Minimize cost to sort the Array by moving elements with cost as the value itself

19 June 2025

0

Given an array arr[] of N positive integers, the task is to find the minimum cost to sort the given array by moving an array element to any position such that the cost of moving that element is the value of that element.

Examples:

Input: arr[] = {7, 1, 2, 3}
Output: 6
Explanation:
Following are the possible set of moves to sort the array with minimum cost:

Move 1 to the front, arr[] = {1, 7, 2, 3}. cost = 1

Move 2 to 2nd place, arr[] = {1, 2, 7, 3}. cost = 2

Move 3 to 3rd place, arr[] = {1, 2, 3, 7}, cost = 3

Therefore, the total cost is (1 + 2 + 3) = 6.

Input: arr[] = {7, 1, 2, 5}
Output: 7

Approach: The given problem can be solved by using Dynamic Programming. The idea is to fixed the array elements that forms the longest non-decreasing subsequence having the maximum sum and perform the given moves to all the remaining array elements. Follow the below steps to solve the problem:

Find the longest non-decreasing subsequence with the maximal sum and store it in a variable, say S.
After the above step, print the value of the (sum of array element – S) as the resultant possible minimum cost of sorting the given array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum of
// non-decreasing subsequence
int maxSumIS(int arr[], int n)
{
    int i, j, max = 0;
    // Stores the maximum sum of subsequence ending at index i
    int dp[n];
    // Initialize dp[] values for all indexes
    for (i = 0; i < n; i++)
        dp[i] = arr[i];
 
    // Compute maximum sum values in bottom up manner
    for (i = 1; i < n; i++)
        for (j = 0; j < i; j++)
            if (arr[i] >= arr[j] && dp[i] < dp[j] + arr[i])
                dp[i] = dp[j] + arr[i];
 
    // Pick maximum of all msis values
    for (i = 0; i < n; i++)
        if (max < dp[i])
            max = dp[i];
 
    // Return the maximum sum as max
    return max;
}
 
// Function to find the minimum cost to
// sort given array in increasing order
int minCostSort(int arr[], int N)
{
    // Find the sum of array
    int sm = 0;
    for (int i = 0; i < N; i++)
        sm += arr[i];
    // Find the maximum sum non-decreasing subsequence
    int res = maxSumIS(arr, N);
    // Return the minimum cost
    return sm - res;
}
 
// Driver Code
int main()
{
    int arr[] = { 7, 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << minCostSort(arr, N);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

C

// C program for the above approach
#include <stdio.h>
 
// Function to find the maximum sum of non-decreasing
// subsequence
int maxSumIS(int arr[], int n)
{
    int i, j, max = 0;
    // Stores the maximum sum of subsequence ending at index i
    int dp[n];
    // Initialize dp[] values for all indexes
    for (i = 0; i < n; i++)
        dp[i] = arr[i];
 
    // Compute maximum sum values in bottom up manner
    for (i = 1; i < n; i++)
        for (j = 0; j < i; j++)
            if (arr[i] >= arr[j] && dp[i] < dp[j] + arr[i])
                dp[i] = dp[j] + arr[i];
 
    // Pick maximum of all msis values
    for (i = 0; i < n; i++)
        if (max < dp[i])
            max = dp[i];
 
    // Return the maximum sum as max
    return max;
}
 
// Function to find the minimum cost to
// sort given array in increasing order
int minCostSort(int arr[], int N)
{
    // Find the sum of array
    int sm = 0;
    for (int i = 0; i < N; i++)
        sm += arr[i];
    // Find the maximum sum non-decreasing subsequence
    int res = maxSumIS(arr, N);
 
    // Return the minimum cost
    return sm - res;
}
 
// Driver Code
int main()
{
    int arr[] = { 7, 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    printf("%d", minCostSort(arr, N));
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

Java

// Java program for the above approach
import java.io.*;
class GFG 
{
   
    // Function to find the maximum sum of
    // non-decreasing subsequence
    static int maxSumIS(int[] arr, int n)
    {
        int i, j, max = 0;
 
        // Stores the maximum sum of
        // subsequence ending at index i
        int[] dp = new int[n];
 
        // Initialize dp[] values for all
        // indexes
        for (i = 0; i < n; i++)
            dp[i] = arr[i];
 
        // Compute maximum sum values
        // in bottom up manner
        for (i = 1; i < n; i++)
            for (j = 0; j < i; j++)
                if (arr[i] >= arr[j]
                    && dp[i] < dp[j] + arr[i])
                    dp[i] = dp[j] + arr[i];
 
        // Pick maximum of all msis values
        for (i = 0; i < n; i++) {
            if (max < dp[i]) {
                max = dp[i];
            }
        }
 
        // Return the maximum sum as max
        return max;
    }
 
    // Function to find the minimum cost to
    // sort given array in increasing order
    static int minCostSort(int[] arr, int N)
    {
        // Find the sum of array
        int sm = 0;
        for (int i = 0; i < N; i++) {
            sm += arr[i];
        }
 
