# Minimum cost to make array size 1 by removing larger of pairs

Given an array of n integers. We need to reduce size of array to one. We are allowed to select a pair of integers and remove the larger one of these two. This decreases the array size by 1. Cost of this operation is equal to value of smaller one. Find out minimum sum of costs of operations needed to convert the array into a single element.**Examples:**

Input: 4 3 2 Output: 4 Explanation: Choose (4, 2) so 4 is removed, new array = {2, 3}. Now choose (2, 3) so 3 is removed. So total cost = 2 + 2 = 4 Input: 3 4 Output: 3 Explanation: choose 3, 4, so cost is 3.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.

The idea is to always pick minimum value as part of the pair and remove larger value. This minimizes cost of reducing array to size 1.

Below is the implementation of the above approach:

## CPP

`// CPP program to find minimum cost to` `// reduce array size to 1,` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to calculate the minimum cost` `int` `cost(` `int` `a[], ` `int` `n)` `{` ` ` `// Minimum cost is n-1 multiplied with` ` ` `// minimum element.` ` ` `return` `(n - 1) * (*min_element(a, a + n));` `}` `// driver program to test the above function.` `int` `main()` `{` ` ` `int` `a[] = { 4, 3, 2 };` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]);` ` ` `cout << cost(a, n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to find minimum cost` `// to reduce array size to 1,` `import` `java.lang.*;` `public` `class` `GFG {` ` ` ` ` `// function to calculate the` ` ` `// minimum cost` ` ` `static` `int` `cost(` `int` `[]a, ` `int` `n)` ` ` `{` ` ` `int` `min = a[` `0` `];` ` ` ` ` `// find the minimum using` ` ` `// for loop` ` ` `for` `(` `int` `i = ` `1` `; i< a.length; i++)` ` ` `{` ` ` `if` `(a[i] < min)` ` ` `min = a[i];` ` ` `}` ` ` ` ` `// Minimum cost is n-1 multiplied` ` ` `// with minimum element.` ` ` `return` `(n - ` `1` `) * min;` ` ` `}` ` ` ` ` `// driver program to test the` ` ` `// above function.` ` ` `static` `public` `void` `main (String[] args)` ` ` `{` ` ` ` ` `int` `[]a = { ` `4` `, ` `3` `, ` `2` `};` ` ` `int` `n = a.length;` ` ` ` ` `System.out.println(cost(a, n));` ` ` `}` `}` `// This code is contributed by parashar.` |

## Python3

`# Python program to find minimum` `# cost to reduce array size to 1` `# function to calculate the` `# minimum cost` `def` `cost(a, n):` ` ` `# Minimum cost is n-1 multiplied` ` ` `# with minimum element.` ` ` `return` `( (n ` `-` `1` `) ` `*` `min` `(a) )` `# driver code` `a ` `=` `[ ` `4` `, ` `3` `, ` `2` `]` `n ` `=` `len` `(a)` `print` `(cost(a, n))` `# This code is contributed by` `# Smitha Dinesh Semwal` |

## C#

`// C# program to find minimum cost to` `// reduce array size to 1,` `using` `System;` `using` `System.Linq;` `public` `class` `GFG {` ` ` ` ` `// function to calculate the minimum cost` ` ` `static` `int` `cost(` `int` `[]a, ` `int` `n)` ` ` `{` ` ` ` ` `// Minimum cost is n-1 multiplied with` ` ` `// minimum element.` ` ` `return` `(n - 1) * a.Min();` ` ` `}` ` ` ` ` `// driver program to test the above function.` ` ` `static` `public` `void` `Main (){` ` ` ` ` `int` `[]a = { 4, 3, 2 };` ` ` `int` `n = a.Length;` ` ` ` ` `Console.WriteLine(cost(a, n));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find minimum cost to` `// reduce array size to 1,` `// function to calculate` `// the minimum cost` `function` `cost(` `$a` `, ` `$n` `)` `{` ` ` ` ` `// Minimum cost is n-1` ` ` `// multiplied with` ` ` `// minimum element.` ` ` `return` `(` `$n` `- 1) * (min(` `$a` `));` `}` ` ` `// Driver Code` ` ` `$a` `= ` `array` `(4, 3, 2);` ` ` `$n` `= ` `count` `(` `$a` `);` ` ` `echo` `cost(` `$a` `, ` `$n` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript program to find minimum cost` `// to reduce array size to 1,` ` ` `// function to calculate the` ` ` `// minimum cost` ` ` `function` `cost(a, n)` ` ` `{` ` ` `let min = a[0];` ` ` ` ` `// find the minimum using` ` ` `// for loop` ` ` `for` `(let i = 1; i< a.length; i++)` ` ` `{` ` ` `if` `(a[i] < min)` ` ` `min = a[i];` ` ` `}` ` ` ` ` `// Minimum cost is n-1 multiplied` ` ` `// with minimum element.` ` ` `return` `(n - 1) * min;` ` ` `}` `// Driver code ` ` ` ` ` `let a = [ 4, 3, 2 ];` ` ` `let n = a.length;` ` ` ` ` `document.write(cost(a, n));` ` ` `</script>` |

**Output: **

4

**Time Complexity** : O(n)

This article is contributed by **Striver**. 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.