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SQRBR - Square Brackets 0(n3) solution

Problem Link

#include <iostream>
using namespace std;

long long int dp[40][40];

int findPermutation(char s[], int i, int j)
{
if(s[i] == '?' &&  s[j] == ']')
return 1;

if(s[j] == '?' && s[i] == '[')
{
return 1;
}

if((s[i] == '[' && s[j] == ']'))
return 1;

if(s[i] == '?' && s[j] == '?')
return 1;

return 0;
}


int main()
{
int d, n, k, pos;
cin >> d;
char arr[40];
while(d)
{
cin >> n >> k;
for(int i = 0; i < 40; i++)
{
for(int j = 0; j < 40; j++)
{
dp[i][j] = 0;
if(i > j){
dp[i][j] = 1;
}
}
}

for(int i = 0; i < 2 * n; i++)
{
arr[i] = '?';
}

arr[2 * n] = '\0';

for(int i = 0; i < k; i++)
{
cin >> pos;
arr[pos - 1] = '[';
}

n = 2 * n;

for(int l = 2; l <= n; l++)
{
for(int i = 0; i <= n - l; i++)
{
int j = i + l - 1;

if(l == 2)
{
dp[i][j] = findPermutation(arr, i, j);
}
else if(l % 2 != 0)
{
dp[i][j] = 0;
}
else
{
for(int k = i + 1; k <= j; k++)
{
dp[i][j] += findPermutation(arr, i, k) * dp[i + 1][k - 1] * dp[k + 1][j];
}
}
}
}


cout << dp[0][n - 1] << endl;
d--;

}
return 0;
}

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