# 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]

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 C++ 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;

}
}

// This code is contributed by Pramod Kumar

Output:

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