From 9085ee6a988091eaaaa6ec76538ef36c2fbdb0fd Mon Sep 17 00:00:00 2001
From: Elena Mokeeva <ryzheboka@gmail.com>
Date: Tue, 12 Oct 2021 14:56:12 +0200
Subject: [PATCH] chore(C): add longest common subsequence (#536)

---
 algorithms/C/README.md                        |   1 +
 .../C/strings/longest-common-subsequence.c    | 106 ++++++++++++++++++
 2 files changed, 107 insertions(+)
 create mode 100644 algorithms/C/strings/longest-common-subsequence.c

diff --git a/algorithms/C/README.md b/algorithms/C/README.md
index 4b814d11..140bdd49 100644
--- a/algorithms/C/README.md
+++ b/algorithms/C/README.md
@@ -39,6 +39,7 @@
 - [Count Words](strings/count-words.c)
 - [Palindrome](strings/palindrome.c)
 - [Permutation of String](strings/Permutation-of-String.c)
+- [Longest Common Subsequence](strings/longest-common-subsequence.c)
 
 ## Tree
 - [Height Of Tree](tree/height-of-a-tree.c)
diff --git a/algorithms/C/strings/longest-common-subsequence.c b/algorithms/C/strings/longest-common-subsequence.c
new file mode 100644
index 00000000..12f33c5d
--- /dev/null
+++ b/algorithms/C/strings/longest-common-subsequence.c
@@ -0,0 +1,106 @@
+/*  Program to find the longest common subsequence of two strings.
+Example: longest common subsequence of "abcde" and "abzyxcde" is "cde".
+
+Worst case time complexety: O(n*m^2), where n is the length of the first string,
+                            and m is the length of the second string.
+                            Example strings for worst case time: "aaaaa" and "aaaaa"
+                            Explanation: for each character in the first string, you 
+                            iterate through the second string. For each character of
+                            the second string, you iterate through the common sequence
+                            that starts at that character. In the worst case,
+                            for each character in the second string, you iterate through
+                            the remaining second string. That's why the algorithm is O(n*m^2) */
+#include <stdio.h>
+#include <string.h>
+#include <stdlib.h>
+
+
+char* longest_common_subsequence(char *string1, char *string2);
+
+int main() 
+{
+    /* Example: Finding longest common subsequence of following Strings
+        "Hello World! Ciao Mondo! Hei Maailma!" 
+        and "Bonjour Le Monde! Hei Maailma! Ciao Mondo!" */
+    
+    char* first_seqence1 = "Hello World! Ciao Mondo! Hei Maailma!";
+    char* second_sequence1 = "Bonjour Le Monde! Hei Maailma! Ciao Mondo!";
+    char* longest_subsequence1 = longest_common_subsequence(first_seqence1, second_sequence1);
+    printf("%s\n", longest_subsequence1);
+    free(longest_subsequence1);
+
+    /* Output: "! Hei Maailma!"*/
+
+    /* Example: Finding longest common subsequence of following Strings
+        "abcde"
+        and "abzyxcde" */
+
+    char* first_seqence2 = "abcde";
+    char* second_sequence2 = "abzyxcde";
+    char* longest_subsequence2 = longest_common_subsequence(first_seqence2, second_sequence2);
+    printf("%s\n", longest_subsequence2);
+    free(longest_subsequence2);
+
+    /* Output: "cde"*/
+
+    /* Example: Finding longest common subsequence of following Strings
+        "aaaaaa"
+        and "aaa" */
+
+    char* first_seqence3 = "aaaaaa";
+    char* second_sequence3 = "aaa";
+    char* longest_subsequence3 = longest_common_subsequence(first_seqence3, second_sequence3);
+    printf("%s\n", longest_subsequence3);
+    free(longest_subsequence3);
+
+    /* Output: "aaa"*/
+
+}
+
+/* Function which compares two character sequences and returns a pointer
+to the longest common subsequence */
+char* longest_common_subsequence(char *string1, char *string2)
+{
+    int len_of_longest = 0; //length of the currently longest common subsequence
+    int index_in_current_subsequence = 0; //index for iteration through a common subsequence
+    char *pointer_to_longest; //pointer to the longest common subsequence inside string1
+    char *original_string2 = string2; //copy of string2
+
+    //iterating through first string
+    // and evaluating the length of substring starting at the current character
+    while(*string1!='\0')
+    {
+        //iterating through string2 to check for common substring
+        while(*string2!='\0')
+        {
+            /* iterating through the common substring, to see how long it is. 
+	    Checking each character, if the subsequence doesn't match any longer,
+            of if one of the strings has ended */
+            while (*(string2+index_in_current_subsequence) == 
+*(string1+index_in_current_subsequence) && *(string2+index_in_current_subsequence) != '\0' &&
+*(string2+index_in_current_subsequence) != '\0')
+            {
+                index_in_current_subsequence++;
+            }
+
+            //saving the substring, if it's longer the the previously longest substring
+            if (index_in_current_subsequence>len_of_longest) 
+            {
+                len_of_longest=index_in_current_subsequence;
+                pointer_to_longest=string1;
+            }
+            index_in_current_subsequence = 0;
+            string2++;
+        }
+        string2 = original_string2; //returning to the begin of string2
+        string1++;
+    }
+
+    //copying the subsequence from string2 into memory (heap)
+    char *subsequence = (char*) malloc((len_of_longest+1)*sizeof(char));
+    strncpy(subsequence, pointer_to_longest, len_of_longest);
+    
+    //Setting the terminating character at the end
+    *(subsequence+len_of_longest)='\0';
+    return subsequence;
+}