需的最近的较小元素

给定一个由N 个非负整数组成的数组arr[] ，任务是找到使所有数组元素相等所需的操作次数。在每个操作中，任何数组元素都可以更改为其最近的较小数组元素。

例子：

Input: arr[] = {2, 5, 4, 3, 5, 4}
Output: 11
Explanation: 
Step 1: Replace all 5s with 4s. Therefore, arr[] = {2, 4, 4, 3, 4, 4}. Count of operations = 2
Step 2: Replace all 4s with 3s. Therefore, arr[] = {2, 3, 3, 3, 3, 3}. Count of operations = 4
Steps 3: Replace all 3s with 2s. Therefore, arr[] = {2, 2, 2, 2, 2, 2}. Count of operations = 5
Therefore, total number of operations = 11

Input : arr[] = {2, 2, 2}
Output : 0

方法：想法是使用排序算法和Map数据结构。请按照以下步骤解决问题：

1. 初始化 Map 以存储每个数组元素的频率，除了最小元素，其中键 (K)是唯一元素的集合，值 (V)是它们的频率。
2. 根据键以降序对 Map 进行排序。
3. 用0初始化两个变量ans和prev_val分别存储当前答案和前缀和。
4. 现在迭代 Map 并将每个元素的相应频率与prev_val一起添加到ans作为ans = ans + (freq) + prev_val
5. 在迭代 Map 时，每次都将prev_val增加当前元素的频率。
6. 完全遍历Map后，将ans打印为最小操作次数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the minimum number of
// decrements to make all elements equals
int MinimumNoOfOperations(int arr[], int n)
{
 
    // Find minimum element
    int min_ele = INT_MAX;
    for (int i = 0; i < n; ++i) {
        min_ele = min(min_ele, arr[i]);
    }
 
    // Stores frequencies of array elements
    map > mp;
 
    // Update frequencies in the Map
    for (int i = 0; i < n; ++i) {
 
        if (arr[i] == min_ele)
            continue;
        else
            mp[arr[i]]++;
    }
 
    // Iterate the map
    map::iterator it;
 
    // Stores the count of
    // decrements at each iteration
    int prev_val = 0;
 
    // Stores the total
    // count of decrements
    int ans = 0;
 
    // Calculate the number of decrements
    for (it = mp.begin(); it != mp.end();
         ++it) {
 
        ans += (it->second + prev_val);
        prev_val += it->second;
    }
 
    // Return minimum operations
    return ans;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 5, 4, 3, 5, 4 };
 
    // Given size
    int size = sizeof(arr)
               / sizeof(arr[0]);
 
    // Function call
    cout << MinimumNoOfOperations(
        arr, size);
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
 
class solution{
    
// Function to print the minimum
// number of decrements to make
// all elements equals
static int MinimumNoOfOperations(int arr[],
                                 int n)
{
  // Find minimum element
  int min_ele = Integer.MAX_VALUE;
   
  for (int i = 0; i < n; ++i)
  {
    min_ele = Math.min(min_ele,
                       arr[i]);
  }
 
  // Stores frequencies of array
  // elements
  TreeMap mp =
          new TreeMap(
          Collections.reverseOrder());
 
  // Update frequencies in
  // the Map
  for (int i = 0; i < n; ++i)
  {
    if (arr[i] == min_ele)
      continue;
    else
      mp.put(arr[i],
      mp.getOrDefault(arr[i], 0) + 1);
  }
 
  // Stores the count of
  // decrements at each
  // iteration
  int prev_val = 0;
 
  // Stores the total
  // count of decrements
  int ans = 0;
 
  // Calculate the number of
  // decrements
  for (Map.Entry it : mp.entrySet())
  {
    ans += (it.getValue() + prev_val);
    prev_val += it.getValue();
  }
 
  // Return minimum operations
  return ans;
}
 
// Driver Code
public static void main(String args[])
{
  // Given array
  int arr[] = {2, 5, 4, 3, 5, 4};
   
  // Given size
  int size = arr.length;
   
