Longest Common Subsequence

Why it belongs on this sheet

This is one of the most important 2D DP problems and a staple of placement prep.

Pattern

Match or skip

Approach

Let dp[i][j] be the LCS length for prefixes ending at i and j. If the characters match, extend diagonally. Otherwise take the better answer after skipping one character from either string.

Java solution

class Solution {
  public int longestCommonSubsequence(String text1, String text2) {
    int[][] dp = new int[text1.length() + 1][text2.length() + 1];

    for (int i = 1; i <= text1.length(); i++) {
      for (int j = 1; j <= text2.length(); j++) {
        if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
          dp[i][j] = dp[i - 1][j - 1] + 1;
        } else {
          dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
        }
      }
    }

    return dp[text1.length()][text2.length()];
  }
}

Complexity

• Time: O(m * n)
• Space: O(m * n)

Interview note

This is the kind of problem where drawing the table once is often worth it before coding.