Number of Substrings Containing All Three Characters
A medium-tier problem at 73% community acceptance, tagged with Hash Table, String, Sliding Window. Reported in interviews at DE Shaw and 1 others.
This problem shows up when DE Shaw and PayPal want to see if you can think beyond brute force on string counting tasks. You're given a string and need to count substrings that contain at least one 'a', one 'b', and one 'c'. The naive approach times out fast. Most candidates either miss the sliding window trick or implement it wrong on the first try. If this problem lands in your live assessment and you blank on the pattern, StealthCoder solves it invisibly and you move forward. The acceptance rate sits at 73 percent, which means plenty of people get it, but plenty also don't.
Companies that ask "Number of Substrings Containing All Three Characters"
Number of Substrings Containing All Three Characters 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 counting problem, not a validation problem. You can't just check if a substring is valid. Instead, you count how many valid substrings end at each position. Sliding window works here, but it's the two-pointer flavor where you maintain a window containing all three characters, then expand right and count. For each right pointer position, every substring that starts from the leftmost valid position up to the current left pointer and ends at right is valid. Common mistake: students implement a standard sliding window that shrinks too aggressively and misses the count. Another trap: confusing this with 'exactly three characters' problems. Hash Table or frequency array tracks character counts. Once you see the pattern, it's fast. If you hit this live and need a safety net, StealthCoder delivers the working solution in seconds.
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Number of Substrings Containing All Three Characters 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.
Number of Substrings Containing All Three Characters interview FAQ
Is this really a medium or harder?+
The 73 percent acceptance rate suggests it leans easier-medium. The trick is pattern recognition, not complex logic. Most difficulty comes from implementing the count correctly on first pass. If you've drilled sliding window recently, you'll see it fast.
Do I need both Hash Table and Sliding Window?+
You need sliding window as the core strategy. Hash Table (or a simple frequency array) is the auxiliary data structure to track which characters are in your current window. You can solve it with just one approach, but both topics listed reflect the full solution shape.
What's the trap that kills most candidates?+
Counting wrong. Students shrink the window correctly but don't realize that once a window is valid, every left-side boundary position up to the current left pointer generates a new valid substring. That's where the count jump happens and where logic errors creep in.
Will De Shaw or PayPal ask this exact problem?+
These two companies have reportedly asked it. Given the acceptance rate and medium difficulty, it's not rare. It's a solid screening filter for substring logic and sliding window fluency. Seeing it reported from real companies means it's worth drilling before an OA there.
How do I practice this without just memorizing?+
Write out a small example by hand. Mark which substrings contain 'a', 'b', and 'c'. Then trace your sliding window logic step by step to match your manual count. The aha moment comes when you see why the count jumps at certain positions, not when you copy a solution.
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