# Longest Common Subsequence (Leetcode 1143) - Coding Interview Question

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:3Explanation:The longest common subsequence is "ace" and its length is 3.

**Example 2:**

Input:text1 = "abc", text2 = "abc"Output:3Explanation:The longest common subsequence is "abc" and its length is 3.

**Example 3:**

Input:text1 = "abc", text2 = "def"Output:0Explanation:There is no such common subsequence, so the result is 0.

**Constraints:**

`1 <= text1.length, text2.length <= 1000`

`text1`

and`text2`

consist of only lowercase English characters.

**Solution:**

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class LongestCommonSubsequence {
Integer[][] dp;
public int longestCommonSubsequence(String s1, String s2) {
return lcsIterative(s1, s2);
// dp = new Integer[s1.length()][s2.length()];
// return lcs(s1, s2, 0, 0);
}
//Dynamic Programming Iterative approach.
int lcsIterative(String s1, String s2) {
int[][] dp = new int[s1.length() + 1][s2.length() + 1];
for (int i = 1; i <= s1.length(); i++) {
for (int j = 1; j <= s2.length(); j++) {
if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
dp[i][j] = 1 + dp[i - 1][j - 1];
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
return dp[s1.length()][s2.length()];
}
int lcs(String s1, String s2, int n, int m) {
if (n == s1.length() || m == s2.length()) {
return 0;
}
if (dp[n][m] != null) {
return dp[n][m];
}
int res = 0;
if (s1.charAt(n) == s2.charAt(m)) {
res = 1 + lcs(s1, s2, n + 1, m + 1);
} else {
res = Math.max(lcs(s1, s2, n + 1, m), lcs(s1, s2, n, m + 1));
}
dp[n][m] = res;
return res;
}
}