# Maximum sum of smallest and second smallest in an array

#### Given an array, find maximum sum of smallest and second smallest elements chosen from all possible subarrays. More formally, if we write all (nC2) subarrays of array of size >=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them.

Examples:

Input : arr[] = [4, 3, 1, 5, 6] Output : 11 Subarrays with smallest and second smallest are, [4, 3] smallest = 3 second smallest = 4 [4, 3, 1] smallest = 1 second smallest = 3 [4, 3, 1, 5] smallest = 1 second smallest = 3 [4, 3, 1, 5, 6] smallest = 1 second smallest = 3 [3, 1] smallest = 1 second smallest = 3 [3, 1, 5] smallest = 1 second smallest = 3 [3, 1, 5, 6] smallest = 1 second smallest = 3 [1, 5] smallest = 1 second smallest = 5 [1, 5, 6] smallest = 1 second smallest = 5 [5, 6] smallest = 5 second smallest = 6 Maximum sum among all above choices is, 5 + 6 = 11 Input : arr[] = {5, 4, 3, 1, 6} Output : 9

A **simple solution** is to generate all subarrays, find sum of smallest and second smallest of every subarray. Finally return maximum of all sums.

An **efficient solution** is based on the observation that this problem reduces to finding a maximum sum of two consecutive elements in array.

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If (x,y) is the pair ,such that (x+y) is the answer , then x and y must be consecutive elements in the array.

#### Proof:

For a subarray with 2 elements , 1st and 2nd smallest elements are those 2 elements.

Now x and y are present in some subarray such thatthey are the endpoints.

Now, x, y must be the smallest 2 elements of that subarray. If there are other elements Z_{1 }, Z_{2}, ……., Z_{K} between x and y, they are greater than or equal to x and y,

#### Case1 :

If there is one element z between x and y , then the smaller subarray with the elements max(x,y) and z , should be the answer , because max(x,y) + z >= x + y

#### Case2:

If there are more than one elements between x and y , then the subarray within x and y will have all consecutive elements (Z_{i} + Z_{i+1}) >= (x+y), so (x,y) pair can’t be the answer.

So, by contradictions, x and y must be consecutive elements in the array.

## CPP

`// C++ program to get max sum with smallest` `// and second smallest element from any subarray` `#include <bits/stdc++.h>` `using` `namespace` `std;` `/* Method returns maximum obtainable sum value` ` ` `of smallest and the second smallest value` ` ` `taken over all possible subarrays */` `int` `pairWithMaxSum(` `int` `arr[], ` `int` `N)` `{` ` ` `if` `(N < 2)` ` ` `return` `-1;` ` ` `// Find two consecutive elements with maximum` ` ` `// sum.` ` ` `int` `res = arr[0] + arr[1];` ` ` `for` `(` `int` `i=1; i<N-1; i++)` ` ` `res = max(res, arr[i] + arr[i+1]);` ` ` `return` `res;` `}` `// Driver code to test above methods` `int` `main()` `{` ` ` `int` `arr[] = {4, 3, 1, 5, 6};` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `);` ` ` `cout << pairWithMaxSum(arr, N) << endl;` ` ` `return` `0;` `}` |

## JAVA

`// Java program to get max sum with smallest` `// and second smallest element from any subarray` `import` `java.lang.*;` `class` `num{` `// Method returns maximum obtainable sum value` `// of smallest and the second smallest value` `// taken over all possible subarrays */` `static` `int` `pairWithMaxSum(` `int` `[] arr, ` `int` `N)` `{` `if` `(N < ` `2` `)` ` ` `return` `-` `1` `;` `// Find two consecutive elements with maximum` `// sum.` `int` `res = arr[` `0` `] + arr[` `1` `];` `for` `(` `int` `i=` `1` `; i<N-` `1` `; i++)` ` ` `res = Math.max(res, arr[i] + arr[i+` `1` `]);` `return` `res;` `}` `// Driver program` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = {` `4` `, ` `3` `, ` `1` `, ` `5` `, ` `6` `};` ` ` `int` `N = arr.length;` ` ` `System.out.println(pairWithMaxSum(arr, N));` `}` `}` `//This code is contributed by` `//Smitha Dinesh Semwal` |

