##### Welcome to Subscribe On Youtube

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];
}
};

• 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 = [ * (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]
}
}