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Find minimum adjustment cost of an array

Given an array of positive integers, replace each element in the array such that the difference between adjacent elements in the array is less than or equal to a given target. We need to minimize the adjustment cost, that is the sum of differences between new and old values. We basically need to minimize ∑|A[i] – Anew[i]| where 0 ≤ i ≤ n-1, n is size of A[] and Anew[] is the array with adjacent difference less that or equal to target.

Assume all elements of the array is less than constant M = 100.

Examples:



Input: arr = [1, 3, 0, 3], target = 1
Output: Minimum adjustment cost is 3
Explanation: One of the possible solutions 
is [2, 3, 2, 3]

Input: arr = [2, 3, 2, 3], target = 1
Output: Minimum adjustment cost is 0
Explanation:  All adjacent elements in the input 
array are already less than equal to given target

Input: arr = [55, 77, 52, 61, 39, 6, 
             25, 60, 49, 47], target = 10
Output: Minimum adjustment cost is 75
Explanation: One of the possible solutions is 
[55, 62, 52, 49, 39, 29, 30, 40, 49, 47]


In order to minimize the adjustment cost ∑|A[i] – Anew[i]| for all index i in the array, |A[i] – Anew[i]| should be as close to zero as possible. Also, |A[i] – Anew[i+1] ]| ≤ Target.

This problem can be solved by dynamic programming.

Let dp[i][j] defines minimal adjustment cost on changing A[i] to j, then the DP relation is defined by –



dp[i][j] = min{dp[i - 1][k]} + |j - A[i]|
           for all k's such that |k - j| ≤ target

Here, 0 ≤ i ≤ n and 0 ≤ j ≤ M where n is number of elements in the array and M = 100. We have to consider all k such that max(j – target, 0) ≤ k ≤ min(M, j + target)

Finally, the minimum adjustment cost of the array will be min{dp[n – 1][j]} for all 0 ≤ j ≤ M.

Below is the implementation of above idea –

C++

// C++ program to find minimum adjustment cost of an array
#include <bits/stdc++.h>
using namespace std;
  
#define M 100
  
// Function to find minimum adjustment cost of an array
int minAdjustmentCost(int A[], int n, int target)
{
    // dp[i][j] stores minimal adjustment cost on changing
    // A[i] to j
    int dp[n][M + 1];
  
    // handle first element of array seperately
    for (int j = 0; j <= M; j++)
        dp[0][j] = abs(j - A[0]);
  
    // do for rest elements of the array
    for (int i = 1; i < n; i++)
    {
        // replace A[i] to j and calculate minimal adjustment
        // cost dp[i][j]
        for (int j = 0; j <= M; j++)
        {
          // initialize minimal adjustment cost to INT_MAX
          dp[i][j] = INT_MAX;
  
          // consider all k such that k >= max(j - target, 0) and
          // k <= min(M, j + target) and take minimum
          for (int k = max(j-target,0); k <= min(M,j+target); k++)
             dp[i][j] = min(dp[i][j], dp[i - 1][k] + abs(A[i] - j));
        }
    }    
  
    // return minimum value from last row of dp table
    int res = INT_MAX; 
    for (int j = 0; j <= M; j++)
        res = min(res, dp[n - 1][j]);
  
    return res;
}
  
// Driver Program to test above functions
int main()
{
    int arr[] = {55, 77, 52, 61, 39, 6, 25, 60, 49, 47};
    int n = sizeof(arr) / sizeof(arr[0]);
    int target = 10;
  
    cout << "Minimum adjustment cost is "
         << minAdjustmentCost(arr, n, target) << endl;
  
    return 0;
}

Java

// Java program to find minimum adjustment cost of an array
import java.io.*;
import java.util.*;
  
class GFG 
{
    public static int M = 100;
      
    // Function to find minimum adjustment cost of an array
    static int minAdjustmentCost(int A[], int n, int target)
    {
        // dp[i][j] stores minimal adjustment cost on changing
        // A[i] to j
        int[][] dp = new int[n][M + 1];
   
        // handle first element of array seperately
        for (int j = 0; j <= M; j++)
            dp[0][j] = Math.abs(j - A[0]);
   
        // do for rest elements of the array
        for (int i = 1; i < n; i++)
        {
            // replace A[i] to j and calculate minimal adjustment
            // cost dp[i][j]
            for (int j = 0; j <= M; j++)
            {
                // initialize minimal adjustment cost to INT_MAX
                dp[i][j] = Integer.MAX_VALUE;
   
                // consider all k such that k >= max(j - target, 0) and
                // k <= min(M, j + target) and take minimum
                int k = Math.max(j-target,0);
                for ( ; k <= Math.min(M,j+target); k++)
                    dp[i][j] = Math.min(dp[i][j], dp[i - 1][k] +  
                                                Math.abs(A[i] - j));
            }
        }    
   
        // return minimum value from last row of dp table
        int res = Integer.MAX_VALUE; 
        for (int j = 0; j <= M; j++)
            res = Math.min(res, dp[n - 1][j]);
   
        return res;
    }
      
    // Driver program
    public static void main (String[] args) 
    {
        int arr[] = {55, 77, 52, 61, 39, 6, 25, 60, 49, 47};
        int n = arr.length;
        int target = 10;
   
        System.out.println("Minimum adjustment cost is "
                    +minAdjustmentCost(arr, n, target));
    }
}
  
