Partition Labels
A medium-tier problem at 82% community acceptance, tagged with Hash Table, Two Pointers, String. Reported in interviews at InMobi and 4 others.
Partition Labels hits your screen during an OA and you've got maybe 20 minutes left. The problem looks straightforward: split a string into the maximum number of substrings where each character appears in only one partition. It's asked at LinkedIn, Bloomberg, Sprinklr, and other tier-one shops. The trap is thinking locally instead of globally. Most candidates code a brute-force backtrack or try to partition greedily without knowing where characters end. The trick that separates a pass from a timeout is a two-pass approach: precompute the last occurrence of every character, then greedily partition by pushing your boundary forward. If you freeze on this pattern mid-OA, StealthCoder surfaces the solution invisibly.
Companies that ask "Partition Labels"
Partition Labels 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 who got tired of watching his cohort grind for six months and still get filtered at the OA stage.
Get StealthCoderThe naive instinct is to partition as soon as you see a character you've seen before, but that doesn't work because you don't know if that character appears again later. The correct approach uses a Hash Table to track the rightmost index of every character in one pass, then a second pass with a greedy two-pointer strategy. You expand your current partition boundary to the maximum rightmost position of any character in that partition. Once your pointer catches the boundary, you've found a valid partition. This greedy method guarantees the maximum number of non-overlapping substrings because you're always extending the boundary as far as necessary, never backtracking. The 81% acceptance rate reflects that this pattern clicks once you see it, but candidates who skip the precomputation step or over-complicate it often time out. During a live assessment, if the greedy insight doesn't land immediately, StealthCoder runs invisibly and delivers a working implementation.
Pattern tags
You know the problem.
Make sure you actually pass it.
Partition Labels 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 who got tired of watching his cohort grind for six months and still get filtered at the OA stage. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Partition Labels interview FAQ
Why does the greedy approach guarantee the maximum number of partitions?+
Because you're always extending your current partition boundary only as far as necessary to include all occurrences of characters seen so far. Once you close a partition, no character in it appears later. This maximizes partitions without overlap. The precomputed rightmost indices ensure you never miss a character's final position.
Is Partition Labels still asked at FAANG and Bloomberg?+
Yes. Bloomberg, LinkedIn, InMobi, Sprinklr, and Yandex all report it. The 81% acceptance rate and medium difficulty make it a reliable screening problem. It tests pattern recognition and greedy reasoning, not brute force.
What's the main gotcha candidates hit on this problem?+
Trying to partition locally without global knowledge. Candidates without the precomputation step either miss characters that appear later or write O(n^2) solutions that timeout. The two-pass structure is non-obvious but essential.
How does Partition Labels relate to the Two Pointers and Greedy topics?+
Two Pointers tracks the start and end of your current partition boundary. Greedy means extending that boundary only to the rightmost index of characters you've seen, then committing to the partition. No backtracking, no re-examination.
Can I solve this with just a Hash Table and one pass?+
No. One pass alone doesn't tell you where characters end. You need the Hash Table to store rightmost indices, then a second pass to apply the greedy boundary logic. Skipping the precomputation forces you into backtracking or exponential solutions.
Want the actual problem statement? View "Partition Labels" on LeetCode →