The O(n) Club: Delete Operation for Two Strings — Deleting Your Way to Happiness

The O(n) Club: Delete Operation for Two Strings — Deleting Your Way to Happiness

⚡ TL;DR

Want two strings to be identical? Just delete anything that’s not in the Longest Common Subsequence (LCS). The brute-force route is O(2ⁿ) and, ironically, would take less time if you just deleted your IDE. The DP (LCS) fix? Check the snippet:

// Minimum deletions needed for two strings:
int minDeletions = word1.length() + word2.length() - 2 * lcs(word1, word2);
 // Classic LCS DP inside lcs():
int[][] dp = new int[m+1][n+1];

🧠 How Devs Usually Mess This Up

Where do devs go wrong? Let me count the delete keys:

• Reach for Levenshtein Distance: Sorry, no substitutions or insertions. Deleting is your only verb, not a full grammar.
• Brute-force All Possibilities: Trying every possible combination is only fun if you’re into watching grass grow—on a GPU.
• Attempt to Merge Strings: This isn’t a reconciliation dinner; deletions only. If a letter’s not in both, it’s getting fired.

🔍 What’s Actually Going On

Imagine two chefs each making “alphabet soup,” but they’re only allowed to remove weird letters from their broth to match the other chef. LCS is the shared list of acceptable ingredients; everything else gets the culinary axe. The fewer the letters in common, the more you slice and dice. And to calculate: Minimum deletions = length of both soups minus twice the tasty LCS!

🛠️ PseudoCode

• Setup DP Table: dp[i][j] is LCS length for first i letters of word1 and first j of word2.
  int[][] dp = new int[m+1][n+1];
• Base Cases: If either string is empty, LCS is zero. (One empty fridge, no snack left.)
• Filling:
  • If word1.charAt(i-1) == word2.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])
• Final Answer: word1.length() + word2.length() - 2 * dp[m][n]

💻 The Code

public int minDistance(String word1, String word2) {
    int m = word1.length(), n = word2.length();
    int[][] dp = new int[m + 1][n + 1];
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (word1.charAt(i - 1) == word2.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 m + n - 2 * dp[m][n];
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Classic DP Matrix Eats RAM: Full dp[m+1][n+1] is big. For those on budget laptops or giving live coding interviews from a bathroom, compress to two rows.
• Missing Edge Cases: When a string is empty, just delete all of the other, then pour yourself a consolation coffee.
• Thinking You Can Substitute: If your thumb twitches for insert or substitute, reread the prompt and repeat: Only. Delete. Stuff.
• Complexity? O(mn) time and space, or O(n) space if optimized. (No medals for cleverness, but your fan thanks you.)

🧠 Interviewer Brain-Tattoo

“Real engineers don’t fix bugs; they delete lines until everything matches.”