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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 <= 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][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]
        }
    }
    
    

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