LCS

Author: sachuverma

DSA Crack Sheet/README.md

@@ -83,6 +83,7 @@
 - [Min Number of Flips](https://practice.geeksforgeeks.org/problems/min-number-of-flips/0# "view question") - [Cpp Solution](./solutions/Min%20Number%20of%20Flips.cpp)
 - [Second most repeated string in a sequence](https://practice.geeksforgeeks.org/problems/second-most-repeated-string-in-a-sequence0534/1# "view question") - [Cpp Solution](./solutions/Second%20most%20repeated%20string%20in%20a%20sequence.cpp)
 - [Minimum Swaps for Bracket Balancing](https://practice.geeksforgeeks.org/problems/minimum-swaps-for-bracket-balancing/0# "view question") - [Cpp Solution](./solutions/Minimum%20Swaps%20for%20Bracket%20Balancing.cpp)
+- [Longest Common Subsequence](https://practice.geeksforgeeks.org/problems/longest-common-subsequence-1587115620/1 "view question") - [Cpp Solution](./solutions/Longest%20Common%20Subsequence.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Searching & Sorting

DSA Crack Sheet/solutions/Longest Common Subsequence.cpp

@@ -0,0 +1,57 @@
+/*
+Longest Common Subsequence
+==========================
+
+Given two sequences, find the length of longest subsequence present in both of them. Both the strings are of uppercase.
+
+Example 1:
+Input:
+A = 6, B = 6
+str1 = ABCDGH
+str2 = AEDFHR
+Output: 3
+Explanation: LCS for input Sequences
+“ABCDGH” and “AEDFHR” is “ADH” of
+length 3.
+
+Example 2:
+Input:
+A = 3, B = 2
+str1 = ABC
+str2 = AC
+Output: 2
+Explanation: LCS of "ABC" and "AC" is
+"AC" of length 2.
+Your Task:
+Complete the function lcs() which takes the length of two strings respectively and two strings as input parameters and returns the length of the longest subsequence present in both of them.
+
+Expected Time Complexity : O(|str1|*|str2|)
+Expected Auxiliary Space: O(|str1|*|str2|)
+
+Constraints:
+1<=size(str1),size(str2)<=100
+*/
+
+int lcs(int x, int y, string s1, string s2)
+{
+  int dp[x + 1][y + 1];
+  for (int i = 0; i <= x; ++i)
+  {
+    for (int j = 0; j <= y; ++j)
+    {
+      if (i == 0 || j == 0)
+        dp[i][j] = 0;
+      else
+      {
+        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[x][y];
+}