使数组排序所需的最小增量或减量操作

给定N个整数的数组arr [] ，任务是通过执行最少数量的操作以不降序对数组进行排序。在单个操作中，可以将数组的元素增加或减少1 。打印所需的最少操作数。
例子：

Input: arr[] = {1, 2, 1, 4, 3} 
Output: 2 
Add 1 to the 3rd element(1) and subtract 1 from 
the 4th element(4) to get {1, 2, 2, 3, 3}
Input: arr[] = {1, 2, 2, 100} 
Output: 0 
Given array is already sorted. 
 

观察：由于我们希望最大程度地减少对数组进行排序所需的操作次数，因此应保持以下内容：

• 一个数字将永远不会减少到小于初始数组的最小值的值。
• 数字永远不会增加到大于初始数组最大值的值。
• 将数字从X更改为Y所需的操作数为abs(X – Y)。

方法：基于以上观察，可以使用动态编程解决此问题。

1. 令DP(i，j)表示当第i^个元素等于j时，以不降序对数组的^第i个元素进行排序所需的最小操作。
2. 现在DP(N，j)的需要对于j的所有可能的值，其中N是阵列的大小来计算的。根据观察， j≥初始数组的最小元素，j≤初始数组的最大元素。
3. 可以很容易地回答i = 1的DP(i，j)中的基本情况。什么是最低程度的业务needes在^第一单元以非递减顺序排序等的^第一单元是等于j?。 DP(1，j)= abs(array [1] – j) 。
4. 现在考虑i> 1的DP(i，j) 。如果将^第i^个元素设置为j，则需要对^第一个i – 1个元素进行排序，第(i – 1)^个元素必须≤j，即DP(i，j)=(DP(i – 1的最小值) ，k) ，其中k从1到j ) + abs(array [i] – j)
5. 使用上述递归关系和基本案例，可以轻松地计算结果。

以下是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum number
// of given operations required
// to sort the array
int getMinimumOps(vector ar)
{
    // Number of elements in the array
    int n = ar.size();
 
    // Smallest element in the array
    int small = *min_element(ar.begin(), ar.end());
 
    // Largest element in the array
    int large = *max_element(ar.begin(), ar.end());
 
    /*
        dp(i, j) represents the minimum number
        of operations needed to make the
        array[0 .. i] sorted in non-decreasing
        order given that ith element is j
    */
    int dp[n][large + 1];
 
    // Fill the dp[]][ array for base cases
    for (int j = small; j <= large; j++) {
        dp[0][j] = abs(ar[0] - j);
    }
 
    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    for (int i = 1; i < n; i++) {
        int minimum = INT_MAX;
        for (int j = small; j <= large; j++) {
 
            /*
            If the ith element is j then we can have
            any value from small to j for the i-1 th
            element
            We choose the one that requires the
            minimum operations
        */
            minimum = min(minimum, dp[i - 1][j]);
            dp[i][j] = minimum + abs(ar[i] - j);
        }
    }
 
    /*
        If we made the (n - 1)th element equal to j
        we required dp(n-1, j) operations
        We choose the minimum among all possible
        dp(n-1, j) where j goes from small to large
    */
    int ans = INT_MAX;
    for (int j = small; j <= large; j++) {
        ans = min(ans, dp[n - 1][j]);
    }
 
    return ans;
}
 
// Driver code
int main()
{
    vector ar = { 1, 2, 1, 4, 3 };
 
    cout << getMinimumOps(ar);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the minimum number
// of given operations required
// to sort the array
static int getMinimumOps(Vector ar)
{
    // Number of elements in the array
    int n = ar.size();
 
    // Smallest element in the array
    int small = Collections.min(ar);
 
    // Largest element in the array
    int large = Collections.max(ar);
 
    /*
        dp(i, j) represents the minimum number
        of operations needed to make the
        array[0 .. i] sorted in non-decreasing
        order given that ith element is j
    */
    int [][]dp = new int[n][large + 1];
 
    // Fill the dp[]][ array for base cases
    for (int j = small; j <= large; j++)
    {
        dp[0][j] = Math.abs(ar.get(0) - j);
    }
 
    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    for (int i = 1; i < n; i++)
    {
        int minimum = Integer.MAX_VALUE;
        for (int j = small; j <= large; j++)
        {
 
            /*
            If the ith element is j then we can have
            any value from small to j for the i-1 th
            element
            We choose the one that requires the
            minimum operations
            */
            minimum = Math.min(minimum, dp[i - 1][j]);
            dp[i][j] = minimum + Math.abs(ar.get(i) - j);
        }
    }
 
    /*
        If we made the (n - 1)th element equal to j
        we required dp(n-1, j) operations
        We choose the minimum among all possible
        dp(n-1, j) where j goes from small to large
    */
    int ans = Integer.MAX_VALUE;
    for (int j = small; j <= large; j++)
    {
        ans = Math.min(ans, dp[n - 1][j]);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    Integer []arr = { 1, 2, 1, 4, 3 };
    Vector ar = new Vector<>(Arrays.asList(arr));
 
