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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 <= 10001 <= 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..(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][0] = dp[0][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];
}
};
-
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; } } ############ class Solution { public int longestCommonSubsequence(String text1, String text2) { int m = text1.length(), n = text2.length(); int[][] f = 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)) { f[i][j] = f[i - 1][j - 1] + 1; } else { f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]); } } } return f[m][n]; } } -
// 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]; } }; -
class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: m, n = len(text1), len(text2) dp = [[0] * (n + 1) for _ in range(m + 1)] for i in range(1, m + 1): for j in range(1, n + 1): if text1[i - 1] == text2[j - 1]: dp[i][j] = dp[i - 1][j - 1] + 1 else: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) return dp[-1][-1] -
func longestCommonSubsequence(text1 string, text2 string) int { m, n := len(text1), len(text2) f := make([][]int, m+1) for i := range f { f[i] = make([]int, n+1) } for i := 1; i <= m; i++ { for j := 1; j <= n; j++ { if text1[i-1] == text2[j-1] { f[i][j] = f[i-1][j-1] + 1 } else { f[i][j] = max(f[i-1][j], f[i][j-1]) } } } return f[m][n] } func max(a, b int) int { if a > b { return a } return b } -
function longestCommonSubsequence(text1: string, text2: string): number { const m = text1.length; const n = text2.length; const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0)); for (let i = 1; i <= m; i++) { for (let j = 1; j <= n; j++) { if (text1[i - 1] === text2[j - 1]) { f[i][j] = f[i - 1][j - 1] + 1; } else { f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]); } } } return f[m][n]; } -
/** * @param {string} text1 * @param {string} text2 * @return {number} */ var longestCommonSubsequence = function (text1, text2) { const m = text1.length; const n = text2.length; const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0)); for (let i = 1; i <= m; ++i) { for (let j = 1; j <= n; ++j) { if (text1[i - 1] == text2[j - 1]) { f[i][j] = f[i - 1][j - 1] + 1; } else { f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]); } } } return f[m][n]; }; -
public class Solution { public int LongestCommonSubsequence(string text1, string text2) { int m = text1.Length, n = text2.Length; int[,] f = new int[m + 1, n + 1]; for (int i = 1; i <= m; ++i) { for (int j = 1; j <= n; ++j) { if (text1[i - 1] == text2[j - 1]) { f[i, j] = f[i - 1, j - 1] + 1; } else { f[i, j] = Math.Max(f[i - 1, j], f[i, j - 1]); } } } return f[m, n]; } } -
impl Solution { pub fn longest_common_subsequence(text1: String, text2: String) -> i32 { let (m, n) = (text1.len(), text2.len()); let (text1, text2) = (text1.as_bytes(), text2.as_bytes()); let mut f = vec![vec![0; n + 1]; m + 1]; for i in 1..=m { for j in 1..=n { f[i][j] = if text1[i - 1] == text2[j - 1] { f[i - 1][j - 1] + 1 } else { f[i - 1][j].max(f[i][j - 1]) } } } f[m][n] } }