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

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: 6 ((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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the number of ways// to insert the bracket pairsint cntWays(string str, int n){    int x = n + 1;    int ways = x * x * (x * x - 1) / 12;    return ways;} // Driver codeint main(){    string str = "ab";    int n = str.length();     cout << cntWays(str, n);     return 0;}
// Java implementation of the approachimport java.util.*; class GFG{ // Function to return the number of ways// to insert the bracket pairsstatic int cntWays(String str, int n){    int x = n + 1;    int ways = x * x * (x * x - 1) / 12;    return ways;} // Driver codepublic 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 implementation of the approach # Function to return the number of ways# to insert the bracket pairsdef cntWays(string, n) :     x = n + 1;    ways = x * x * (x * x - 1) // 12;    return ways; # Driver codeif __name__ == "__main__" :     string = "ab";    n = len(string);     print(cntWays(string, n)); # This code is contributed by AnkitRai01
// C# implementation of the approachusing System;     class GFG{ // Function to return the number of ways// to insert the bracket pairsstatic int cntWays(String str, int n){    int x = n + 1;    int ways = x * x * (x * x - 1) / 12;    return ways;} // Driver codepublic static void Main(String []args){    String str = "ab";    int n = str.Length;     Console.WriteLine(cntWays(str, n));}} // This code is contributed by Rajput-Ji
<script> // Javascript implementation of the approach // Function to return the number of ways// to insert the bracket pairsfunction cntWays(str, n){    var x = n + 1;    var ways = x * x * (x * x - 1) / 12;    return ways;} // Driver codevar str = "ab";var n = str.length;document.write(cntWays(str, n)); // This code is contributed by rutvik_56.</script>
6
 
Time Complexity: O(1)
Auxiliary Space: O(1)
ankthon
Rajput-Ji
rutvik_56
subhammahato348
Parentheses-Problems
Mathematical
Strings
Strings
Mathematical
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