Description
Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.
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.
- For example,
"ace" is a subsequence of "abcde".
A common subsequence of two strings is a subsequence that is common to both strings.
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, text2.length <= 1000text1 and text2 consist of only lowercase English characters.
Solutions
Solution 1: Dynamic Programming
We define f[i][j] as the length of the longest common subsequence of the first i characters of text1 and the first j characters of text2. Therefore, the answer is f[m][n], where m and n are the lengths of text1 and text2, respectively.
If the ith character of text1 and the jth character of text2 are the same, then f[i][j] = f[i - 1][j - 1] + 1; if the ith character of text1 and the jth character of text2 are different, then f[i][j] = max(f[i - 1][j], f[i][j - 1]). The state transition equation is:
f[i][j] =
\begin{cases}
f[i - 1][j - 1] + 1, & if text1[i - 1] = text2[j - 1] \
max(f[i - 1][j], f[i][j - 1]), & if text1[i - 1] ≠ text2[j - 1]
\end{cases}
The time complexity is O(m × n), and the space complexity is O(m × n). Here, m and n are the lengths of text1 and text2, respectively.
PythonJavaC++GoTypeScriptRustJavaScriptC#Kotlin
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m, n = len(text1), len(text2)
f = [[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]:
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](code-box)
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];
}
}(code-box)
class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {
int m = text1.size(), n = text2.size();
int f[m + 1][n + 1];
memset(f, 0, sizeof f);
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] = max(f[i - 1][j], f[i][j - 1]);
}
}
}
return f[m][n];
}
};(code-box)
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]
}(code-box)
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];
}(code-box)
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]
}
}(code-box)
/**
* @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];
};(code-box)
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];
}
}(code-box)
class Solution {
fun longestCommonSubsequence(text1: String, text2: String): Int {
val m = text1.length
val n = text2.length
val f = Array(m + 1) { IntArray(n + 1) }
for (i in 1..m) {
for (j in 1..n) {
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]
}
}(code-box)
