dp init #6

Author: akashchouhan16

DynamicProgramming/LCS_Problem/DPsoln_LCS.cpp

@@ -0,0 +1,78 @@
+#include "../headers.hpp"
+/*
+Sample Test Case : 
+2
+Enter Two Strings : hegstu  hgxsttu
+Enter Two Strings : abedfhr abcdgh
+OUTPUT using Bottom Up Approach : 
+case #1 : 5
+case #2 : 4
+OUTPUT using Top Down Approach :  
+case #1 : 5
+case #2 : 4
+*/
+int LCS_memoized(string str1, string str2, int n, int m, int **dp)
+{
+    if (n == 0 || m == 0)
+        return 0;
+
+    if (dp[n][m] != -1)
+        return dp[n][m];
+
+    if (str1[n - 1] == str2[m - 1])
+        return dp[n][m] = (1 + LCS_memoized(str1, str2, n - 1, m - 1, dp));
+    else
+        return dp[n][m] = max(LCS_memoized(str1, str2, n - 1, m, dp), LCS_memoized(str1, str2, n, m - 1, dp));
+}
+
+int LCS_topDown(string s1, string s2, int n, int m)
+{
+    int dp[n + 1][m + 1];
+    // Init the dp for base case -> n==0 || m==0 must return 0;
+    for (int i = 0; i <= m; i++)
+        dp[0][i] = 0;
+    for (int i = 0; i <= n; i++)
+        dp[i][0] = 0;
+
+    for (int i = 1; i <= n; i++)
+    {
+        for (int j = 1; j <= m; j++)
+        {
+            if (s1[i - 1] == s2[j - 1])
+                dp[i][j] = (1 + dp[i - 1][j - 1]);
+            else
+            {
+                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
+            }
+        }
+    }
+    return dp[n][m];
+}
+int main()
+{
+    vi output1, output2;
+    // Output1 stores solns for the Bottom Up approach.
+    // Output2 stores solns for the Top Down approach.
+    tests(t)
+    {
+        string str1, str2;
+        cout << "Enter Two Strings : ";
+        cin >> str1 >> str2;
+        int n = str1.size(), m = str2.size();
+
+        int **dp = new int *[n + 1];
+        loop(i, n + 1) dp[i] = new int[m + 1];
+
+        loop(i, n + 1)
+            loop(j, m + 1)
+                dp[i][j] = -1;
+
+        output1.pb(LCS_memoized(str1, str2, n, m, dp));
+        output2.pb(LCS_topDown(str1, str2, n, m));
+    }
+    cout << "OUTPUT using Bottom Up Approach : \n";
+    loop(i, output1.size()) cout << "case #" << i + 1 << " : " << output1[i] << endl;
+    cout << "OUTPUT using Top Down Approach : \n";
+    loop(i, output2.size()) cout << "case #" << i + 1 << " : " << output2[i] << endl;
+    return 0;
+}

DynamicProgramming/LCS_Problem/recursive_LongestCommonSubsequence.cpp

@@ -0,0 +1,37 @@
+#include "../headers.hpp"
+
+/* Sample Test Case :
+2
+Enter Two Strings : abcdgh abedfhr
+Enter Two Strings : axyezc sxecoq
+case #1 : 4
+case #2 : 3
+*/
+int LCS(string str1, string str2, int n, int m)
+{
+    if (n == 0 || m == 0)
+        return 0;
+
+    if (str1[n - 1] == str2[m - 1])
+        return (1 + LCS(str1, str2, n - 1, m - 1));
+    else
+    {
+        return max(LCS(str1, str2, n - 1, m), LCS(str1, str2, n, m - 1));
+    }
+}
+
+int main()
+{
+    vi output;
+    tests(t)
+    {
+        string str1, str2;
+        cout << "Enter Two Strings : ";
+        cin >> str1 >> str2;
+        int n = str1.size(), m = str2.size();
+
+        output.pb(LCS(str1, str2, n, m));
+    }
+    loop(i, output.size()) cout << "case #" << i + 1 << " : " << output[i] << endl;
+    return 0;
+}