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

• Difficulty Level : Easy
• Last Updated : 28 Jul, 2022

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. ```

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 ` `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

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## Javascript

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Output

`4`

Time Complexity: O(N), as we are using a min function which will cost O(N).
Auxiliary Space: O(1), as we are not using any extra space.

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