Balanced expressions such that given positions have opening brackets

Given an integer n and an array of positions ‘position[]’ (1 <= position[i] <= 2n), find the number of ways of proper bracket expressions that can be formed of length 2n such that given positions have opening bracket. Examples :

Input : n = 3, position[] = [2}
Output : 3 
Explanation : 
The proper bracket sequences of length 6 and 
opening bracket at position 2 are:
[ [ ] ] [ ] 
[ [ [ ] ] ]
[ [ ] [ ] ]

Input : n = 2, position[] = {1, 3}
Output : 1
Explanation: The only possibility is:
[ ] [ ]



Approach : This problem can be solved by Dynamic programming..

Let DPi, j be the number of valid ways of filling the first i positions such that there are j more brackets of type ‘[‘ than of type ‘]’. Valid ways would mean that it is the prefix of a matched bracket expression and that the locations at which enforced ‘[‘ brackets are enforced, have been satisfied. It is easy to see that DP2N, 0 is the final answer.

The base case of the DP is, DP0, 0=1. We need to fill the first position with a ‘[‘ bracket, and there is only way to do this.

If the position has a opening bracket sequence which can be marked by a hash array, then the recurrence occurs as :

if(j != 0) dpi, j = dpi-1, j-1
else  dpi, j =  0; 

If the position has no opening bracket sequence, then recurrence happens as :

if(j != 0) dpi, j = dpi - 1, j - 1 + dpi - 1, j + 1
 else  dpi, j = dpi - 1, j + 1

The answer will be DP2n, 0

Given below is the CPP implementation of the above approach :

C++

// CPP code to find number of ways of
// arranging bracket with proper expressions
#include <bits/stdc++.h>
using namespace std;
  
#define N 1000
  
// function to calculate the number
// of proper bracket sequence
long long arrangeBraces(int n, int pos[], int k)
{
  
    // hash array to mark the
    // positions of opening brackets
    bool h[N];
  
    // dp 2d array
    int dp[N][N];
  
    memset(h, 0, sizeof h);
    memset(dp, 0, sizeof dp);
  
    // mark positions in hash array
    for (int i = 0; i < k; i++)
        h[pos[i]] = 1;
  
    // first position marked as 1
    dp[0][0] = 1;
  
    // iterate and formulate the recurrences
    for (int i = 1; i <= 2 * n; i++) {
        for (int j = 0; j <= 2 * n; j++) {
  
            // if position has a opening bracket
            if (h[i]) {
                if (j != 0)
                    dp[i][j] = dp[i - 1][j - 1];
                else
                    dp[i][j] = 0;
            }
            else {
                if (j != 0)
                    dp[i][j] = dp[i - 1][j - 1] +
                               dp[i - 1][j + 1];
                else
                    dp[i][j] = dp[i - 1][j + 1];
            }
        }
    }
  
    // return answer
    return dp[2 * n][0];
}
  
// driver code
int main()
{
    int n = 3;
  
    // positions where opening braces
    // will be placed
    int pos[] = { 2 };
    int k = sizeof(pos)/sizeof(pos[0]);
  
    cout << arrangeBraces(n, pos, k);
    return 0;
}


Output :

3

Time Complexity: O(n^2)
Auxiliary Space: O(n^2)



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