Find an equal point in a string of brackets

Last Updated : 25 Mar, 2025

Given a string of brackets, the task is to find an index k which decides the number of opening brackets is equal to the number of closing brackets.
The string must be consists of only opening and closing brackets i.e. ‘(‘ and ‘)’.

An equal point is an index such that the number of opening brackets before it is equal to the number of closing brackets from and after.
If multiple such points exist then return the first valid index and if no such index exists then return -1.

Examples:

Input: str = “(())))(“
Output: 4
Explanation: After index 4, string splits into (()) and ))(. The number of opening brackets in the first part is equal to the number of closing brackets in the second part.

Input: str = “))”
Output: -1
Explanation: NO such index exists such that the count of opening brackets before that is equal to the count of closing brackets from and after that.

Untitled

Table of Content

[Naive Approach] Using Nested Loops – O(n^2) Time and O(1) Space
[Expected Approach] Using One variable – O(n) Time and O(1) Space

[Naive Approach] Using Nested Loops – O(n^2) Time and O(1) Space

We iterates through each possible index , counts the number of opening brackets ('(') before this index, and counts the number of closing brackets (')') after this index .
If the counts are equal, we returns the index .
If no such index is found, it returns -1.

C++

#include <bits/stdc++.h>
using namespace std;
int findEqlPoint(string s) {
    int n = s.size();
    for (int i = 0; i < n; i++) {
        int openCnt = 0, closeCnt = 0;
        
        //count opening brackets before i index and closing brackets after i index
        for (int j = 0; j < i; j++){
            if (s[j] == '(')
                openCnt++;
        }
        for (int j = i; j < n; j++){
            if (s[j] == ')') 
                closeCnt++;
        }
        if (openCnt == closeCnt) 
            return i;
    }
    return -1;
}

int main() {
    string s = "(())))(";
    cout << findEqlPoint(s) << endl;  
    return 0;
}

Java

public class Main {

    public static int findEqlPoint(String s) {
        int n = s.length();
        for (int i = 0; i < n; i++) {
            int openCnt = 0, closeCnt = 0;

            // Count opening brackets before i index and closing brackets after i index
            for (int j = 0; j < i; j++) {
                if (s.charAt(j) == '(')
                    openCnt++;
            }
            for (int j = i; j < n; j++) {
                if (s.charAt(j) == ')') 
                    closeCnt++;
            }
            if (openCnt == closeCnt) return i;
        }
        return -1;
    }

    public static void main(String[] args) {
        String s ="(())))(";
        System.out.println(findEqlPoint(s)); 
    }
}

Python

def findEqlPoint(s):
    n = len(s)
    for i in range(0, n):
        openCnt, closeCnt = 0, 0

        # Count opening brackets before i index and closing brackets after i index
        for j in range(i):
            if s[j] == '(':
                openCnt += 1
        for j in range(i, n):
            if s[j] == ')':
                closeCnt += 1
        if openCnt == closeCnt:
            return i
    return -1

s = "(())))("
print(findEqlPoint(s))

using System;

class Program {

    public static int FindEqlPoint(string s) {
        int n = s.Length;
        for (int i = 0; i < n; i++) {
            int openCnt = 0, closeCnt = 0;

            // Count opening brackets before i index and closing brackets after i index
            for (int j = 0; j < i; j++) {
                if (s[j] == '(')
                    openCnt++;
            }
            for (int j = i; j < n; j++) {
                if (s[j] == ')') 
                    closeCnt++;
            }
            if (openCnt == closeCnt) 
                return i;
        }
        return -1;
    }

    static void Main() {
        string s = "(())))(";
        Console.WriteLine(FindEqlPoint(s));  
    }
}

JavaScript

function findEqlPoint(s) {
    let n = s.length;
    for (let i = 0; i < n; i++) {
        let openCnt = 0, closeCnt = 0;

        // Count opening brackets before i index and closing brackets after i index
        for (let j = 0; j < i; j++) {
            if (s[j] === '(') 
                openCnt++;
        }
        for (let j = i; j < n; j++) {
            if (s[j] === ')') 
                closeCnt++;
        }
        if (openCnt === closeCnt) 
            return i;
    }
    return -1;
}

let s ="(())))(";
console.log(findEqlPoint(s));

Output

[Expected Approach] Using One variable – O(n) Time and O(1) Space

We first count the total number of closing brackets and then iterate over the string.
We update the counts of opening and closing brackets at every index as we traverse the string, and when they are equal, we return the index .
if no such index exists we return -1.

