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Queries to find minimum sum of array elements from either end of an array
• Difficulty Level : Easy
• Last Updated : 29 Apr, 2021

Given an array arr[] consisting of N distinct integers and an array queries[] consisting of Q queries, the task is for each query is to find queries[i] in the array and calculate the minimum of sum of array elements from the start and end of the array up to queries[i].

Examples:

Input: arr[] = {2, 3, 6, 7, 4, 5, 30}, Q = 2, queries[] = {6, 5}
Output: 11 27
Explanation:
Query 1: Sum from start = 2 + 3 + 6 = 11. Sum from end = 30 + 5 + 4 + 7 + 6 = 52. Therefore, 11 is the required answer.
Query 2: Sum from start = 27. Sum from end = 35. Therefore, 27 is the required answer.

Input: arr[] = {1, 2, -3, 4}, Q = 2, queries[] = {4, 2}
Output: 4 2

Naive Approach: The simplest approach is to traverse the array upro queries[i] for each query and calculate sum from the end as well as from the start of the array. Finally, print the minimum of the sums obtained.

Below is the implementation of the above approach:

## C++

 `// C++ implementation``// of the above approach` `#include ``using` `namespace` `std;` `// Function to calclate the minimum``// sum from either end of the arrays``// for the given queries``void` `calculateQuery(``int` `arr[], ``int` `N,``                    ``int` `query[], ``int` `M)``{``    ``// Traverse the query[] array``    ``for` `(``int` `i = 0; i < M; i++) {``        ``int` `X = query[i];` `        ``// Stores sum from start``        ``// and end of the array``        ``int` `sum_start = 0, sum_end = 0;` `        ``// Calculate distance from start``        ``for` `(``int` `j = 0; j < N; j++) {` `            ``sum_start += arr[j];``            ``if` `(arr[j] == X)``                ``break``;``        ``}` `        ``// Calculate distance from end``        ``for` `(``int` `j = N - 1; j >= 0; j--) {` `            ``sum_end += arr[j];``            ``if` `(arr[j] == X)``                ``break``;``        ``}` `        ``cout << min(sum_end, sum_start) << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 6, 7, 4, 5, 30 };``    ``int` `queries[] = { 6, 5 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `M = ``sizeof``(queries)``            ``/ ``sizeof``(queries);` `    ``calculateQuery(arr, N, queries, M);` `    ``return` `0;``}`

## Java

 `// Java implementation of the``// above approach``import` `java.util.*;``import` `java.lang.*;` `public` `class` `GFG``{` `  ``// Function to calclate the minimum``  ``// sum from either end of the arrays``  ``// for the given queries``  ``static` `void` `calculateQuery(``int` `arr[], ``int` `N,``                             ``int` `query[], ``int` `M)``  ``{` `    ``// Traverse the query[] array``    ``for` `(``int` `i = ``0``; i < M; i++)``    ``{``      ``int` `X = query[i];` `      ``// Stores sum from start``      ``// and end of the array``      ``int` `sum_start = ``0``, sum_end = ``0``;` `      ``// Calculate distance from start``      ``for` `(``int` `j = ``0``; j < N; j++)``      ``{``        ``sum_start += arr[j];``        ``if` `(arr[j] == X)``          ``break``;``      ``}` `      ``// Calculate distance from end``      ``for` `(``int` `j = N - ``1``; j >= ``0``; j--)``      ``{``        ``sum_end += arr[j];``        ``if` `(arr[j] == X)``          ``break``;``      ``}``      ``System.out.print(Math.min(sum_end, sum_start) + ``" "``);``    ``}``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main (String[] args)``  ``{``    ``int` `arr[] = { ``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``30` `};``    ``int` `queries[] = { ``6``, ``5` `};``    ``int` `N = arr.length;``    ``int` `M = queries.length;``    ``calculateQuery(arr, N, queries, M);``  ``}``}` `// This code is contributed by jana_sayantan.`

