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Number of ways to insert two pairs of parentheses into a string of N characters

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  • Last Updated : 10 Mar, 2022

Given a string str of length N, the task is to find the number of ways to insert only 2 pairs of parentheses into the given string such that the resultant string is still valid.
Examples: 
 

Input: str = “ab” 
Output:
((a))b, ((a)b), ((ab)), (a)(b), (a(b)), a((b)) 
which are a total of 6 ways.
Input: str = “aab” 
Output: 20 
 

 

Approach: it can be observed that for the lengths of the string 1, 2, 3, …, N a series will be formed as 1, 6, 20, 50, 105, 196, 336, 540, … whose Nth term is (N + 1)2 * ((N + 1)2 – 1) / 12.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of ways
// to insert the bracket pairs
int cntWays(string str, int n)
{
    int x = n + 1;
    int ways = x * x * (x * x - 1) / 12;
    return ways;
}
 
// Driver code
int main()
{
    string str = "ab";
    int n = str.length();
 
    cout << cntWays(str, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the number of ways
// to insert the bracket pairs
static int cntWays(String str, int n)
{
    int x = n + 1;
    int ways = x * x * (x * x - 1) / 12;
    return ways;
}
 
// Driver code
public static void main(String []args)
{
    String str = "ab";
    int n = str.length();
 
    System.out.println(cntWays(str, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to return the number of ways
# to insert the bracket pairs
def cntWays(string, n) :
 
    x = n + 1;
    ways = x * x * (x * x - 1) // 12;
    return ways;
 
# Driver code
if __name__ == "__main__" :
 
    string = "ab";
    n = len(string);
 
    print(cntWays(string, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
class GFG
{
 
// Function to return the number of ways
// to insert the bracket pairs
static int cntWays(String str, int n)
{
    int x = n + 1;
    int ways = x * x * (x * x - 1) / 12;
    return ways;
}
 
// Driver code
public static void Main(String []args)
{
    String str = "ab";
    int n = str.Length;
 
    Console.WriteLine(cntWays(str, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the number of ways
// to insert the bracket pairs
function cntWays(str, n)
{
    var x = n + 1;
    var ways = x * x * (x * x - 1) / 12;
    return ways;
}
 
// Driver code
var str = "ab";
var n = str.length;
document.write(cntWays(str, n));
 
// This code is contributed by rutvik_56.
</script>

Output: 

6

 

Time Complexity: O(1)

Auxiliary Space: O(1)


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