Palindrome Partitioning
A medium-tier problem at 72% community acceptance, tagged with String, Dynamic Programming, Backtracking. Reported in interviews at Bloomberg and 6 others.
Palindrome Partitioning is a medium-difficulty string problem that appears in interviews at Amazon, Meta, Bloomberg, Uber, and Yahoo. The task sounds simple on the surface: partition a string into all possible subsets where each subset is a palindrome. But the backtracking logic and the need to validate palindromes efficiently trip up candidates who haven't seen the pattern before. With a 72% acceptance rate, it's not a gimme, and it's the kind of problem where blanking on the recursion structure or forgetting memoization can cost you minutes you don't have. If this problem hits your live OA and you freeze, StealthCoder surfaces a working solution instantly, invisible to the proctor.
Companies that ask "Palindrome Partitioning"
Palindrome Partitioning 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 because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround.
Get StealthCoderThe trick is recognizing this as a backtracking problem, not a dynamic programming one (though DP can optimize it). You recurse on the string, at each step deciding where to cut. For each position, you try every possible palindrome starting at that position, then recurse on the remainder. Most candidates either build strings inefficiently or fail to cache palindrome checks upfront. The obvious brute-force approach without memoization on palindrome validation will TLE on longer inputs. The pattern: precompute which substrings are palindromes using DP, then backtrack through the string, pruning non-palindrome branches. Common pitfall is modifying and restoring the result list correctly across recursive calls. If you nail the backtracking structure and palindrome caching, this becomes a medium-complexity problem. If you don't, you're recomputing palindromes and wasting cycles. StealthCoder is the hedge for candidates who drilled the pattern but forgot the optimization under pressure.
Pattern tags
You know the problem.
Make sure you actually pass it.
Palindrome Partitioning 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 because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Palindrome Partitioning interview FAQ
Is Palindrome Partitioning still asked at Meta and Amazon?+
Yes. Both appear in the company list for this problem and it remains a frequently asked interview question. The problem sits at the intersection of string manipulation and backtracking, both core skills tested at those companies. Acceptance rate of 72% suggests it's neither a screening knockout nor a freebie.
What's the key insight I'm missing if I TLE?+
You're likely recomputing palindrome checks during recursion. Pre-build a 2D DP table where dp[i][j] tells you if substring from index i to j is a palindrome. This costs O(n^2) upfront but saves you from rechecking during backtracking. Without it, you'll timeout on strings longer than 12-15 characters.
How does this relate to Backtracking vs Dynamic Programming?+
This is a backtracking problem fundamentally. You explore all valid partitions. DP enters as an optimization: precompute palindromes to avoid redundant work. Some solutions use DP with memoization on remaining string length, but the classic approach is backtracking with a palindrome lookup table.
What's the most common mistake on this problem?+
Forgetting to restore the result list after each recursive call. You build a partition path, recurse, then pop the last element before trying the next palindrome cut. Missing the pop means your final result contains duplicate or malformed partitions.
Can I solve this without precomputing palindromes?+
Technically yes, but it's slow. You can check palindromes on the fly using a helper function, but you'll recheck the same substrings multiple times. For a 10-character string, you might call the palindrome check hundreds of times. Precomputation costs memory but runs in O(n^2 + output size) instead of much worse.
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