## Python3

 `# Python 3 implementation``# of the above approach` `# Function to calclate the minimum``# sum from either end of the arrays``# for the given queries`  `def` `calculateQuery(arr, N, query, M):` `    ``# Traverse the query[] array``    ``for` `i ``in` `range``(M):``        ``X ``=` `query[i]` `        ``# Stores sum from start``        ``# and end of the array``        ``sum_start ``=` `0``        ``sum_end ``=` `0` `        ``# Calculate distance from start``        ``for` `j ``in` `range``(N):` `            ``sum_start ``+``=` `arr[j]``            ``if` `(arr[j] ``=``=` `X):``                ``break` `        ``# Calculate distance from end``        ``for` `j ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``):` `            ``sum_end ``+``=` `arr[j]``            ``if` `(arr[j] ``=``=` `X):``                ``break``        ``print``(``min``(sum_end, sum_start), end``=``" "``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``30``]``    ``queries ``=` `[``6``, ``5``]``    ``N ``=` `len``(arr)``    ``M ``=` `len``(queries)` `    ``calculateQuery(arr, N, queries, M)` `    ``# This code is contributed by chitranayal.`

## C#

 `// C# implementation``// of the above approach``using` `System;``class` `GFG {` `  ``// Function to calclate the minimum``  ``// sum from either end of the arrays``  ``// for the given queries``  ``static` `void` `calculateQuery(``int``[] arr, ``int` `N,``                             ``int``[] query, ``int` `M)``  ``{``    ` `    ``// Traverse the query[] array``    ``for` `(``int` `i = 0; i < M; i++) {``      ``int` `X = query[i];` `      ``// Stores sum from start``      ``// and end of the array``      ``int` `sum_start = 0, sum_end = 0;` `      ``// Calculate distance from start``      ``for` `(``int` `j = 0; j < N; j++) {` `        ``sum_start += arr[j];``        ``if` `(arr[j] == X)``          ``break``;``      ``}` `      ``// Calculate distance from end``      ``for` `(``int` `j = N - 1; j >= 0; j--) {` `        ``sum_end += arr[j];``        ``if` `(arr[j] == X)``          ``break``;``      ``}` `      ``Console.Write(Math.Min(sum_end, sum_start) + ``" "``);``    ``}``  ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] arr = { 2, 3, 6, 7, 4, 5, 30 };``    ``int``[] queries = { 6, 5 };``    ``int` `N = arr.Length;``    ``int` `M = queries.Length;` `    ``calculateQuery(arr, N, queries, M);``  ``}``}` `// This code is contributed by divyesh072019.`

## Javascript

 ``
Output:
`11 27`

Time Complexity: O(Q * N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using extra space and the concept of prefix sum and suffix sum and to answer each query in constant computational complexity. The idea is to preprocess prefix and suffix sums for each index. Follow the steps below to solve the problem:

• Initialize two variables, say prefix and suffix.
• Initialize an Unordered Map, say mp, to map array elements as keys to pairs in which the first value gives the prefix sum and the second value gives the suffix sum.
• Traverse the array and keep adding arr[i] to prefix and store it in the map with arr[i] as key and prefix as value.
• Traverse the array in reverse and keep adding arr[i] to suffix and store in the map with arr[i] as key and suffix as value.
• Now traverse the array queries[] and for each query queries[i], print the value of the minimum of mp[queries[i]].first and mp[queries[i]].second as the result.

Below is the implementation of the above approach:

## C++

 `// C++ implementation``// of the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum sum``// for the given queries``void` `calculateQuery(``int` `arr[], ``int` `N,``                    ``int` `query[], ``int` `M)``{` `    ``// Stores prefix and suffix sums``    ``int` `prefix = 0, suffix = 0;` `    ``// Stores pairs of prefix and suffix sums``    ``unordered_map<``int``, pair<``int``, ``int``> > mp;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Add element to prefix``        ``prefix += arr[i];` `        ``// Store prefix for each element``        ``mp[arr[i]].first = prefix;``    ``}` `    ``// Traverse the array in reverse``    ``for` `(``int` `i = N - 1; i >= 0; i--) {` `        ``// Add element to suffix``        ``suffix += arr[i];` `        ``// Storing suffix for each element``        ``mp[arr[i]].second = suffix;``    ``}` `    ``// Travere the array queries[]``    ``for` `(``int` `i = 0; i < M; i++) {` `        ``int` `X = query[i];` `        ``// Minimum of suffix``        ``// and prefix sums``        ``cout << min(mp[X].first,``                    ``mp[X].second)``             ``<< ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 6, 7, 4, 5, 30 };``    ``int` `queries[] = { 6, 5 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `M = ``sizeof``(queries) / ``sizeof``(queries);` `    ``calculateQuery(arr, N, queries, M);` `    ``return` `0;``}`

