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] – A_new[i]| where 0 ? i ? n-1, n is size of A[] and A_new[] is the array with adjacent difference less than or equal to the 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] – A_new[i]| for all index i in the array, |A[i] – A_new[i]| should be as close to zero as possible. Also, |A[i] – A_new[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 the 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.

Algorithm:

• Create a 2D array with the initializations dp[n][M+1] to record the least adjustment cost of changing A[i] to j, where n is the array’s length and M is its maximum value.
• Calculate the smallest adjustment cost of changing A[0] to j for the first element of the array, dp[0][j], using the formula dp[0][j] = abs (j – A[0]).
• Replace A[i] with j in the remaining array elements, dp[i][j], and use the formula dp[i][j] = min(dp[i-1][k] + abs(A[i] – j)), where k takes all feasible values between max(j-target,0) and min(M,j+target), to get the minimal adjustment cost.
• As the minimum adjustment cost, give the lowest number from the last row of the dp table. 

Below is the implementation of the 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 separately
    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 separately
        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 separately
    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
        // separately
        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.

Javascript

<script>
    // Javascript program to find minimum adjustment cost of an array
    let 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
        let dp = new Array(n);
        for (let i = 0; i < n; i++)
        {
            dp[i] = new Array(n);
            for (let j = 0; j <= M; j++)
            {
                dp[i][j] = 0;
            }
        }

        // handle first element of array separately
        for (let j = 0; j <= M; j++)
            dp[0][j] = Math.abs(j - A[0]);

        // do for rest elements of the array
        for (let i = 1; i < n; i++)
        {
            // replace A[i] to j and calculate minimal adjustment
            // cost dp[i][j]
            for (let j = 0; j <= M; j++)
            {
                // initialize minimal adjustment cost to INT_MAX
                dp[i][j] = Number.MAX_VALUE;

                // consider all k such that k >= max(j - target, 0) and
                // k <= min(M, j + target) and take minimum
                let 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
        let res = Number.MAX_VALUE;
        for (let j = 0; j <= M; j++)
            res = Math.min(res, dp[n - 1][j]);

        return res;
    }

    let arr = [55, 77, 52, 61, 39, 6, 25, 60, 49, 47];
    let n = arr.length;
    let target = 10;

    document.write("Minimum adjustment cost is "
                       +minAdjustmentCost(arr, n, target));

  // This code is contributed by decode2207.
</script>

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 separately
    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

Time Complexity: O(n*m²)
Auxiliary Space: O(n *m)

Efficient approach : Space optimization

In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

• Create a 1D vector dp of size m+1.
• Set a base case by initializing the values of DP.
• Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
• Now Create a temporary 1d vector prev_dp used to store the current values from previous computations.
• After every iteration assign the value of prev_dp to dp for further iteration.
• Initialize a variable res to store the final answer and update it by iterating through the Dp.
• At last return and print the final answer stored in res.

Implementation: 

C++

#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)
{
    int dp[M + 1]; // Array to store the minimum adjustment costs for each value

    for (int j = 0; j <= M; j++)
        dp[j] = abs(j - A[0]); // Initialize the first row with the absolute differences

    for (int i = 1; i < n; i++) // Iterate over the array elements
    {
        int prev_dp[M + 1];
        memcpy(prev_dp, dp, sizeof(dp)); // Store the previous row's minimum costs

        for (int j = 0; j <= M; j++) // Iterate over the possible values
        {
            dp[j] = INT_MAX; // Initialize the current value with maximum cost

            // Find the minimum cost by considering the range of previous values
            for (int k = max(j - target, 0); k <= min(M, j + target); k++)
                dp[j] = min(dp[j], prev_dp[k] + abs(A[i] - j));
        }
    }

    int res = INT_MAX;
    for (int j = 0; j <= M; j++)
        res = min(res, dp[j]); // Find the minimum cost in the last row

    return res; // Return the minimum adjustment cost
}

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

import java.util.Arrays;

public class MinimumAdjustmentCost {
    static final int M = 100;

    // Function to find the minimum adjustment cost of an array
    static int minAdjustmentCost(int[] A, int n, int target) {
        int[] dp = new int[M + 1];

        // Initialize the first row with absolute differences
        for (int j = 0; j <= M; j++) {
            dp[j] = Math.abs(j - A[0]);
        }

        // Iterate over the array elements
        for (int i = 1; i < n; i++) {
            int[] prev_dp = Arrays.copyOf(dp, dp.length); // Store the previous row's minimum costs

