From 86ef526697cf8ceb5b64135c4ba25d075b8cfa1b Mon Sep 17 00:00:00 2001
From: begeekmyfriend <begeekmyfriend@gmail.com>
Date: Wed, 16 Aug 2017 09:51:19 +0800
Subject: [PATCH] Edit distance

Signed-off-by: begeekmyfriend <begeekmyfriend@gmail.com>
---
 072_edit_distance/Makefile        |   2 +
 072_edit_distance/edit_distance.c | 102 ++++++++++++++++++++++++++++++
 2 files changed, 104 insertions(+)
 create mode 100644 072_edit_distance/Makefile
 create mode 100644 072_edit_distance/edit_distance.c

diff --git a/072_edit_distance/Makefile b/072_edit_distance/Makefile
new file mode 100644
index 0000000..b8aad45
--- /dev/null
+++ b/072_edit_distance/Makefile
@@ -0,0 +1,2 @@
+all:
+	gcc -O2 -o test edit_distance.c
diff --git a/072_edit_distance/edit_distance.c b/072_edit_distance/edit_distance.c
new file mode 100644
index 0000000..c854187
--- /dev/null
+++ b/072_edit_distance/edit_distance.c
@@ -0,0 +1,102 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+#include <string.h>
+
+/*
+ *  Dynamic Programming
+ *
+ *    Definitaion
+ *
+ *        m[i][j] is minimal distance from word1[0..i] to word2[0..j]
+ *
+ *    So, 
+ *
+ *        1) if word1[i] == word2[j], then m[i][j] == m[i-1][j-1].
+ *
+ *        2) if word1[i] != word2[j], then we need to find which one below is minimal:
+ *
+ *             min( m[i-1][j-1], m[i-1][j],  m[i][j-1] )
+ *            
+ *             and +1 - current char need be changed.
+ *
+ *        Let's take a look  m[1][2] :  "a" => "ab" 
+ *
+ *               +---+  +---+                                         
+ *        ''=> a | 1 |  | 2 | '' => ab                                
+ *               +---+  +---+                                         
+ *                                                                    
+ *               +---+  +---+                                         
+ *        a => a | 0 |  | 1 | a => ab                                 
+ *               +---+  +---+                                         
+ *                                                            
+ *        To know the minimal distance `a => ab`, we can get it from one of the following cases:
+ *
+ *            1) delete the last char in word1,  minDistance( '' => ab ) + 1                                                             
+ *            2) delete the last char in word2,  minDistance(  a => a ) + 1                                                             
+ *            3) change the last char,           minDistance( '' => a ) + 1                                                             
+ *                                                            
+ *  
+ *    For Example:
+ *
+ *        word1="abb", word2="abccb"
+ *
+ *        1) Initialize the DP matrix as below:
+ *
+ *               "" a b c c b
+ *            "" 0  1 2 3 4 5
+ *            a  1
+ *            b  2
+ *            b  3
+ *
+ *        2) Dynamic Programming
+ *
+ *               "" a b c c b
+ *            ""  0 1 2 3 4 5
+ *            a   1 0 1 2 3 4 
+ *            b   2 1 0 1 2 3
+ *            b   3 2 1 1 2 2
+ *
+ */ 
+
+static int minDistance(char* word1, char* word2)
+{
+    int i, j;
+    int len1 = strlen(word1);
+    int len2 = strlen(word2);
+    int *table = malloc((len1 + 1) * (len2 + 1) * sizeof(int));
+    int **dp = malloc((len1 + 1) * sizeof(int *));
+    for (i = 0; i < len1 + 1; i++) {
+        dp[i] = table + i * (len2 + 1);
+    }
+
+    for (i = 0; i < len2 + 1; i++) {
+        dp[0][i] = i;
+    }
+    for (i = 0; i < len1 + 1; i++) {
+        dp[i][0] = i;
+    }
+
+    for (i = 1; i < len1 + 1; i++) {
+        for (j = 1; j < len2 + 1; j++) {
+            if (word1[i - 1] == word2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1];
+            } else {
+                int min = dp[i - 1][j] > dp[i][j - 1] ? dp[i][j - 1] : dp[i - 1][j];
+                dp[i][j] = min > dp[i - 1][j - 1] ? dp[i - 1][j - 1] : min;
+                dp[i][j] += 1;
+            }
+        }
+    }
+    return dp[len1][len2];
+}
+
+int main(int argc, char **argv)
+{
+    if (argc != 3) {
+        fprintf(stderr, "Usage: ./test word1 word2\n");
+        exit(-1);
+    }
+    printf("%d\n", minDistance(argv[1], argv[2]));
+    return 0;
+}