LeetCode-in-All
1143. Longest Common Subsequence
Medium
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.
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 <= text1.length, text2.length <= 1000text1andtext2consist of only lowercase English characters.
Solution
#include <stdio.h>
#include <string.h>
int longestCommonSubsequence(char* text1, char* text2) {
int n = strlen(text1);
int m = strlen(text2);
// Create a 2D array to store the lengths of longest common subsequences
int dp[n + 1][m + 1];
// Initialize the dp array
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
dp[i][j] = 0; // Initialize all values to 0
}
}
// Fill the dp array using bottom-up dynamic programming
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (text1[i - 1] == text2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = (dp[i - 1][j] > dp[i][j - 1]) ? dp[i - 1][j] : dp[i][j - 1];
}
}
}
// The length of the longest common subsequence is stored in dp[n][m]
return dp[n][m];
}