Two equal sum segment range queries

• Difficulty Level : Medium
• Last Updated : 12 May, 2021

Given an array arr[] consisting of N positive integers, and some queries consisting of a range [L, R], the task is to find whether the sub-array from the given index range can be divided into two contiguous parts of non-zero length and equal sum.
Examples:

Input: arr[] = {1, 1, 2, 3}, q[] = {{0, 1}, {1, 3}, {1, 2}}
Output:
Yes
Yes
No
q[0]: The sub-array can be split into {1}, {1}.
q[1]: The sub-array can be split into {1, 2}, {3}.
q[2]: The sub-array can’t be split into two equal segments.
Input: arr[] = {2, 1, 3, 4, 1, 2}, q[] = {{0, 5}, {1, 3}}
Output:
No
Yes

A simple solution will be to iterate through the entire range and calculate the sum of the range. Then, we will iterate through the entire array again. We will sum up the elements starting from the index ‘L’. If at any step, we find that the current sum is half of the total sum of the range, the sub-array represented by the range can be broken into two equal halves. It can take upto O(n) time to answer a query using this approach.
A better solution is using a prefix-sum array. First, we create a prefix-sum array p_arr of arr. Now, using ‘p_arr’, we can determine the sum of all the elements in the range ‘L’ to ‘R’ in O(1) time. Once, we have our sum, we need to determine if there exists an index ‘i’ from L to R-1 such that the sum of all the numbers between L to i of the original array is half the sum of the range. For, that we can simply insert all the values in the prefix-sum array ‘p_arr’ in an unordered map.

If the value of sum is even and p_arr[l-1] + sum/2 exists in the map, the array can be split into two segments of equal sum.

Thus, the time complexity of answering a query becomes O(1).
Below is the implementation of the above approach:

C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to find the required prefix sum``void` `prefixSum(``int``* p_arr, ``int``* arr, ``int` `n)``{``    ``p_arr[0] = arr[0];``    ``for` `(``int` `i = 1; i < n; i++)``        ``p_arr[i] = arr[i] + p_arr[i - 1];``}` `// Function to hash all the values of prefix``// sum array in an unordered map``void` `hashPrefixSum(``int``* p_arr, ``int` `n, unordered_set<``int``>& q)``{``    ``for` `(``int` `i = 0; i < n; i++)``        ``q.insert(p_arr[i]);``}` `// Function to check if a range``// can be divided into two equal parts``void` `canDivide(``int``* p_arr, ``int` `n,``               ``unordered_set<``int``>& q, ``int` `l, ``int` `r)``{``    ``// To store the value of sum``    ``// of entire range``    ``int` `sum;` `    ``if` `(l == 0)``        ``sum = p_arr[r];``    ``else``        ``sum = p_arr[r] - p_arr[l - 1];` `    ``// If value of sum is odd``    ``if` `(sum % 2 == 1) {``        ``cout << ``"No"` `<< endl;``        ``return``;``    ``}` `    ``// To store p_arr[l-1]``    ``int` `beg = 0;` `    ``if` `(l != 0)``        ``beg = p_arr[l - 1];` `    ``// If the value exists in the map``    ``if` `(q.find(beg + sum / 2) != q.end())``        ``cout << ``"Yes"` `<< endl;``    ``else``        ``cout << ``"No"` `<< endl;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 1, 2, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// prefix-sum array``    ``int` `p_arr[n];` `    ``prefixSum(p_arr, arr, n);` `    ``// Map to store the values of prefix-sum``    ``unordered_set<``int``> q;` `    ``hashPrefixSum(p_arr, n, q);` `    ``// Perform queries``    ``canDivide(p_arr, n, q, 0, 1);``    ``canDivide(p_arr, n, q, 1, 3);``    ``canDivide(p_arr, n, q, 1, 2);` `    ``return` `0;``}`

Java

 `// Java implementation of the approach``import` `java.util.*;``class` `GFG {` `// Function to find the required prefix sum``static` `void` `prefixSum(``int``[] p_arr, ``int``[] arr, ``int` `n)``{``    ``p_arr[``0``] = arr[``0``];``    ``for` `(``int` `i = ``1``; i < n; i++)``        ``p_arr[i] = arr[i] + p_arr[i - ``1``];``}`` ` `// Function to q all the values of prefix``// sum array in an unordered map``static` `void` `qPrefixSum(``int``[]p_arr, ``int` `n, HashSetq)``{``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``q.add(p_arr[i]);``}`` ` `// Function to check if a range``// can be divided into two equal parts``static` `void` `canDivide(``int``[] p_arr, ``int` `n,``               ``HashSetq, ``int` `l, ``int` `r)``{``    ``// To store the value of sum``    ``// of entire range``    ``int` `sum;`` ` `    ``if` `(l == ``0``)``        ``sum = p_arr[r];``    ``else``        ``sum = p_arr[r] - p_arr[l - ``1``];`` ` `    ``// If value of sum is odd``    ``if` `(sum % ``2` `== ``1``) {``        ``System.out.println(``"No"``);``        ``return``;``    ``}`` ` `    ``// To store p_arr[l-1]``    ``int` `beg = ``0``;`` ` `    ``if` `(l != ``0``)``        ``beg = p_arr[l - ``1``];`` ` `    ``// If the value exists in the map``    ``if``(q.contains(beg + sum / ``2``) && (beg + sum / ``2``)!=(``int``)q.toArray()[ q.size()-``1` `] )``        ``System.out.println(``"Yes"``);``    ``else``        ``System.out.println(``"No"``);``}`` ` `// Driver code`` ``public` `static` `void` `main(String[] args) {``   ``int` `arr[] = { ``1``, ``1``, ``2``, ``3` `};``    ``int` `n = arr.length;`` ` `    ``// prefix-sum array``    ``int` `p_arr[] = ``new` `int``[n];`` ` `    ``prefixSum(p_arr, arr, n);`` ` `    ``// Map to store the values of prefix-sum``    ``HashSet q = ``new` `HashSet<>();`` ` `    ``qPrefixSum(p_arr, n, q);`` ` `    ``// Perform queries``    ``canDivide(p_arr, n, q, ``0``, ``1``);``    ``canDivide(p_arr, n, q, ``1``, ``3``);``    ``canDivide(p_arr, n, q, ``1``, ``2``);``    ``}``}` `// This code contributed by Rajput-Ji`

