# Find the smallest contiguous sum pair in an Array

• Last Updated : 26 Nov, 2022

Given an array arr[] containing N distinct integers, the task is to find a contiguous pair such that the sum of both elements in the pair is minimum.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: (1, 2)
Explanation:
Here, contiguous pairs with their sum are (1, 2) = 3, (2, 3) = 5, (3, 4) = 7 and the minimum is 3.
Input: arr[] = {4, 9, -3, 2, 0}
Output: (-3, 2)
Explanation:
Here, contiguous pairs with their sum are (4, 9) = 13, (9, -3) = 6, (-3, 2) = -1, (2, 0) = 2

Approach:
To solve the problem mentioned above, we have to consider all the contiguous pairs and find their sum. The pair having the smallest(minimum) sum is the required answer.
Below is the implementation of the above approach:

## C++

 `//C++ program to find the smallest``// sum contiguous pair``#include``using` `namespace` `std;`` ` `// Function to find the smallest sum``// contiguous pair``vector<``int``> smallestSumpair(``int` `arr[], ``int` `n)``{` `    ``// Contiguous pair``    ``vector<``int``>pair;` `    ``// Initialize minimum sum``    ``// with maximum value``    ``int` `min_sum = INT_MAX;` `    ``for``(``int` `i = 1; i < n; i++)``    ``{` `        ``// Checking for minimum value``        ``if``( min_sum > (arr[i] + arr[i - 1]))``        ``{``            ``min_sum = arr[i] + arr[i - 1];``            ``if` `(pair.empty())``            ``{` `                ``// Add to pair``                ``pair.push_back(arr[i - 1]);``                ``pair.push_back(arr[i]);``            ``}``            ``else``            ``{` `                ``// Updating pair``                ``pair[0] = arr[i - 1];``                ``pair[1] = arr[i];``            ``}``        ``}``    ``}``    ``return` `pair;``}`` ` `// Driver code``int` `main()``{``   ``int` `arr[] = {4, 9, -3, 2, 0};``   ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``  ` `   ``vector<``int``>pair = smallestSumpair(arr, n);``   ``cout << pair[0] << ``" "` `<< pair[1];``}` `// This code is contributed by chitranayal`

## Java

 `// Java program to find the smallest``// sum contiguous pair``import` `java.util.*;` `class` `GFG{``    ` `// Function to find the smallest sum``// contiguous pair``public` `static` `Vector smallestSumpair(``int``[] arr,``                                              ``int` `n)``{``    ` `    ``// Stores the contiguous pair``    ``Vector pair = ``new` `Vector();``    ` `    ``// Initialize minimum sum``    ``int` `min_sum = Integer.MAX_VALUE, i;``    ` `    ``for``(i = ``1``; i < n; i++)``    ``{``        ` `        ``// Checking for minimum value``        ``if` `(min_sum > (arr[i] + arr[i - ``1``]))``        ``{``            ``min_sum = arr[i] + arr[i - ``1``];``            ` `            ``if` `(pair.isEmpty())``            ``{``                ` `                ``// Add to pair``                ``pair.add(arr[i - ``1``]);``                ``pair.add(arr[i]);``            ``}``            ``else``            ``{``                ` `                ``// Updating pair``                ``pair.set(``0``, arr[i - ``1``]);``                ``pair.set(``1``, arr[i]);``            ``}``        ``}``    ``}``    ``return` `pair;``}` `// Driver Code        ``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``4``, ``9``, -``3``, ``2``, ``0` `};``    ``int` `N = arr.length;``    ` `    ``Vector pair = ``new` `Vector();``    ``pair = smallestSumpair(arr, N);``    ` `    ``System.out.println(pair.get(``0``) + ``" "` `+``                       ``pair.get(``1``));``}``}` `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 program to find the smallest``# sum contiguous pair` `# importing sys``import` `sys` `# Function to find the smallest sum``# contiguous pair`  `def` `smallestSumpair(arr, n):` `    ``# Contiguous pair``    ``pair ``=` `[]` `    ``#  Initialize minimum sum``    ``# with maximum value``    ``min_sum ``=` `sys.maxsize` `    ``for` `i ``in` `range``(``1``, n):` `        ``# checking for minimum value``        ``if` `min_sum > (arr[i] ``+` `arr[i``-``1``]):``            ``min_sum ``=` `arr[i] ``+` `arr[i``-``1``]` `            ``if` `pair ``=``=` `[]:` `                ``# Add to pair``                ``pair.append(arr[i``-``1``])``                ``pair.append(arr[i])``            ``else``:` `                ``# Updating pair``                ``pair[``0``] ``=` `arr[i``-``1``]``                ``pair[``1``] ``=` `arr[i]` `    ``return` `pair`  `# Driver code``arr ``=` `[``4``, ``9``, ``-``3``, ``2``, ``0``]``n ``=` `len``(arr)``pair ``=` `smallestSumpair(arr, n)``print``(pair[``0``], pair[``1``])`

## C#

 `// C# program to find the smallest``// sum contiguous pair``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the smallest sum``// contiguous pair``public` `static` `ArrayList smallestSumpair(``int``[] arr,``                                        ``int` `n)``{``    ` `    ``// Stores the contiguous pair``    ``ArrayList pair = ``new` `ArrayList();``    ` `    ``// Initialize minimum sum``    ``int` `min_sum = ``int``.MaxValue, i;``    ` `    ``for``(i = 1; i < n; i++)``    ``{``        ` `        ``// Checking for minimum value``        ``if` `(min_sum > (arr[i] + arr[i - 1]))``        ``{``            ``min_sum = arr[i] + arr[i - 1];``            ` `            ``if` `(pair.Count == 0)``            ``{``                ` `                ``// Add to pair``                ``pair.Add(arr[i - 1]);``                ``pair.Add(arr[i]);``            ``}``            ``else``            ``{``                ` `                ``// Updating pair``                ``pair[0] = arr[i - 1];``                ``pair[1] = arr[i];``            ``}``        ``}``    ``}``    ``return` `pair;``}` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``    ``int` `[]arr = { 4, 9, -3, 2, 0 };``    ``int` `N = arr.Length;``    ` `    ``ArrayList pair = ``new` `ArrayList();``    ``pair = smallestSumpair(arr, N);``    ` `    ``Console.Write(pair[0] + ``" "` `+ pair[1]);``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output

`-3 2`

Time Complexity: O(n)

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