        // Find the maximum sum non-decreasing
        // subsequence
        int res = maxSumIS(arr, N);
 
        // Return the minimum cost
        return sm - res;
    }
 
    // Driver Code
    public static void main(String []args)
    {
        int[] arr = { 7, 1, 2, 3 };
        int N = arr.length;
        System.out.print(minCostSort(arr, N));
    }
}
 
// This code is contributed by shivanisinghss2110

Python3

# python program for the above approach
 
# Function to find the maximum sum of
# non-decreasing subsequence
def maxSumIS(arr, n):
 
    max = 0
 
    # Stores the maximum sum of
    # subsequence ending at index i
    dp = [0 for _ in range(n)]
 
    # Initialize dp[] values for all
    # indexes
    for i in range(0, n):
        dp[i] = arr[i]
 
    # Compute maximum sum values
    # in bottom up manner
    for i in range(1, n):
        for j in range(0, i):
            if (arr[i] >= arr[j] and dp[i] < dp[j] + arr[i]):
                dp[i] = dp[j] + arr[i]
 
    # Pick maximum of all msis values
    for i in range(0, n):
        if (max < dp[i]):
            max = dp[i]
 
    # Return the maximum sum as max
    return max
 
# Function to find the minimum cost to
# sort given array in increasing order
def minCostSort(arr, N):
 
    # Find the sum of array
    sm = 0
    for i in range(0, N):
        sm += arr[i]
 
    # Find the maximum sum non-decreasing
    # subsequence
    res = maxSumIS(arr, N)
 
    # Return the minimum cost
    return sm - res
 
# Driver Code
if __name__ == "__main__":
 
    arr = [7, 1, 2, 3]
    N = len(arr)
    print(minCostSort(arr, N))
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG 
{
   
    // Function to find the maximum sum of
    // non-decreasing subsequence
    static int maxSumIS(int[] arr, int n)
    {
        int i, j, max = 0;
 
        // Stores the maximum sum of
        // subsequence ending at index i
        int[] dp = new int[n];
 
        // Initialize dp[] values for all
        // indexes
        for (i = 0; i < n; i++)
            dp[i] = arr[i];
 
        // Compute maximum sum values
        // in bottom up manner
        for (i = 1; i < n; i++)
            for (j = 0; j < i; j++)
                if (arr[i] >= arr[j]
                    && dp[i] < dp[j] + arr[i])
                    dp[i] = dp[j] + arr[i];
 
        // Pick maximum of all msis values
        for (i = 0; i < n; i++) {
            if (max < dp[i]) {
                max = dp[i];
            }
        }
 
        // Return the maximum sum as max
        return max;
    }
 
    // Function to find the minimum cost to
    // sort given array in increasing order
    static int minCostSort(int[] arr, int N)
    {
        // Find the sum of array
        int sm = 0;
        for (int i = 0; i < N; i++) {
            sm += arr[i];
        }
 
        // Find the maximum sum non-decreasing
        // subsequence
        int res = maxSumIS(arr, N);
 
        // Return the minimum cost
        return sm - res;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 7, 1, 2, 3 };
        int N = arr.Length;
        Console.WriteLine(minCostSort(arr, N));
    }
}
 
// This code is contributed by ukasp.

Javascript

<script>
       // JavaScript Program to implement
       // the above approach
 
 
       // Function to find the maximum sum of
       // non-decreasing subsequence
       function maxSumIS(arr, n) {
           let i, j, max = 0;
 
           // Stores the maximum sum of
           // subsequence ending at index i
           let dp = new Array(n);
 
           // Initialize dp[] values for all
           // indexes
           for (i = 0; i < n; i++)
               dp[i] = arr[i];
 
           // Compute maximum sum values
           // in bottom up manner
           for (i = 1; i < n; i++)
               for (j = 0; j < i; j++)
                   if (arr[i] >= arr[j]
                       && dp[i] < dp[j] + arr[i])
                       dp[i] = dp[j] + arr[i];
 
           // Pick maximum of all msis values
           for (i = 0; i < n; i++) {
               if (max < dp[i]) {
                   max = dp[i];
               }
           }
 
           // Return the maximum sum as max
           return max;
       }
 
       // Function to find the minimum cost to
       // sort given array in increasing order
       function minCostSort(arr, N) {
           // Find the sum of array
           let sm = 0;
           for (let i = 0; i < N; i++) {
               sm += arr[i];
           }
 
           // Find the maximum sum non-decreasing
           // subsequence
           let res = maxSumIS(arr, N);
 
           // Return the minimum cost
           return sm - res;
       }
 
       // Driver Code
       let arr = [7, 1, 2, 3];
       let N = arr.length;
       document.write(minCostSort(arr, N));
 
 
    // This code is contributed by Potta Lokesh
 
   </script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

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