Count ways to split array into two equal sum subarrays by changing sign of any one array element
Last Updated :
23 Jul, 2025
Given an array arr[] consisting of N integers, the task is to count ways to split array into two subarrays of equal sum by changing the sign of any one array element.
Examples:
Input: arr[] = {2, 2, -3, 3}
Output: 2
Explanation:
Changing arr[0] = 2 to arr[0] = -2, the array becomes {-2, 2, -3, 3}. Only 1 possible split is {-2, 2} and {-3, 3}.
Changing arr[1] = 2 to arr[1] = -2, the array becomes {2, -2, -3, 3}. Only 1 possible split is {-2, 2} and {-3, 3}.
Changing arr[2] = -3 to arr[2] = 3, the array becomes {2, 2, 3, 3}. No way to split the array.
Changing arr[3] = 3 to arr[2] = -3, the array becomes {2, 2, -3, -3}. No way to split the array.
Therefore, the total number of ways to split = 1 + 1 + 0 + 0 = 2.
Input: arr[] = {2, 2, 1, -3, 3}
Output: 0
Naive Approach: The simplest approach to solve the problem is to traverse the array and change the sign of every array element one by one and count the number of ways to split array into two equal sum subarrays for every alteration. Finally, print the sum of all possible ways.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to store the prefix sum and suffix sum for every array indices to find the sum of the splitter subarrays in O(1) computational complexity. Follow the steps below to solve the problem:
- Initialize a variable, say count, to store the number of ways to split the array.
- Initialize two variables, say prefixSum and suffixSum, with 0, to store the prefix and suffix sums of both the arrays.
- Initialize two Maps prefixCount and suffixCount to store the count of elements in prefix and suffix arrays.
- Traverse the array arr[] and update the frequency of each element in suffixCount.
- Traverse the array arr[] and perform the following steps:
- Insert arr[i] into the prefixCount map and remove it from suffixCount.
- Add arr[i] to prefixSum and set suffixSum to the difference of the total sum of the array and prefixSum.
- Store the difference between the sum of the subarrays, i.e. prefixSum – suffixSum in a variable, say diff.
- The count of ways to split at the ith index is calculated by:
- If diff is odd, then the array cannot be split.
- If diff is even, then add the value (prefixCount[diff 1="2" language="/"][/diff] + suffixCount[ -diff / 2]) to count.
- After completing the above steps, the value of count gives the total count of possible splits.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count ways of splitting
// the array in two subarrays of equal
// sum by changing sign of any 1 element
int countSubArraySignChange(int arr[], int N)
{
// Stores the count of elements
// in prefix and suffix of array
unordered_map<int, int> prefixCount;
unordered_map<int, int> suffixCount;
// Stores the total sum of array
int total = 0;
// Traverse the array
for (int i = N - 1; i >= 0; i--) {
total += arr[i];
// Increase the frequency of
// current element in suffix
suffixCount[arr[i]]++;
}
// Stores prefix sum upto
// an index
int prefixSum = 0;
// Stores sum of suffix
// from an index
int suffixSum = 0;
// Stores the count of ways to
// split the array in 2 subarrays
// having equal sum
int count = 0;
// Traverse the array
for (int i = 0; i < N - 1; i++) {
// Modify prefix sum
prefixSum += arr[i];
// Add arr[i] to prefix Map
prefixCount[arr[i]]++;
// Calculate suffix sum by
// subtracting prefix sum
// from total sum of elements
suffixSum = total - prefixSum;
// Remove arr[i] from suffix Map
suffixCount[arr[i]]--;
// Store the difference
// between the subarrays
int diff = prefixSum - suffixSum;
// Check if diff is even or not
if (diff % 2 == 0) {
// Count number of ways to
// split array at index i such
// that subarray sums are same
int x = prefixCount[diff / 2]
+ suffixCount[-diff / 2];
// Update the count
count = count + x;
}
}
// Return the count
