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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 <bits/stdc++.h>
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, HashSet<Integer>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<Integer>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) {
        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<Integer> 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




<script>
 
// Javascript implementation of the approach
 
// Function to find the required prefix sum
function prefixSum(p_arr, arr, n)
{
    p_arr[0] = arr[0];
    for (var 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
function hashPrefixSum(p_arr, n, q)
{
    for (var i = 0; i < n; i++)
        q.add(p_arr[i]);
}
 
// Function to check if a range
// can be divided into two equal parts
function canDivide(p_arr, n, q, l, r)
{
    // To store the value of sum
    // of entire range
    var 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) {
        document.write( "No" );
        return;
    }
 
    // To store p_arr[l-1]
    var beg = 0;
 
    if (l != 0)
        beg = p_arr[l - 1];
 
    // If the value exists in the map
    if (q.has((beg + sum / 2)))
        document.write( "Yes<br>" );
    else
        document.write( "No<br>" );
}
 
// Driver code
var arr = [1, 1, 2, 3 ];
var n = arr.length;
 
// prefix-sum array
var p_arr = Array(n);
prefixSum(p_arr, arr, n);
 
// Map to store the values of prefix-sum
var q = new Set();
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);
 
</script>
Output: 
Yes
Yes
No

 

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