## Python3

`# Python 3 program to get max` `# sum with smallest and second` `# smallest element from any` `# subarray` `# Method returns maximum obtainable` `# sum value of smallest and the` `# second smallest value taken` `# over all possible subarrays` `def` `pairWithMaxSum(arr, N):` ` ` `if` `(N < ` `2` `):` ` ` `return` `-` `1` ` ` ` ` `# Find two consecutive elements with` ` ` `# maximum sum.` ` ` `res ` `=` `arr[` `0` `] ` `+` `arr[` `1` `]` ` ` ` ` `for` `i ` `in` `range` `(` `1` `, N` `-` `1` `):` ` ` `res ` `=` `max` `(res, arr[i] ` `+` `arr[i ` `+` `1` `])` ` ` ` ` `return` `res` ` ` `# Driver code` `arr ` `=` `[` `4` `, ` `3` `, ` `1` `, ` `5` `, ` `6` `]` `N ` `=` `len` `(arr)` `print` `(pairWithMaxSum(arr, N))` `# This code is contributed by Smitha Dinesh Semwal` |

## C#

`// C# program to get max sum with smallest` `// and second smallest element from any subarray` `using` `System;` `class` `GFG {` `// Method returns maximum obtainable sum value` `// of smallest and the second smallest value` `// taken over all possible subarrays` `static` `int` `pairWithMaxSum(` `int` `[]arr, ` `int` `N)` `{` ` ` `if` `(N < 2)` ` ` `return` `-1;` `// Find two consecutive elements` `// with maximum sum.` `int` `res = arr[0] + arr[1];` `for` `(` `int` `i = 1; i < N - 1; i++)` ` ` `res = Math.Max(res, arr[i] + arr[i + 1]);` `return` `res;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `[]arr = {4, 3, 1, 5, 6};` ` ` `int` `N = arr.Length;` ` ` `Console.Write(pairWithMaxSum(arr, N));` `}` `}` `// This code is contributed by Nitin Mittal.` |

## PHP

`<?php` `// PHP program to get max sum with smallest` `// and second smallest element from any subarray` `/* Method returns maximum` ` ` `obtainable sum value` ` ` `of smallest and the` ` ` `second smallest value` ` ` `taken over all possible` ` ` `subarrays */` `function` `pairWithMaxSum( ` `$arr` `, ` `$N` `)` `{` ` ` `if` `(` `$N` `< 2)` ` ` `return` `-1;` ` ` ` ` `// Find two consecutive` ` ` `// elements with maximum` ` ` `// sum.` ` ` `$res` `= ` `$arr` `[0] + ` `$arr` `[1];` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$N` `- 1; ` `$i` `++)` ` ` `$res` `= max(` `$res` `, ` `$arr` `[` `$i` `] +` ` ` `$arr` `[` `$i` `+ 1]);` ` ` ` ` `return` `$res` `;` `}` ` ` `// Driver Code` ` ` `$arr` `= ` `array` `(4, 3, 1, 5, 6);` ` ` `$N` `= ` `count` `(` `$arr` `);` ` ` `echo` `pairWithMaxSum(` `$arr` `, ` `$N` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`// javascript program to get max sum with smallest` `// and second smallest element from any subarray` ` ` `// Method returns maximum obtainable sum value` `// of smallest and the second smallest value` `// taken over all possible subarrays` `function` `pairWithMaxSum(arr, N)` `{` ` ` `if` `(N < 2)` ` ` `return` `-1;` ` ` `// Find two consecutive elements` `// with maximum sum.` `var` `res = arr[0] + arr[1];` `for` `(` `var` `i = 1; i < N - 1; i++)` ` ` `res = Math.max(res, arr[i] + arr[i + 1]);` ` ` `return` `res;` `}` ` ` `// Driver code` ` ` `var` `arr = [4, 3, 1, 5, 6]` ` ` `var` `N = arr.length;` ` ` `document.write(pairWithMaxSum(arr, N));` `// This code is contributed by bunnyram19.` |

Output:

11

Time Complexity : O(n)

Thanks to Md Mishfaq Ahmed for suggesting this approach.

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