Find minimum adjustment cost of an array in C++

Concept

With respect of a given array of positive integers, we replace each element in the array so that the difference between adjacent elements in the array is either less than or equal to a given target. Now, our task to minimize the adjustment cost, that is the sum of differences between new and old values. So, we basically need to minimize Σ|A[i] – A_new[i]| where 0 ≤ i ≤ n-1, n is denoted as size of A[] and A_new[] is denoted as the array with adjacent difference less than or equal to target. Let all elements of the array is smaller than constant M = 100.

Input

arr = [56, 78, 53, 62, 40, 7, 26, 61, 50, 48], target = 20

Output

Minimum adjustment cost is 35

Method

For minimizing the adjustment cost Σ|A[i] – A_new[i]|, for all index i in the array, we remember that |A[i] – A_new[i]|should be as close to zero as possible. It should be noted that also,

|A[i] – A_new[i+1] ]| ≤ Target.

Here, this problem can be solved by dynamic programming(DP).

Assume dp1[i][j] represents minimal adjustment cost on changing A[i] to j, then the DP relation is defined by –

dp1[i][j] = min{dp1[i - 1][k]} + |j - A[i]|

for all k's such that |k - j| ≤ target

In this case, 0 ≤ i ≤ n and 0 ≤ j ≤ M where n is number of elements in the array and M = 100. So, all k values are considered in this way such that max(j – target, 0) ≤ k ≤ min(M, j + target) At last, the minimum adjustment cost of the array will be min{dp1[n – 1][j]} for all 0 ≤ j ≤ M.

Example

 Live Demo

// C++ program to find minimum adjustment cost of an array
#include <bits/stdc++.h>
using namespace std;
#define M1 100
//Shows function to find minimum adjustment cost of an array
int minAdjustmentCost(int A1[], int n1, int target1){
   // dp1[i][j] stores minimal adjustment cost on changing
   // A1[i] to j
   int dp1[n1][M1 + 1];
   // Tackle first element of array separately
   for (int j = 0; j <= M1; j++)
      dp1[0][j] = abs(j - A1[0]);
   // Perform for rest elements of the array
   for (int i = 1; i < n1; i++){
      // Now replace A1[i] to j and calculate minimal adjustment
      // cost dp1[i][j]
      for (int j = 0; j <= M1; j++){
         // We initialize minimal adjustment cost to INT_MAX
         dp1[i][j] = INT_MAX;
         // We consider all k such that k >= max(j - target1, 0)
         and
         // k <= min(M1, j + target1) and take minimum
      for (int k = max(j-target1,0); k <= min(M1,j+target1);
         k++)
         dp1[i][j] = min(dp1[i][j], dp1[i - 1][k] + abs(A1[i] -j));
      }
   }
   //Now return minimum value from last row of dp table
   int res1 = INT_MAX;
   for (int j = 0; j <= M1; j++)
      res1 = min(res1, dp1[n1 - 1][j]);
   return res1;
}
// Driver Program to test above functions
int main(){
   int arr1[] = {56, 78, 53, 62, 40, 7, 26, 61, 50, 48};
   int n1 = sizeof(arr1) / sizeof(arr1[0]);
   int target1 = 20;
   cout << "Minimum adjustment cost is "
   << minAdjustmentCost(arr1, n1, target1) << endl;
   return 0;
}

Output

Minimum adjustment cost is 35