Given two strings text1 and text2, return the length of their longest common subsequence.
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. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
If there is no common subsequence, return 0.
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.
Solution
This is LCS, a Dynamic Programming (DP) problem, which can break down into smaller, simpler subproblems, and so on. DP problem usually reuse solutions to lower level subproblems.
We define dp[m+1][n+1] represents a set of longest common subsequect of prefiex Xi and Yj. given that:
dp[0][0] = 0;
if X[i - 1]==Y[j -1] (current character is the same), dp[i][j] = 1 + d[i - 1][j - 1]
if X[i-1]!=Y[j-1], dp[i][j] = max(dp[i-1],j],dp[i][j-1])
repeat until computed the whole string, dp[m][n] is the answer.
For example:
Complexity Analysis
Time Complexity:O(N*M)
Space Complexity:O(N*M)
N - the length of text1
M - the length of text2
Code
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
if (text1.equals(text2)) return text1.length();
int m = text1.length();
int n = text2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
dp[i][j] = 1 + dp[i - 1][j - 1];
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m][n];
}
}