    System.out.println(getMinimumOps(ar));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the minimum number
# of given operations required
# to sort the array
def getMinimumOps(ar):
     
    # Number of elements in the array
    n = len(ar)
 
    # Smallest element in the array
    small = min(ar)
 
    # Largest element in the array
    large = max(ar)
 
    """
        dp(i, j) represents the minimum number
        of operations needed to make the
        array[0 .. i] sorted in non-decreasing
        order given that ith element is j
    """
    dp = [[ 0 for i in range(large + 1)]
              for i in range(n)]
 
    # Fill the dp[]][ array for base cases
    for j in range(small, large + 1):
        dp[0][j] = abs(ar[0] - j)
    """
    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    """
    for i in range(1, n):
        minimum = 10**9
        for j in range(small, large + 1):
             
        # """
        #     /*
        #     If the ith element is j then we can have
        #     any value from small to j for the i-1 th
        #     element
        #     We choose the one that requires the
        #     minimum operations
        # """
            minimum = min(minimum, dp[i - 1][j])
            dp[i][j] = minimum + abs(ar[i] - j)
    """
    /*
        If we made the (n - 1)th element equal to j
        we required dp(n-1, j) operations
        We choose the minimum among all possible
        dp(n-1, j) where j goes from small to large
    */
    """
    ans = 10**9
    for j in range(small, large + 1):
        ans = min(ans, dp[n - 1][j])
 
    return ans
 
# Driver code
ar = [1, 2, 1, 4, 3]
 
print(getMinimumOps(ar))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Linq;
using System.Collections.Generic;            
     
class GFG
{
 
// Function to return the minimum number
// of given operations required
// to sort the array
static int getMinimumOps(List ar)
{
    // Number of elements in the array
    int n = ar.Count;
 
    // Smallest element in the array
    int small = ar.Min();
 
    // Largest element in the array
    int large = ar.Max();
 
    /*
        dp(i, j) represents the minimum number
        of operations needed to make the
        array[0 .. i] sorted in non-decreasing
        order given that ith element is j
    */
    int [,]dp = new int[n, large + 1];
 
    // Fill the dp[], array for base cases
    for (int j = small; j <= large; j++)
    {
        dp[0, j] = Math.Abs(ar[0] - j);
    }
 
    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    for (int i = 1; i < n; i++)
    {
        int minimum = int.MaxValue;
        for (int j = small; j <= large; j++)
        {
 
            /*
            If the ith element is j then we can have
            any value from small to j for the i-1 th
            element
            We choose the one that requires the
            minimum operations
            */
            minimum = Math.Min(minimum, dp[i - 1, j]);
            dp[i, j] = minimum + Math.Abs(ar[i] - j);
        }
    }
 
    /*
        If we made the (n - 1)th element equal to j
        we required dp(n-1, j) operations
        We choose the minimum among all possible
        dp(n-1, j) where j goes from small to large
    */
    int ans = int.MaxValue;
    for (int j = small; j <= large; j++)
    {
        ans = Math.Min(ans, dp[n - 1, j]);
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 1, 4, 3 };
    List ar = new List(arr);
 
    Console.WriteLine(getMinimumOps(ar));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the approach
 
// Function to return the minimum number
// of given operations required
// to sort the array
function getMinimumOps(ar)
{
    // Number of elements in the array
    var n = ar.length;
 
    // Smallest element in the array
    var small = Math.min.apply(Math,ar);
 
    // Largest element in the array
    var large = Math.max.apply(Math,ar);
 
    /*
        dp(i, j) represents the minimum number
        of operations needed to make the
        array[0 .. i] sorted in non-decreasing
        order given that ith element is j
    */
    var dp = new Array(n);
    var i,j;
    for(i=0;i<dp.length;i++)
        dp[i] = new Array(large+1);
 
    // Fill the dp[]][ array for base cases
    for (j = small; j <= large; j++) {
        dp[0][j] = Math.abs(ar[0] - j);
    }
 
    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    for (i = 1; i < n; i++) {
        var minimum = 2147483647;
        for (j = small; j <= large; j++) {
 
            /*
            If the ith element is j then we can have
            any value from small to j for the i-1 th
            element
            We choose the one that requires the
            minimum operations
        */
            minimum = Math.min(minimum, dp[i - 1][j]);
            dp[i][j] = minimum + Math.abs(ar[i] - j);
        }
    }
 
    /*
        If we made the (n - 1)th element equal to j
        we required dp(n-1, j) operations
        We choose the minimum among all possible
        dp(n-1, j) where j goes from small to large
    */
    var ans = 2147483647;
    for (j = small; j <= large; j++) {
        ans = Math.min(ans, dp[n - 1][j]);
    }
 
    return ans;
}
 
// Driver code
    var ar = [1, 2, 1, 4, 3];
 
    document.write(getMinimumOps(ar));
 

输出：

2

复杂度分析：上述方法的时间复杂度为O(N * R)，其中N是数组中元素的数量，R =最大–数组中最小的元素+ 1。

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