Distinct Subsequences

DP/distinct_subsequences.cpp

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+/*
+PROBLEM STATEMENT:-
+Given two strings let's say s and t
+return the number of distinct subsequences of s which equals t
+for example:
+s= babgbag ,t= bag
+we can extract t from s by following ways
+
+index: 0 1 2 3 4 5 6
+string:b a b g b a g
+postions (0,1,3) (0,1,6) (0,5,6) (2,5,6) (4,5,6)
+hence answer=5
+*/
+#include<bits/stdc++.h>
+using namespace std;
+int distinct(string x,string y,int n,int m){
+vector<vector<unsigned long long>> dp(n+1,vector<unsigned long long>(m+1,0));
+if (m > n)
+        return 0;
+//initialising dp 2D-vector
+for (int i = 0; i <= n; i++)
+        dp[i][0] = 1;//t is empty
+for (int j = 1; j <= m; j++)
+        dp[0][j] = 0;//s is empty
+for(int i=1;i!=n+1;i++){
+for(int j=1;j!=m+1;j++)
+    {if(x[i-1]==y[j-1])
+        dp[i][j]=dp[i-1][j-1]+dp[i-1][j];//if matches then two cases it may be part of that subsequence or may not be.
+        //hence adding both cases also i for string s and j for t
+    else
+        dp[i][j]=dp[i-1][j];//if dosen't match then ignore that char of s
+    }
+}
+    return dp[n][m];
+}
+int main(){
+    string s,t;
+    cin>>s>>t;
+    cout<<distinct(s,t,s.size(),t.size())<<endl;
+}
+//TIME COMPLEXITY: O(N^2)