# Minimum sum by choosing minimum of pairs from array

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; ` `} ` |

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## Java

`// Java program to minimize the ` `// cost of array minimization ` `import` `java.util.Arrays; ` ` ` `public` `class` `GFG { ` ` ` `// Returns minimum possible ` `// sum in array B[] ` ` ` `static` `int` `minSum(` `int` `[] A, ` `int` `n) { ` ` ` `int` `min_val = Arrays.stream(A).min().getAsInt(); ` ` ` `return` `(min_val * (n - ` `1` `)); ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `static` `public` `void` `main(String[] args) { ` ` ` `int` `[] A = {` `3` `, ` `6` `, ` `2` `, ` `8` `, ` `7` `, ` `5` `}; ` ` ` `int` `n = A.length; ` ` ` `System.out.println((minSum(A, n))); ` ` ` ` ` `} ` `} ` `// This code is contributed by Rajput-Ji ` |

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## 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)) ` |

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## C#

`// C# program to minimize the ` `// cost of array minimization ` `using` `System; ` `using` `System.Linq; ` ` ` `public` `class` `GFG ` `{ ` ` ` `// Returns minimum possible ` `// sum in array B[] ` `static` `int` `minSum(` `int` `[]A, ` `int` `n) ` `{ ` ` ` `int` `min_val = A.Min(); ` ` ` `return` `(min_val * (n - 1)); ` `} ` ` ` ` ` `// Driver Code ` ` ` `static` `public` `void` `Main() ` ` ` `{ ` ` ` `int` `[]A = {3, 6, 2, 8, 7, 5}; ` ` ` `int` `n = A.Length; ` ` ` `Console.WriteLine(minSum(A, n)); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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## PHP

`<?php ` `// PHP program to minimize the ` `// cost of array minimization ` ` ` `// Returns minimum possible ` `// sum in array B[] ` `function` `minSum(` `$A` `, ` `$n` `) ` `{ ` ` ` `$min_val` `= min(` `$A` `); ` ` ` `return` `(` `$min_val` `* (` `$n` `- 1)); ` `} ` ` ` ` ` `// Driver Code ` ` ` `$A` `= ` `array` `(3, 6, 2, 8, 7, 5); ` ` ` `$n` `= ` `count` `(` `$A` `); ` ` ` `echo` `minSum(` `$A` `, ` `$n` `); ` ` ` `// This code is contributed by vt_m. ` `?> ` |

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**Output:**

10

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

This article is contributed by **Shivam Pradhan (anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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