tags: `leetcode` `medium` `String` `Dynamic Programming`

1143. Longest Common Subsequence

Description

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

Examples

Example 1:

Input: text1 = “abcde”, text2 = “ace”
Output: 3
Explanation: The longest common subsequence is “ace” and its length is 3.

Example 2:

Input: text1 = “abc”, text2 = “abc”
Output: 3
Explanation: The longest common subsequence is “abc” and its length is 3.

Example 3:

Input: text1 = “abc”, text2 = “def”
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

$1 \leq text1.length, text2.length \leq 1000$
text1 and text2 consist of only lowercase English characters.

Code

#define MAX(x, y) ( ((x)>(y)) ? (x) : (y))
int longestCommonSubsequence(char * text1, char * text2){
    int length1 = strlen(text1), length2 = strlen(text2);
    int **lcs = malloc((length1 + 1) * sizeof(int*));
    for(int i = 0 ; i <= length1 ; i++)
        lcs[i] = calloc((length2 + 1), sizeof(int));

    for(int row = 1 ; row <= length1 ; row++){
        for(int col = 1 ; col <= length2 ; col++){
            if(text1[row - 1] == text2[col - 1])
                lcs[row][col] = lcs[row - 1][col - 1]+1;
            else
                lcs[row][col] = MAX(lcs[row - 1][col], lcs[row][col - 1]);
        }
    }
    return lcs[length1][length2];
};

Complexity

Space	Time
$O(length_1*length_2)$	$O(length_1*length_2)$

$length_i$: length of input $text_i$

Result

Runtime: 12 ms, 89.58%
Memory: 12.2 MB, 19.58%

tags: leetcode medium String Dynamic Programming