# Minimum value of “max + min” in a subarray

• Difficulty Level : Easy
• Last Updated : 13 May, 2021

Given a array of n positive elements we need to find the lowest possible sum of max and min elements in a subarray given that size of subarray should be greater than equal to 2.
Examples:

```Input : arr[] = {1 12 2 2}
Output : 4
Sum of 2+2 of subarray [2, 2]

Input  : arr[] = {10 20 30 40 23 45}
Output : 30
Sum of 10+20 of subarray[10, 20]```

A simple solution is to generate all subarrays, compute sum of maximum and minimum and finally return lowest sum.
An efficient solution is based on the fact that adding any element to a subarray would not increase sum of maximum and minimum. Consider the array [a1, a2, a3, a4, a5….an] Each ai will be minimum of some subarray [al, ar] such that i lies between [l, r] and all elements in the subarray are greater than or equal to ai. The cost of such subarray would be ai + max(subarray). Since the max of an array will never decrease on adding elements to the array, It will only increase if we add larger elements so It is always optimal to consider only those subarrays having length 2.
In short consider all subarrays of length 2 and compare sum and take the minimum one which will reduce the time complexity by O(N) now we have to run the loop only once.

## C++

 `// CPP program to find sum of maximum and``// minimum in any subarray of an array of``// positive numbers.``#include ``using` `namespace` `std;` `int` `maxSum(``int` `arr[], ``int` `n)``{``    ``if` `(n < 2)``        ``return` `-1;``    ``int` `ans = arr[0] + arr[1];``    ``for` `(``int` `i = 1; i + 1 < n; i++)``        ``ans = min(ans, (arr[i] + arr[i + 1]));``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = {1, 12, 2, 2};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << maxSum(arr, n);``    ``return` `0;``}`

## Java

 `// java program to find sum of maximum and``// minimum in any subarray of an array of``// positive numbers.``import` `java.io.*;` `class` `GFG {``    ` `    ``static` `int` `maxSum(``int` `arr[], ``int` `n)``    ``{``        ``if` `(n < ``2``)``            ``return` `-``1``;``        ``int` `ans = arr[``0``] + arr[``1``];``        ``for` `(``int` `i = ``1``; i + ``1` `< n; i++)``            ``ans = Math.min(ans, (arr[i]``                           ``+ arr[i + ``1``]));``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = {``1``, ``12``, ``2``, ``2``};``        ``int` `n = arr.length;``        ` `        ``System.out.println( maxSum(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## Python 3

 `# Python 3 program to find sum of maximum``# and minimum in any subarray of an array``# of positive numbers.` `def` `maxSum(arr, n):``    ``if` `(n < ``2``):``        ``return` `-``1``    ``ans ``=` `arr[``0``] ``+` `arr[``1``]``    ``for` `i ``in` `range``(``1``, n ``-` `1``, ``1``):``        ``ans ``=` `min``(ans, (arr[i] ``+` `arr[i ``+` `1``]))``    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``12``, ``2``, ``2``]``    ``n ``=` `len``(arr)``    ``print``(maxSum(arr, n))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to find sum of maximum and``// minimum in any subarray of an array of``// positive numbers.``using` `System ;` `class` `GFG {``    ` `    ``static` `int` `maxSum(``int` `[]arr, ``int` `n)``    ``{``        ``if` `(n < 2)``            ``return` `-1;``        ``int` `ans = arr[0] + arr[1];``        ``for` `(``int` `i = 1; i + 1 < n; i++)``            ``ans = Math.Min(ans, (arr[i]``                        ``+ arr[i + 1]));``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = {1, 12, 2, 2};``        ``int` `n = arr.Length;``        ` `        ``Console.WriteLine( maxSum(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

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## Javascript

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Output:
`4`

Time Complexity : O(n)
Auxiliary Space : O(1)

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