return count;
}
// Driver Code
int main()
{
int arr[] = { 2, 2, -3, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << countSubArraySignChange(arr, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count ways of splitting
// the array in two subarrays of equal
// sum by changing sign of any 1 element
static int countSubArraySignChange(int arr[], int N)
{
// Stores the count of elements
// in prefix and suffix of array
HashMap<Integer,Integer> prefixCount = new HashMap<Integer,Integer>();
HashMap<Integer,Integer> suffixCount = new HashMap<Integer,Integer>();
// Stores the total sum of array
int total = 0;
// Traverse the array
for (int i = N - 1; i >= 0; i--)
{
total += arr[i];
// Increase the frequency of
// current element in suffix
if(suffixCount.containsKey(arr[i])){
suffixCount.put(arr[i], suffixCount.get(arr[i]) + 1);
}
else{
suffixCount.put(arr[i], 1);
}
}
// Stores prefix sum upto
// an index
int prefixSum = 0;
// Stores sum of suffix
// from an index
int suffixSum = 0;
// Stores the count of ways to
// split the array in 2 subarrays
// having equal sum
int count = 0;
// Traverse the array
for (int i = 0; i < N - 1; i++)
{
// Modify prefix sum
prefixSum += arr[i];
// Add arr[i] to prefix Map
if(prefixCount.containsKey(arr[i]))
{
prefixCount.put(arr[i], prefixCount.get(arr[i])+1);
}
else
{
prefixCount.put(arr[i], 1);
}
// Calculate suffix sum by
// subtracting prefix sum
// from total sum of elements
suffixSum = total - prefixSum;
// Remove arr[i] from suffix Map
if(suffixCount.containsKey(arr[i]))
{
suffixCount.put(arr[i], suffixCount.get(arr[i]) - 1);
}
// Store the difference
// between the subarrays
int diff = prefixSum - suffixSum;
// Check if diff is even or not
if (diff % 2 == 0)
{
// Count number of ways to
// split array at index i such
// that subarray sums are same
int x = (prefixCount.containsKey(diff / 2)?prefixCount.get(diff / 2):0)
+ (suffixCount.containsKey(-diff / 2)?suffixCount.get(-diff / 2):0);
// Update the count
count = count + x;
}
}
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 2, -3, 3 };
int N = arr.length;
// Function Call
System.out.print(countSubArraySignChange(arr, N));
}
}
// This code is contributed by 29AjayKumar
# Python3 program for the above approach
# Function to count ways of splitting
# the array in two subarrays of equal
# sum by changing sign of any 1 element
def countSubArraySignChange(arr, N):
# Stores the count of elements
# in prefix and suffix of array
prefixCount = {}
suffixCount = {}
# Stores the total sum of array
total = 0
# Traverse the array
for i in range(N - 1, -1, -1):
total += arr[i]
# Increase the frequency of
# current element in suffix
suffixCount[arr[i]] = suffixCount.get(arr[i], 0) + 1
# Stores prefix sum upto
# an index
prefixSum = 0
# Stores sum of suffix
# from an index
suffixSum = 0
# Stores the count of ways to
# split the array in 2 subarrays
# having equal sum
count = 0
# Traverse the array
for i in range(N - 1):
# Modify prefix sum
prefixSum += arr[i]
# Add arr[i] to prefix Map
prefixCount[arr[i]] = prefixCount.get(arr[i], 0) + 1
# Calculate suffix sum by
# subtracting prefix sum
# from total sum of elements
suffixSum = total - prefixSum
# Remove arr[i] from suffix Map
suffixCount[arr[i]] -= 1
# Store the difference
# between the subarrays
diff = prefixSum - suffixSum
# Check if diff is even or not
if (diff % 2 == 0):
# Count number of ways to
# split array at index i such
# that subarray sums are same
y, z = 0, 0
if -diff//2 in suffixCount:
y = suffixCount[-dff//2]
if diff//2 in prefixCount:
z = prefixCount[diff/2]
x = z+ y
# Update the count
count = count + x
# Return the count
return count
# Driver Code
if __name__ == '__main__':
arr=[2, 2, -3, 3]
N = len(arr)
# Function Call
print(countSubArraySignChange(arr, N))
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count ways of splitting
// the array in two subarrays of equal
