Given an array A[] of n-elements. We need to select two adjacent elements and delete the larger of them and store smaller of them to another array say B[]. We need to perform this operation till array A[] contains only single element. Finally, we have to construct the array B[] in such a way that total sum of its element is minimum. Print the total sum of array B[].

Examples:

Input : A[] = {3, 4} Output : 3 Input : A[] = {2, 4, 1, 3} Output : 3

There is an easy trick to solve this question and that is always choose the smallest element of array A[] and its adjacent, delete the adjacent element and copy smallest one to array B[]. Again for next iteration we have same smallest element and any random adjacent element which is to be deleted. After n-1 operations all of elements of A[] got deleted except the smallest one and at the same time array B[] contains “n-1” elements and all are equal to smallest element of array A[].

Thus total sum of array B[] is equal to **smallest * (n-1)**.

## C++

// CPP program to minimize the cost // of array minimization #include <bits/stdc++.h> using namespace std; // Returns minimum possible sum in // array B[] int minSum(int A[], int n) { int min_val = *min_element(A, A+n); return (min_val * (n-1)); } // driver function int main() { int A[] = { 3, 6, 2, 8, 7, 5}; int n = sizeof(A)/ sizeof (A[0]); cout << minSum(A, n); return 0; }

## Python

# Python code for minimum cost of # array minimization # Function defintion for minCost def minSum(A): # find the minimum element of A[] min_val = min(A); # return the answer return min_val * (len(A)-1) # driver code A = [7, 2, 3, 4, 5, 6] print (minSum(A))

Output:

10

Time Complexity : O(n) in finding the smallest element of the array.

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