Delete Operation for Two Strings
A medium-tier problem at 64% community acceptance, tagged with String, Dynamic Programming. Reported in interviews at ByteDance and 0 others.
You're given two strings and need to find the minimum number of deletions to make them equal. ByteDance asks this one. It's a dynamic programming problem that looks simple until you realize the naive greedy approach fails. The pattern: find the longest common subsequence, then the answer is the sum of both lengths minus twice the LCS length. Most candidates either try greedy deletion or get bogged down in 2D DP state management. If this hits your live assessment and you blank on the LCS connection, StealthCoder solves it in seconds, invisible to the proctor.
Companies that ask "Delete Operation for Two Strings"
Delete Operation for Two Strings is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share.
Get StealthCoderThe trick is recognizing this as a longest common subsequence problem in disguise. You don't delete characters greedily from position zero. Instead, you find what both strings share, then delete everything else. Build a 2D DP table where dp[i][j] represents the LCS length of the first i characters of string one and first j characters of string two. When characters match, dp[i][j] = dp[i-1][j-1] + 1. When they don't, take the max of skipping either character. Common pitfall: candidates try to track deletion positions directly instead of computing LCS first. The formula is len(s1) + len(s2) - 2 * lcs_length. Time is O(m*n), space is O(m*n) unless you optimize. StealthCoder is the hedge for the one problem you didn't drill.
Pattern tags
You know the problem.
Make sure you actually pass it.
Delete Operation for Two Strings recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Delete Operation for Two Strings interview FAQ
Is this just asking me to delete until strings match?+
Not quite. You delete the minimum total characters from both strings so they become identical. The optimal strategy isn't obvious because greedy deletion fails. You need to find what's common first (the longest common subsequence), then delete around it.
How does this relate to edit distance?+
Both use 2D DP, but edit distance counts substitutions and insertions. This problem only allows deletions and has no cost weight. It's simpler: compute LCS, then the answer is both lengths minus twice the LCS. No substitution logic needed.
What's the actual trick so I don't blank?+
The insight is that the minimum deletions equals the characters not in the longest common subsequence. Build a standard LCS DP table, find its length, then apply the formula: len(s1) + len(s2) - 2*lcs_length. That's the answer.
Does ByteDance have a preference between space-optimized and standard DP?+
Input data doesn't specify. Standard 2D DP (O(m*n) space) is clearer and passes. If asked to optimize, you can reduce space to O(min(m, n)) by keeping only two rows, but that's rarely required.
Is this problem asked often in recent reports?+
ByteDance reports it. The acceptance rate is about 63 percent, which suggests it's not trivial but standard interview material. Most candidates who know the LCS pattern solve it cleanly.
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