            // Iterate over the possible values
            for (int j = 0; j <= M; j++) {
                dp[j] = Integer.MAX_VALUE; // Initialize the current value with maximum cost

                // Find the minimum cost by considering the range of previous values
                for (int k = Math.max(j - target, 0); k <= Math.min(M, j + target); k++) {
                    dp[j] = Math.min(dp[j], prev_dp[k] + Math.abs(A[i] - j));
                }
            }
        }

        int res = Integer.MAX_VALUE;
        for (int j = 0; j <= M; j++) {
            res = Math.min(res, dp[j]); // Find the minimum cost in the last row
        }

        return res; // Return the minimum adjustment cost
    }

    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));
    }
}

Python3

def min_adjustment_cost(A, n, target):
    M = 100
    dp = [0] * (M + 1)

    # Initialize the first row of dp with absolute differences
    for j in range(M + 1):
        dp[j] = abs(j - A[0])

    # Iterate over the array elements
    for i in range(1, n):
        prev_dp = dp[:]  # Store the previous row's minimum costs
        for j in range(M + 1):
            dp[j] = float('inf')  # Initialize the current value with maximum cost

            # Find the minimum cost by considering the range of previous values
            for k in range(max(j - target, 0), min(M, j + target) + 1):
                dp[j] = min(dp[j], prev_dp[k] + abs(A[i] - j))

    res = float('inf')
    for j in range(M + 1):
        res = min(res, dp[j])  # Find the minimum cost in the last row

    return res

if __name__ == "__main__":
    arr = [55, 77, 52, 61, 39, 6, 25, 60, 49, 47]
    n = len(arr)
    target = 10

    print("Minimum adjustment cost is", min_adjustment_cost(arr, n, target))

C#

using System;

class Program
{
    const int M = 100;

    // Function to find minimum adjustment cost of an array
    static int MinAdjustmentCost(int[] A, int n, int target)
    {
        int[] dp = new int[M + 1]; // Array to store the minimum adjustment costs for each value

        for (int j = 0; j <= M; j++)
        {
            dp[j] = Math.Abs(j - A[0]); // Initialize the first row with the absolute differences
        }

        for (int i = 1; i < n; i++) // Iterate over the array elements
        {
            int[] prevDp = (int[])dp.Clone(); // Store the previous row's minimum costs

            for (int j = 0; j <= M; j++) // Iterate over the possible values
            {
                dp[j] = int.MaxValue; // Initialize the current value with maximum cost

                // Find the minimum cost by considering the range of previous values
                for (int k = Math.Max(j - target, 0); k <= Math.Min(M, j + target); k++)
                {
                    dp[j] = Math.Min(dp[j], prevDp[k] + Math.Abs(A[i] - j));
                }
            }
        }

        int res = int.MaxValue;
        for (int j = 0; j <= M; j++)
        {
            res = Math.Min(res, dp[j]); // Find the minimum cost in the last row
        }

        return res; // Return the minimum adjustment cost
    }

    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));
    }
}

Javascript

const M = 100;

// Function to find minimum adjustment cost of an array
function minAdjustmentCost(A, n, target) {
    let dp = new Array(M + 1); // Array to store the minimum adjustment costs for each value

    for (let j = 0; j <= M; j++)
        dp[j] = Math.abs(j - A[0]); // Initialize the first row with the absolute differences

    for (let i = 1; i < n; i++) // Iterate over the array elements
    {
        let prev_dp = [...dp]; // Store the previous row's minimum costs

        for (let j = 0; j <= M; j++) // Iterate over the possible values
        {
            dp[j] = Number.MAX_VALUE; // Initialize the current value with maximum cost

            // Find the minimum cost by considering the range of previous values
            for (let k = Math.max(j - target, 0); k <= Math.min(M, j + target); k++)
                dp[j] = Math.min(dp[j], prev_dp[k] + Math.abs(A[i] - j));
        }
    }

    let res = Number.MAX_VALUE;
    for (let j = 0; j <= M; j++)
        res = Math.min(res, dp[j]); // Find the minimum cost in the last row

    return res; // Return the minimum adjustment cost
}

let arr = [55, 77, 52, 61, 39, 6, 25, 60, 49, 47];
let n = arr.length;
let target = 10;

console.log("Minimum adjustment cost is " + minAdjustmentCost(arr, n, target));
// This code is contributed by Kanchan Agarwal

Output

Minimum adjustment cost is 75

Time Complexity: O(n*m²)
Auxiliary Space: O(m)

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