Longest Common Subsequence

Problem

Given two strings text1 and text2, return the length of their longest common subsequence (LCS). A subsequence keeps characters in their original relative order but may skip any number of them (it need not be contiguous). A common subsequence is one that appears in both strings.

If there is no common subsequence, return 0.

Examples

• text1 = "abcde", text2 = "ace" → 3 — "ace" is a subsequence of both.
• text1 = "abc", text2 = "abc" → 3 — identical strings; the whole string is the LCS.
• text1 = "abc", text2 = "def" → 0 — no shared characters at all.

Constraints

• 1 <= text1.length, text2.length <= 1000
• Both strings consist of lowercase English letters.

Lengths up to 1000 each mean the expected solution is O(m × n) — up to a million cells.

Think about it first

TL;DR

Classic LCS DP over prefix pairs (i, j) — O(m × n) time, O(min(m, n)) space after rolling to two rows.

Approach 1 — Brute-force recursion

Compare the last characters of the two prefixes. If they are equal, they extend the LCS: add 1 and recurse on both shorter prefixes. If they differ, drop the last character of one string or the other and keep whichever result is larger.

def longestCommonSubsequence(text1: str, text2: str) -> int:

    def dfs(i: int, j: int) -> int:
        if i == 0 or j == 0:
            return 0
        if text1[i - 1] == text2[j - 1]:
            return 1 + dfs(i - 1, j - 1)
        return max(dfs(i - 1, j), dfs(i, j - 1))

    return dfs(len(text1), len(text2))

Complexity: O(2^(m+n)) time worst case, O(m+n) recursion depth.

With two length-1000 strings and few matches, the recursion branches into an exponentially large tree, so this times out.

Approach 2 — Top-down memoization

Only the pair (i, j) varies, giving (m+1)(n+1) distinct subproblems. Cache each result and the recursion tree reduces to filling that grid once.

from functools import lru_cache


def longestCommonSubsequence(text1: str, text2: str) -> int:

    @lru_cache(maxsize=None)
    def dfs(i: int, j: int) -> int:
        if i == 0 or j == 0:
            return 0
        if text1[i - 1] == text2[j - 1]:
            return 1 + dfs(i - 1, j - 1)
        return max(dfs(i - 1, j), dfs(i, j - 1))

    return dfs(len(text1), len(text2))

Complexity: O(m × n) time and space.

Approach 3 — Bottom-up 2-D table

dp[i][j] is the length of the LCS of text1[:i] and text2[:j]. Each cell depends only on its diagonal, top, and left neighbors:

flowchart LR
    D["dp[i-1][j-1]"] -->|"chars match: +1"| C["dp[i][j]"]
    T["dp[i-1][j]"] -->|"mismatch: max"| C
    L["dp[i][j-1]"] -->|"mismatch: max"| C

Because every cell reads only cells above and to the left, filling row by row (left to right) guarantees the inputs are ready.

Recurrence:

dp[i][0] = 0,  dp[0][j] = 0                 # empty prefix shares nothing
if text1[i-1] == text2[j-1]:
    dp[i][j] = dp[i-1][j-1] + 1             # extend the diagonal match
else:
    dp[i][j] = max(dp[i-1][j], dp[i][j-1])  # best of dropping one char

def longestCommonSubsequence(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[m][n]

Walkthrough on text1 = "abcde", text2 = "ace":

The diagonal increments at a, c, and e accumulate to dp[5][3] = 3.

Complexity: O(m × n) time, O(m × n) space.

Approach 4 — Space-optimized two rows

Each cell reads only the previous row and the current row’s left neighbor, so two rows suffice. Keep prev and fill curr.

def longestCommonSubsequence(text1: str, text2: str) -> int:
    m, n = len(text1), len(text2)
    prev = [0] * (n + 1)
    for i in range(1, m + 1):
        curr = [0] * (n + 1)
        for j in range(1, n + 1):
            if text1[i - 1] == text2[j - 1]:
                curr[j] = prev[j - 1] + 1
            else:
                curr[j] = max(prev[j], curr[j - 1])
        prev = curr
    return prev[n]

Complexity: O(m × n) time, O(n) space. (Iterate over the longer string as the outer loop and the shorter as inner for O(min(m, n)) space.)

Common pitfalls

• Using max on the diagonal for a match — when characters match you must take dp[i-1][j-1] + 1; taking max(dp[i-1][j], dp[i][j-1]) there silently drops valid pairings.
• Confusing subsequence with substring — LCS allows gaps; the longest common substring problem uses a different recurrence (reset to 0 on mismatch).
• Index vs length off-by-one — dp[i][j] uses characters text1[i-1] and text2[j-1]; mixing 0-based and 1-based indexing here is the classic bug.
• In the rolled version, allocate a fresh curr (or clear it) each row so stale values don’t leak in.

Pattern takeaway

LCS is the standard 2-D string DP: a match takes the diagonal plus one, a mismatch takes the max of the top and left neighbors. The same grid underlies edit distance, string interleaving, and diff tools. When a problem asks you to align two sequences while preserving order, use the (i, j) prefix table.