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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] – Anew[i]| where 0 ≤ i ≤ n-1, n is denoted as size of A[] and Anew[] 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] – Anew[i]|, for all index i in the array, we remember that |A[i] – Anew[i]|should be as close to zero as possible. It should be noted that also,
|A[i] – Anew[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
// 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
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