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++
#include <bits/stdc++.h>
using namespace std;
int cntWays(string str, int n)
{
int x = n + 1;
int ways = x * x * (x * x - 1) / 12;
return ways;
}
int main()
{
string str = "ab" ;
int n = str.length();
cout << cntWays(str, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int cntWays(String str, int n)
{
int x = n + 1 ;
int ways = x * x * (x * x - 1 ) / 12 ;
return ways;
}
public static void main(String []args)
{
String str = "ab" ;
int n = str.length();
System.out.println(cntWays(str, n));
}
}
|
Python3
def cntWays(string, n) :
x = n + 1 ;
ways = x * x * (x * x - 1 ) / / 12 ;
return ways;
if __name__ = = "__main__" :
string = "ab" ;
n = len (string);
print (cntWays(string, n));
|
C#
using System;
class GFG
{
static int cntWays(String str, int n)
{
int x = n + 1;
int ways = x * x * (x * x - 1) / 12;
return ways;
}
public static void Main(String []args)
{
String str = "ab" ;
int n = str.Length;
Console.WriteLine(cntWays(str, n));
}
}
|
Javascript
<script>
function cntWays(str, n)
{
var x = n + 1;
var ways = x * x * (x * x - 1) / 12;
return ways;
}
var str = "ab" ;
var n = str.length;
document.write(cntWays(str, n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)