Formatted question description: https://leetcode.ca/all/1143.html

# 1143. Longest Common Subsequence (Medium)

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.


Constraints:

• 1 <= text1.length <= 1000
• 1 <= text2.length <= 1000
• The input strings consist of lowercase English characters only.

Related Topics:
Dynamic Programming

Similar Questions:

## Solution 1. DP

Let dp[i][j] be the length of the longest common subsequence of s[0..(i-1)] and t[0..(j-1)].

dp[i][j] = 1 + dp[i-1][j-1]                If s[i-1] == t[j-1]
= max(dp[i-1][j], dp[i][j-1])     If s[i-1] != t[j-1]
dp[i] = dp[i] = 0

// OJ: https://leetcode.com/problems/longest-common-subsequence/

// Time: O(MN)
// Space: O(MN)
class Solution {
public:
int longestCommonSubsequence(string s, string t) {
int M = s.size(), N = t.size();
vector<vector<int>> dp(M + 1, vector<int>(N + 1));
for (int i = 1; i <= M; ++i) {
for (int j = 1; j <= N; ++j) {
if (s[i - 1] == t[j - 1]) dp[i][j] = 1 + dp[i - 1][j - 1];
else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[M][N];
}
};


## Solution 2. DP + Space Optimization

Since dp[i][j] is only dependent on dp[i-1][j-1], dp[i-1][j] and dp[i][j-1], we can reduce the space of dp array from N * N to 1 * N with a temporary variable storing dp[i-1][j-1].

// OJ: https://leetcode.com/problems/longest-common-subsequence/

// Time: O(MN)
// Space: O(min(M, N))
class Solution {
public:
int longestCommonSubsequence(string s, string t) {
int M = s.size(), N = t.size();
if (M < N) swap(M, N), swap(s, t);
vector<int> dp(N + 1);
for (int i = 1; i <= M; ++i) {
int prev = 0;
for (int j = 1; j <= N; ++j) {
int cur = dp[j];
if (s[i - 1] == t[j - 1]) dp[j] = 1 + prev;
else dp[j] = max(dp[j], dp[j - 1]);
prev = cur;
}
}
return dp[N];
}
};


Java

class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int length1 = text1.length(), length2 = text2.length();
int[][] dp = new int[length1 + 1][length2 + 1];
for (int i = 1; i <= length1; i++) {
char c1 = text1.charAt(i - 1);
for (int j = 1; j <= length2; j++) {
char c2 = text2.charAt(j - 1);
if (c1 == c2)
dp[i][j] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
int commonLength = dp[length1][length2];
return commonLength;
}
}