Merge pull request #598 from AbhiC7721/devincept

Added SQRBR problem

Author: Saviour1001

Dynamic Programming/Square_brackets.cpp

@@ -0,0 +1,80 @@
+/*Problem statement => https://www.spoj.com/problems/SQRBR/  
+  You are given:
+
+  -> a positive integer n,
+  -> an integer k, 1<=k<=n,
+  -> an increasing sequence of k integers 0 < s1 < s2 < ... < sk <= 2n.
+What is the number of proper bracket expressions of length 2n with opening brackets appearing in positions s1, s2,...,sk?
+*/
+
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+ 
+typedef string                 STR;
+typedef long long              LL;
+typedef vector<LL>             VLL;
+
+#define all(a)             (a).begin(),(a).end()
+#define dec(n)             cout<<fixed<<setprecision(n);
+#define f(i,a,b)           for (i = a; i < b; i++)
+#define fr(i,a,b)          for (i = a; i > b; i--)
+#define rep(i,n)           for (i = 0 ; i < n; i++)
+#define repr(i,n)          for (i = n - 1; i >= 0; i--)
+#define fsort(a)            sort(a.begin(),a.end())
+#define rsort(a)           sort(a.rbegin(),a.rend())      
+ 
+const LL MOD     = 1000000007;
+ 
+LL sq_brackets(LL n, LL k, VLL pos)
+{
+  vector<bool>h(2*n+1, 0);
+  LL i;
+  rep(i, k) h[pos[i]]=1; // hash array ti store the postions where opening bracket is must
+  // dp array
+  vector<vector<LL>>dp(2*n+1, vector<LL>(2*n+1, 0));
+  // first position only one possible combination
+  dp[0][0] = 1%MOD;
+  // dp[i][j] represents i positions such that there are j more '[' brackets than ']' 
+  for(LL i=1;i<=2*n;i++)
+  {
+    for(LL j=0;j<=2*n;j++)
+    {
+      //if that position has an opening bracket
+      if(h[i])
+      {
+        // if diff(j) is not 0 then ith position has '[' hence dp[i][j]=dp[i-1][j-1]
+        if(j!=0)
+          dp[i][j] = dp[i-1][j-1]%MOD;
+        else
+          dp[i][j]=0; // no possibility if j==0
+      }
+      else
+      {
+        // both opening and closing brackets possible if h[i]!=1
+        if(j!=0)
+          dp[i][j] = (dp[i-1][j-1]%MOD + dp[i-1][j+1]%MOD)%MOD;
+        else
+          dp[i][j] = dp[i-1][j+1]%MOD;
+      }
+    } 
+  }
+  return dp[2*n][0];
+}
+int main()
+{  
+  int t=1;
+  cin>>t;
+  while(t--)
+  {
+    LL n,k,i;
+    cin>>n>>k;
+    VLL pos(k);
+    rep(i,k)
+      cin>>pos[i];
+    LL ans = sq_brackets(n, k, pos);
+    cout<<ans<<"\n";
+  }
+   return 0;
+ }