  // Function call
  System.out.println(
  MinimumNoOfOperations(arr, size));
}
}
 
// This code is contributed by bgangwar59

Python3

# Python3 program for the above approach
import sys
 
# Function to print the minimum number of
# decrements to make all elements equals
def MinimumNoOfOperations(arr, n):
     
    # Find minimum element
    min_ele = sys.maxsize
     
    for i in range(n):
        min_ele = min(min_ele, arr[i])
 
    # Stores frequencies of array elements
    mp = {}
 
    # Update frequencies in the Map
    for i in range(n):
        if (arr[i] == min_ele):
            continue
        else:
            mp[arr[i]] = mp.get(arr[i], 0) + 1
             
    # Stores the count of
    # decrements at each iteration
    prev_val = 0
 
    # Stores the total
    # count of decrements
    ans = 0
 
    # Calculate the number of decrements
    for it in mp:
        ans += (mp[it] + prev_val)
        prev_val += mp[it]
 
    # Return minimum operations
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    # Given array
    arr = [ 2, 5, 4, 3, 5, 4 ]
 
    # Given size
    size = len(arr)
 
    # Function call
    print(MinimumNoOfOperations(arr, size))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System.Collections.Generic;
using System;
using System.Linq;
 
class GFG{
    
// Function to print the minimum
// number of decrements to make
// all elements equals
static int MinimumNoOfOperations(int []arr,
                                 int n)
{
   
  // Find minimum element
  int min_ele = 1000000000;
   
  for(int i = 0; i < n; ++i)
  {
      min_ele = Math.Min(min_ele, arr[i]);
  }
 
  // Stores frequencies of array
  // elements
  Dictionary mp = new Dictionary();
 
  // Update frequencies in
  // the Map
  for(int i = 0; i < n; ++i)
  {
    if (arr[i] == min_ele)
      continue;
    else
    {
      if (mp.ContainsKey(arr[i]) == true)
        mp[arr[i]] += 1;
      else
        mp[arr[i]] = 1;   
    }
  }
 
  // Stores the count of
  // decrements at each
  // iteration
  int prev_val = 0;
 
  // Stores the total
  // count of decrements
  int ans = 0;
 
  // Calculate the number of
  // decrements
  var val = mp.Keys.ToList();
  foreach(var key in val)
  {
    ans += (mp[key] + prev_val);
    prev_val += mp[key];
  }
   
  // Return minimum operations
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
   
  // Given array
  int []arr = { 2, 5, 4, 3, 5, 4 };
   
  // Given size
  int size = arr.Length;
   
  // Function call
  Console.WriteLine(MinimumNoOfOperations(
    arr, size));
}
}
 
// This code is contributed by Stream_Cipher

Javascript

// Javascript program for the above approach
 
// Function to print the minimum number of
// decrements to make all elements equals
function MinimumNoOfOperations(arr, n)
{
 
    // Find minimum element
    var min_ele = 1000000000;
    for (var i = 0; i < n; ++i) {
        min_ele = Math.min(min_ele, arr[i]);
    }
 
    // Stores frequencies of array elements
    var mp = new Map();
 
    // Update frequencies in the Map
    for (var i = 0; i < n; ++i) {
 
        if (arr[i] == min_ele)
            continue;
        else
        {
            if(mp.has(arr[i]))
            {
                mp.set(arr[i], mp.get(arr[i])+1);
            }
            else
            {
                mp.set(arr[i], 1);
            }
        }
    }
 
    // Stores the count of
    // decrements at each iteration
    var prev_val = 0;
 
    // Stores the total
    // count of decrements
    var ans = 0;
    var keys = [];
 
    mp.forEach((value, key) => {
        keys.push(key);
    });
    keys.sort((a,b)=>b-a);
    // Calculate the number of decrements
    keys.forEach(value => {
        ans += (mp.get(value) + prev_val);
        prev_val += mp.get(value);
    });
 
    // Return minimum operations
    return ans;
}
 
// Driver Code
// Given array
var arr = [2, 5, 4, 3, 5, 4];
// Given size
var size = arr.length
// Function call
document.write( MinimumNoOfOperations(
    arr, size));
 

输出：

11

时间复杂度： O(N)，其中 N 是数组的大小。
辅助空间： O(N)

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