Longest Common Subsequence

The problem

Given two strings, return the length of their longest common subsequence. A subsequence is a sequence that appears in the same relative order but not necessarily contiguous.

Looking at the final character of both prefixes reveals a strict choice: if they match, they must belong to an optimal subsequence extending whatever came before; if they differ, at least one of them is useless for this specific step, so drop one and take the better outcome.

Full solution

Space-optimized rolling rows: the 2D LCS grid only ever needs the previous row, so two rows (rolled into one variable swap) replace the full O(m*n) matrix.

def longest_common_subsequence(text1: str, text2: str) -> int:
    # Keep the shorter string along the row so the rolling array stays small.
    if len(text1) < len(text2):
        text1, text2 = text2, text1
    prev = [0] * (len(text2) + 1)
    for i in range(1, len(text1) + 1):
        curr = [0] * (len(text2) + 1)
        for j in range(1, len(text2) + 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[len(text2)]

function longestCommonSubsequence(text1, text2) {
  if (text1.length < text2.length) {
    [text1, text2] = [text2, text1];
  }
  let prev = new Array(text2.length + 1).fill(0);
  for (let i = 1; i <= text1.length; i++) {
    const curr = new Array(text2.length + 1).fill(0);
    for (let j = 1; j <= text2.length; j++) {
      if (text1[i - 1] === text2[j - 1]) {
        curr[j] = prev[j - 1] + 1;
      } else {
        curr[j] = Math.max(prev[j], curr[j - 1]);
      }
    }
    prev = curr;
  }
  return prev[text2.length];
}

The theory behind it

Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems

What Dynamic Programming is

Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.

When to reach for it

Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.

How the pattern works

Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.

What each operation costs

Which roles need this problem

Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.

Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.