Python3

 `# Python3 implementation of the approach` `# Function to find the required prefix Sum``def` `prefixSum(p_arr, arr, n):` `    ``p_arr[``0``] ``=` `arr[``0``]``    ``for` `i ``in` `range``(``1``, n):``        ``p_arr[i] ``=` `arr[i] ``+` `p_arr[i ``-` `1``]` `# Function to hash all the values of``# prefix Sum array in an unordered map``def` `hashPrefixSum(p_arr, n, q):` `    ``for` `i ``in` `range``(n):``        ``q[p_arr[i]] ``=` `1` `# Function to check if a range``# can be divided into two equal parts``def` `canDivide(p_arr, n, q, l, r):``    ` `    ``# To store the value of Sum``    ``# of entire range``    ``Sum` `=` `0` `    ``if` `(l ``=``=` `0``):``        ``Sum` `=` `p_arr[r]``    ``else``:``        ``Sum` `=` `p_arr[r] ``-` `p_arr[l ``-` `1``]` `    ``# If value of Sum is odd``    ``if` `(``Sum` `%` `2` `=``=` `1``):``        ``print``(``"No"``)``        ``return``    ` `    ``# To store p_arr[l-1]``    ``beg ``=` `0` `    ``if` `(l !``=` `0``):``        ``beg ``=` `p_arr[l ``-` `1``]` `    ``# If the value exists in the map``    ``if` `(beg ``+` `Sum` `/``/` `2` `in` `q.keys()):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)` `# Driver code``arr ``=` `[``1``, ``1``, ``2``, ``3``]``n ``=` `len``(arr)` `# prefix-Sum array``p_arr ``=` `[``0` `for` `i ``in` `range``(n)]` `prefixSum(p_arr, arr, n)` `# Map to store the values``# of prefix-Sum``q ``=` `dict``()` `hashPrefixSum(p_arr, n, q)` `# Perform queries``canDivide(p_arr, n, q, ``0``, ``1``)``canDivide(p_arr, n, q, ``1``, ``3``)``canDivide(p_arr, n, q, ``1``, ``2``)` `# This code is contributed``# by mohit kumar`

C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to find the required prefix sum``static` `void` `prefixSum(``int``[] p_arr, ``int``[] arr, ``int` `n)``{``    ``p_arr[0] = arr[0];``    ``for` `(``int` `i = 1; i < n; i++)``        ``p_arr[i] = arr[i] + p_arr[i - 1];``}` `// Function to q all the values of prefix``// sum array in an unordered map``static` `void` `qPrefixSum(``int``[]p_arr, ``int` `n, HashSet<``int``>q)``{``    ``for` `(``int` `i = 0; i < n; i++)``        ``q.Add(p_arr[i]);``}` `// Function to check if a range``// can be divided into two equal parts``static` `void` `canDivide(``int``[] p_arr, ``int` `n,``            ``HashSet<``int``>q, ``int` `l, ``int` `r)``{``    ``// To store the value of sum``    ``// of entire range``    ``int` `sum;` `    ``if` `(l == 0)``        ``sum = p_arr[r];``    ``else``        ``sum = p_arr[r] - p_arr[l - 1];` `    ``// If value of sum is odd``    ``if` `(sum % 2 == 1)``    ``{``        ``Console.WriteLine(``"No"``);``        ``return``;``    ``}` `    ``// To store p_arr[l-1]``    ``int` `beg = 0;` `    ``if` `(l != 0)``        ``beg = p_arr[l - 1];` `    ``// If the value exists in the map``    ``if``(q.Contains(beg + sum / 2) )``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 1, 2, 3 };``    ``int` `n = arr.Length;` `    ``// prefix-sum array``    ``int` `[]p_arr = ``new` `int``[n];` `    ``prefixSum(p_arr, arr, n);` `    ``// Map to store the values of prefix-sum``    ``HashSet<``int``> q = ``new` `HashSet<``int``> ();` `    ``qPrefixSum(p_arr, n, q);` `    ``// Perform queries``    ``canDivide(p_arr, n, q, 0, 1);``    ``canDivide(p_arr, n, q, 1, 3);``    ``canDivide(p_arr, n, q, 1, 2);``}``}` `// This code has been contributed by 29AjayKumar`

Javascript

 ``

Output:

```Yes
Yes
No```

My Personal Notes arrow_drop_up