## Java

 `// Java implementation``// of the above approach``import` `java.util.*;` `class` `GFG``{``  ``static` `class` `pair``  ``{``    ``E first;``    ``P second;``    ``public` `pair(E first, P second) ``    ``{``      ``this``.first = first;``      ``this``.second = second;``    ``}   ``  ``}` `  ``// Function to find the minimum sum``  ``// for the given queries``  ``@SuppressWarnings``({ ``"unchecked"``, ``"rawtypes"` `})``  ``static` `void` `calculateQuery(``int` `arr[], ``int` `N,``                             ``int` `query[], ``int` `M)``  ``{` `    ``// Stores prefix and suffix sums``    ``int` `prefix = ``0``, suffix = ``0``;` `    ``// Stores pairs of prefix and suffix sums``    ``HashMap mp = ``new` `HashMap<>();` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < N; i++) {` `      ``// Add element to prefix``      ``prefix += arr[i];` `      ``// Store prefix for each element``      ``mp.put(arr[i], ``new` `pair(prefix,``0``));``    ``}` `    ``// Traverse the array in reverse``    ``for` `(``int` `i = N - ``1``; i >= ``0``; i--) {` `      ``// Add element to suffix``      ``suffix += arr[i];` `      ``// Storing suffix for each element``      ``mp.put(arr[i], ``new` `pair(mp.get(arr[i]).first,suffix));``    ``}` `    ``// Travere the array queries[]``    ``for` `(``int` `i = ``0``; i < M; i++) {` `      ``int` `X = query[i];` `      ``// Minimum of suffix``      ``// and prefix sums``      ``System.out.print(Math.min((``int``)mp.get(X).first,``                                ``(``int``)mp.get(X).second)``                       ``+ ``" "``);``    ``}``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``30` `};``    ``int` `queries[] = { ``6``, ``5` `};``    ``int` `N = arr.length;``    ``int` `M = queries.length;` `    ``calculateQuery(arr, N, queries, M);` `  ``}``}` `// This code is contributed by 29AjayKumar`

## C#

 `// C# implementation``// of the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG {` `  ``// Function to find the minimum sum``  ``// for the given queries``  ``static` `void` `calculateQuery(``int``[] arr, ``int` `N,``                             ``int``[] query, ``int` `M)``  ``{` `    ``// Stores prefix and suffix sums``    ``int` `prefix = 0, suffix = 0;` `    ``// Stores pairs of prefix and suffix sums``    ``Dictionary<``int``, Tuple<``int``,``int``>> mp =``      ``new` `Dictionary<``int``, Tuple<``int``,``int``>>();` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++)``    ``{` `      ``// Add element to prefix``      ``prefix += arr[i];` `      ``// Store prefix for each element``      ``mp[arr[i]] = ``new` `Tuple<``int``,``int``>(prefix, 0);``    ``}` `    ``// Traverse the array in reverse``    ``for` `(``int` `i = N - 1; i >= 0; i--)``    ``{` `      ``// Add element to suffix``      ``suffix += arr[i];` `      ``// Storing suffix for each element``      ``mp[arr[i]] = ``new` `Tuple<``int``,``int``>(mp[arr[i]].Item1, suffix);``    ``}` `    ``// Travere the array queries[]``    ``for` `(``int` `i = 0; i < M; i++)``    ``{``      ``int` `X = query[i];` `      ``// Minimum of suffix``      ``// and prefix sums``      ``Console.Write(Math.Min(mp[X].Item1, mp[X].Item2) + ``" "``);``    ``}``  ``}` `  ``// Driver code``  ``static` `void` `Main() {``    ``int``[] arr = { 2, 3, 6, 7, 4, 5, 30 };``    ``int``[] queries = { 6, 5 };``    ``int` `N = arr.Length;``    ``int` `M = queries.Length;` `    ``calculateQuery(arr, N, queries, M);``  ``}``}` `// This code is contributed by divyeshrabadiya07.`
Output:
`11 27`

Time Complexity: O(M + N)
Auxiliary Space: O(N)

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