C++

#include <bits/stdc++.h>
using namespace std;

int findEqlPoint(string s) {
    int n = s.size(), openCnt = 0, closeCnt = 0;
    
    // pre calculate the no of closing brackets
    for (int i = 0; i < n; ++i){
        if (s[i] == ')')
             closeCnt++;
    }
    
    for (int i = 0; i < n; i++) {
        if (openCnt == closeCnt) 
            return i;
        
        // no of opening brackets before index i+1    
        if (s[i] == '(') 
            openCnt++;
        
         //no of closing brackets at and after index i+1  
        if (s[i] == ')')
            closeCnt--;
    }
    return -1;
}

int main() {
    string s = "))";
    cout << findEqlPoint(s) << endl; 
    return 0;
}

Java

public class Main {

    public static int findEqlPoint(String s) {
        int n = s.length(), openCnt = 0, closeCnt = 0;

        // Pre calculate the number of closing brackets
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == ')')
                closeCnt++;
        }
        
        for (int i = 0; i < n; i++) {
            if (openCnt == closeCnt) 
                return i;
            
            // Number of opening brackets before index i+1
            if (s.charAt(i) == '(')
                openCnt++;

            // Number of closing brackets at and after index i+1
            if (s.charAt(i) == ')')
                closeCnt--;

        }

        return -1;
    }

    public static void main(String[] args) {
        String s = "))";
        System.out.println(findEqlPoint(s)); 
    }
}

Python

def findEqlPoint(s):
    n = len(s)
    openCnt, closeCnt = 0, 0

    # Pre calculate the number of closing brackets
    for i in range(n):
        if s[i] == ')':
            closeCnt += 1
 
    for i in range(n):

        if openCnt == closeCnt:
            return i 
            
             # Number of opening brackets before index i+1
        if s[i] == '(':
            openCnt += 1

        # Number of closing brackets at and after index i+1
        if s[i] == ')':
            closeCnt -= 1

    return -1


s = "))"
print(findEqlPoint(s))

using System;

class Program {

    public static int FindEqlPoint(string s) {
        int n = s.Length, openCnt = 0, closeCnt = 0;

        // Pre calculate the number of closing brackets
        for (int i = 0; i < n; ++i) {
            if (s[i] == ')')
                closeCnt++;
        }
        
        for (int i = 0; i < n; i++) {
            if (openCnt == closeCnt) 
                return i;
            // Number of opening brackets before index i+1
            if (s[i] == '(')
                openCnt++;

            // Number of closing brackets at and after index i+1
            if (s[i] == ')')
                closeCnt--;
        }

        return -1;
    }

    static void Main() {
        string s = "))";
        Console.WriteLine(FindEqlPoint(s));  
    }
}

JavaScript

function findEqlPoint(s) {
    let n = s.length;
    let openCnt = 0, closeCnt = 0;

    // Pre calculate the number of closing brackets
    for (let i = 0; i < n; ++i) {
        if (s[i] === ')')
        closeCnt++;
    }
    
    for (let i = 0; i < n; i++) {
        if (openCnt === closeCnt) 
            return i;
        // Number of opening brackets before index i+1
        if (s[i] == '(')
            openCnt++;

        // Number of closing brackets at and after index i+1
        if (s[i] == ')')
            closeCnt--;
    }

    return -1;
}


let s = "))";
console.log(findEqlPoint(s));

Output

-1

Maximum number of 2x2 squares that can be fit inside a right isosceles triangle

Sahil Chhabra (akku) and Sankararaman K

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Find an equal point in a string of brackets

[Naive Approach] Using Nested Loops – O(n^2) Time and O(1) Space

[Expected Approach] Using One variable – O(n) Time and O(1) Space

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