// sum by changing sign of any 1 element
static int countSubArraySignChange(int []arr, int N)
{
// Stores the count of elements
// in prefix and suffix of array
Dictionary<int,int> prefixCount = new Dictionary<int,int>();
Dictionary<int,int> suffixCount = new Dictionary<int,int>();
// Stores the total sum of array
int total = 0;
// Traverse the array
for (int i = N - 1; i >= 0; i--)
{
total += arr[i];
// Increase the frequency of
// current element in suffix
if(suffixCount.ContainsKey(arr[i])){
suffixCount[arr[i]] = suffixCount[arr[i]] + 1;
}
else{
suffixCount.Add(arr[i], 1);
}
}
// Stores prefix sum upto
// an index
int prefixSum = 0;
// Stores sum of suffix
// from an index
int suffixSum = 0;
// Stores the count of ways to
// split the array in 2 subarrays
// having equal sum
int count = 0;
// Traverse the array
for (int i = 0; i < N - 1; i++)
{
// Modify prefix sum
prefixSum += arr[i];
// Add arr[i] to prefix Map
if(prefixCount.ContainsKey(arr[i]))
{
prefixCount[arr[i]] = prefixCount[arr[i]] + 1;
}
else
{
prefixCount.Add(arr[i], 1);
}
// Calculate suffix sum by
// subtracting prefix sum
// from total sum of elements
suffixSum = total - prefixSum;
// Remove arr[i] from suffix Map
if(suffixCount.ContainsKey(arr[i]))
{
suffixCount[arr[i]] = suffixCount[arr[i]] - 1;
}
// Store the difference
// between the subarrays
int diff = prefixSum - suffixSum;
// Check if diff is even or not
if (diff % 2 == 0)
{
// Count number of ways to
// split array at index i such
// that subarray sums are same
int x = (prefixCount.ContainsKey(diff / 2)?prefixCount[diff / 2]:0)
+ (suffixCount.ContainsKey(-diff / 2)?suffixCount[-diff / 2]:0);
// Update the count
count = count + x;
}
}
// Return the count
return count;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 2, -3, 3 };
int N = arr.Length;
// Function Call
Console.Write(countSubArraySignChange(arr, N));
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript program for the above approach
// Function to count ways of splitting
// the array in two subarrays of equal
// sum by changing sign of any 1 element
function countSubArraySignChange(arr,N)
{
// Stores the count of elements
// in prefix and suffix of array
let prefixCount = new Map();
let suffixCount = new Map();
// Stores the total sum of array
let total = 0;
// Traverse the array
for (let i = N - 1; i >= 0; i--)
{
total += arr[i];
// Increase the frequency of
// current element in suffix
if(suffixCount.has(arr[i])){
suffixCount.set(arr[i], suffixCount.get(arr[i]) + 1);
}
else{
suffixCount.set(arr[i], 1);
}
}
// Stores prefix sum upto
// an index
let prefixSum = 0;
// Stores sum of suffix
// from an index
let suffixSum = 0;
// Stores the count of ways to
// split the array in 2 subarrays
// having equal sum
let count = 0;
// Traverse the array
for (let i = 0; i < N - 1; i++)
{
// Modify prefix sum
prefixSum += arr[i];
// Add arr[i] to prefix Map
if(prefixCount.has(arr[i]))
{
prefixCount.set(arr[i], prefixCount.get(arr[i])+1);
}
else
{
prefixCount.set(arr[i], 1);
}
// Calculate suffix sum by
// subtracting prefix sum
// from total sum of elements
suffixSum = total - prefixSum;
// Remove arr[i] from suffix Map
if(suffixCount.has(arr[i]))
{
suffixCount.set(arr[i], suffixCount.get(arr[i]) - 1);
}
// Store the difference
// between the subarrays
let diff = prefixSum - suffixSum;
// Check if diff is even or not
if (diff % 2 == 0)
{
// Count number of ways to
// split array at index i such
// that subarray sums are same
let x = (prefixCount.has(diff / 2)?
prefixCount.get(diff / 2):0)
+ (suffixCount.has(-diff / 2)?
suffixCount.get(-diff / 2):0);
// Update the count
count = count + x;
}
}
// Return the count
return count;
}
// Driver Code
let arr=[2, 2, -3, 3];
let N = arr.length;
// Function Call
document.write(countSubArraySignChange(arr, N));
// This code is contributed by patel2127
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem