最小范围增量操作以对数组进行排序

给定一个包含N个元素的数组。允许在数组上执行以下任意多次移动：

• 选择任意L和R，并将L到R范围内的所有数字加1。

任务是找到以非降序对数组进行排序所需的最少数量的此类移动。

例子：

Input : arr[] = {1, 2, 3, 4}
Output : 0
The array is already sorted

Input : arr[] = {3, 2, 1}
Output : 2
Step 1: L=1 and R=2 (0-based)
Step 2: L=2 and R=2
Resultant array [3, 3, 3]

考虑一个排序的数组，增加数组的所有元素仍将导致一个排序的数组。

因此，想法是从最后一个索引开始，从右遍历数组的元素，并跟踪最小元素。如果在任何时候发现元素的顺序在增加，请从当前元素中减去右边的min元素，以计算移动数。

下面是上述方法的实现：

C++

// C++ program to find minimum range
// increments to sort an array
  
#include 
using namespace std;
  
// Function to find minimum range
// increments to sort an array
int minMovesToSort(int arr[], int n)
{
    int moves = 0;
  
    int i, mn = arr[n - 1];
  
    for (i = n - 2; i >= 0; i--) {
  
        // If current element is found greater than
        // last element
        // Increment all terms in
        // range i+1 to n-1
        if (arr[i] > mn)
            moves += arr[i] - mn;
  
        mn = arr[i]; // Minimum in range i to n-1
    }
  
    return moves;
}
  
// Driver Code
int main()
{
    int arr[] = { 3, 5, 2, 8, 4 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << minMovesToSort(arr, n);
  
    return 0;
}

Java

// Java program to find minimum range
// increments to sort an array
  
  
import java.io.*;
  
class GFG {
  
// Function to find minimum range
// increments to sort an array
static int minMovesToSort(int arr[], int n)
{
    int moves = 0;
  
    int i, mn = arr[n - 1];
  
    for (i = n - 2; i >= 0; i--) {
  
        // If current element is found greater than
        // last element
        // Increment all terms in
        // range i+1 to n-1
        if (arr[i] > mn)
            moves += arr[i] - mn;
  
        mn = arr[i]; // Minimum in range i to n-1
    }
  
    return moves;
}
  
// Driver Code
  
  
    public static void main (String[] args) {
        int arr[] = { 3, 5, 2, 8, 4 };
  
    int n = arr.length;
  
    System.out.println( minMovesToSort(arr, n));
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 program to find minimum range
# increments to sort an array
  
# Function to find minimum range
# increments to sort an array
def minMovesToSort(arr, n) :
  
    moves = 0
  
    mn = arr[n - 1]
      
    for i in range(n - 1, -1, -1) :
  
        # If current element is found 
        # greater than last element
        # Increment all terms in
        # range i+1 to n-1
        if (arr[i] > mn) :
            moves += arr[i] - mn
  
        mn = arr[i] # Minimum in range i to n-1
      
    return moves
  
# Driver Code
if __name__ == "__main__" :
  
    arr = [ 3, 5, 2, 8, 4 ]
  
    n = len(arr)
  
    print(minMovesToSort(arr, n))
  
# This code is contributed by Ryuga

C#

// C# program to find minimum range 
// increments to sort an array 
using System;
  
class GFG
{
// Function to find minimum range 
// increments to sort an array 
static int minMovesToSort(int []arr,
                          int n) 
{ 
    int moves = 0; 
  
    int i, mn = arr[n - 1]; 
  
    for (i = n - 2; i >= 0; i--) 
    { 
  
        // If current element is found 
        // greater than last element 
        // Increment all terms in 
        // range i+1 to n-1 
        if (arr[i] > mn) 
            moves += arr[i] - mn; 
  
        mn = arr[i]; // Minimum in range 
                     // i to n-1 
    } 
    return moves; 
} 
  
// Driver Code 
static public void Main ()
{
    int []arr = { 3, 5, 2, 8, 4 }; 
    int n = arr.Length; 
      
    Console.WriteLine(minMovesToSort(arr, n)); 
} 
} 
  
// This code is contributed by ajit

PHP

= 0; $i--)
    {
  
        // If current element is found 
        // greater than last element
        // Increment all terms in
        // range i+1 to n-1
        if ($arr[$i] > $mn)
            $moves += $arr[$i] - $mn;
  
        $mn = $arr[$i]; // Minimum in range i to n-1
    }
  
    return $moves;
}
  
// Driver Code
$arr = array(3, 5, 2, 8, 4 );
  
$n = sizeof($arr);
  
echo minMovesToSort($arr, $n);
  
// This code is contributed 
// by Akanksha Rai
?>

输出：

7

时间复杂度： O(N)