// This code is contributed by Pramod Kumar

Python3

# Python3 program to find minimum
# adjustment cost of an array 
M = 100
  
# Function to find minimum
# adjustment cost of an array
def minAdjustmentCost(A, n, target):
      
    # dp[i][j] stores minimal adjustment 
    # cost on changing A[i] to j 
    dp = [[0 for i in range(M + 1)] 
             for i in range(n)]
               
    # handle first element
    # of array seperately
    for j in range(M + 1):
        dp[0][j] = abs(j - A[0])
  
    # do for rest elements 
    # of the array 
    for i in range(1, n):
          
        # replace A[i] to j and 
        # calculate minimal adjustment
        # cost dp[i][j] 
        for j in range(M + 1):
              
            # initialize minimal adjustment
            # cost to INT_MAX
            dp[i][j] = 100000000
              
            # consider all k such that
            # k >= max(j - target, 0) and
            # k <= min(M, j + target) and 
            # take minimum
            for k in range(max(j - target, 0), 
                           min(M, j + target) + 1):
                dp[i][j] = min(dp[i][j], dp[i - 1][k] + 
                                        abs(A[i] - j))
                                          
    # return minimum value from 
    # last row of dp table
    res = 10000000
    for j in range(M + 1):
        res = min(res, dp[n - 1][j])
    return res
  
# Driver Code 
arr= [55, 77, 52, 61, 39
       6, 25, 60, 49, 47]
n = len(arr)
target = 10
print("Minimum adjustment cost is"
       minAdjustmentCost(arr, n, target), 
                              sep = ' ')
  
# This code is contributed 
# by sahilshelangia

C#

// C# program to find minimum adjustment
// cost of an array
using System;
  
class GFG {
      
    public static int M = 100;
      
    // Function to find minimum adjustment
    // cost of an array
    static int minAdjustmentCost(int []A, int n,
                                     int target)
    {
          
        // dp[i][j] stores minimal adjustment
        // cost on changing A[i] to j
        int[,] dp = new int[n,M + 1];
  
        // handle first element of array
        // seperately
        for (int j = 0; j <= M; j++)
            dp[0,j] = Math.Abs(j - A[0]);
  
        // do for rest elements of the array
        for (int i = 1; i < n; i++)
        {
            // replace A[i] to j and calculate
            // minimal adjustment cost dp[i][j]
            for (int j = 0; j <= M; j++)
            {
                // initialize minimal adjustment
                // cost to INT_MAX
                dp[i,j] = int.MaxValue;
  
                // consider all k such that 
                // k >= max(j - target, 0) and
                // k <= min(M, j + target) and
                // take minimum
                int k = Math.Max(j - target, 0);
                  
                for ( ; k <= Math.Min(M, j +
                                   target); k++)
                    dp[i,j] = Math.Min(dp[i,j],
                                   dp[i - 1,k]
                         + Math.Abs(A[i] - j));
            }
        
  
        // return minimum value from last
        // row of dp table
        int res = int.MaxValue; 
        for (int j = 0; j <= M; j++)
            res = Math.Min(res, dp[n - 1,j]);
  
        return res;
    }
      
    // Driver program
    public static void Main () 
    {
        int []arr = {55, 77, 52, 61, 39,
                        6, 25, 60, 49, 47};
        int n = arr.Length;
        int target = 10;
  
        Console.WriteLine("Minimum adjustment"
                                 + " cost is "
         + minAdjustmentCost(arr, n, target));
    }
}
  
// This code is contributed by Sam007.

PHP

<?php
// PHP program to find minimum 
// adjustment cost of an array
  
$M = 100;
  
// Function to find minimum 
// adjustment cost of an array
function minAdjustmentCost( $A, $n, $target)
{
      
    // dp[i][j] stores minimal 
    // adjustment cost on changing
    // A[i] to j
    global $M;
    $dp = array(array());
  
    // handle first element 
    // of array seperately
    for($j = 0; $j <= $M; $j++)
        $dp[0][$j] = abs($j - $A[0]);
  
    // do for rest 
    // elements of the array
    for($i = 1; $i < $n; $i++)
    {
          
        // replace A[i] to j and 
        // calculate minimal adjustment
        // cost dp[i][j]
        for($j = 0; $j <= $M; $j++)
        {
              
            // initialize minimal adjustment
            // cost to INT_MAX
            $dp[$i][$j] = PHP_INT_MAX;
      
            // consider all k such that 
            // k >= max(j - target, 0) and
            // k <= min(M, j + target) and
            // take minimum
            for($k = max($j - $target, 0);
                $k <= min($M, $j + $target);
                                       $k++)
                $dp[$i][$j] = min($dp[$i][$j], 
                              $dp[$i - 1][$k] + 
                              abs($A[$i] - $j));
        }
    
  
    // return minimum value 
    // from last row of dp table
    $res = PHP_INT_MAX; 
    for($j = 0; $j <= $M; $j++)
        $res = min($res, $dp[$n - 1][$j]);
  
    return $res;
}
  
    // Driver Code
    $arr = array(55, 77, 52, 61, 39, 
                 6, 25, 60, 49, 47);
    $n = count($arr);
    $target = 10;
  
    echo "Minimum adjustment cost is "
        , minAdjustmentCost($arr, $n, $target);
  
// This code is contributed by anuj_67.
?>


Output:

Minimum adjustment cost is 75

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This article is attributed to GeeksforGeeks.org

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