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:
 Longest Palindromic Subsequence (Medium)
 Delete Operation for Two Strings (Medium)
 Shortest Common Supersequence (Hard)
Solution 1. DP
Let dp[i][j]
be the length of the longest common subsequence of s[0..(i1)]
and t[0..(j1)]
.
dp[i][j] = 1 + dp[i1][j1] If s[i1] == t[j1]
= max(dp[i1][j], dp[i][j1]) If s[i1] != t[j1]
dp[i][0] = dp[0][i] = 0
// OJ: https://leetcode.com/problems/longestcommonsubsequence/
// 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[i1][j1]
, dp[i1][j]
and dp[i][j1]
, we can reduce the space of dp
array from N * N
to 1 * N
with a temporary variable storing dp[i1][j1]
.
// OJ: https://leetcode.com/problems/longestcommonsubsequence/
// 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; } }

// OJ: https://leetcode.com/problems/longestcommonsubsequence/ // 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]; } };

print("Todo!")