Number of Wonderful Substrings
A medium-tier problem at 67% community acceptance, tagged with Hash Table, String, Bit Manipulation. Reported in interviews at DE Shaw and 0 others.
Number of Wonderful Substrings is a medium-difficulty string problem that shows up in DE Shaw interviews. You're given a string of lowercase letters and need to count substrings where every character appears an even number of times (including zero). The acceptance rate hovers around 67%, which means most people who see it solve it, but if you blank on the bitmask trick during your live assessment, you're sunk. StealthCoder runs invisibly during screen share and surfaces the working solution in seconds if you hit this problem and freeze.
Companies that ask "Number of Wonderful Substrings"
Number of Wonderful Substrings 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. Made for the engineer who has done the work but might still blank with a webcam pointed at him.
Get StealthCoderThe naive approach is an O(n^3) brute force that counts characters in every substring. That fails on length. The real solution uses a bitmask to track character parity as you iterate through the string. Each bit represents whether a character has appeared an odd or even number of times. Two substrings with the same bitmask state mean the characters between them all appear an even number of times. Hash the bitmasks and count collisions. This drops complexity to O(n) and requires only basic bit manipulation and hash table lookups. Most candidates either miss the bitmask insight entirely or botch the XOR logic under pressure. If this problem hits your live OA and you blank on the pattern, StealthCoder solves it in seconds.
Pattern tags
You know the problem.
Make sure you actually pass it.
Number of Wonderful Substrings 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. Made for the engineer who has done the work but might still blank with a webcam pointed at him. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Number of Wonderful Substrings interview FAQ
Why does the bitmask approach work?+
Each bit represents a character's parity (odd or even count). If two positions have the same bitmask, all characters between them appeared an even number of times. XOR the current character's bit with your running bitmask to toggle its parity. Hash all bitmasks and count how many substrings end at each position with matching prefix states.
Is this still asked at companies like DE Shaw?+
Yes. De Shaw appears in the input data for this problem. It's a classic medium-difficulty coding interview question that tests both string iteration and bit manipulation. The acceptance rate of 67% suggests it's a fair, solvable problem if you know the pattern.
What's the trap in the brute force?+
Checking every substring's character counts explicitly is O(n^3) or O(n^2) at best. On strings longer than 100 characters, you'll time out. The hash table and bitmask reduce it to O(n) because you compute parity once as you scan, not repeatedly for each substring.
How does bit manipulation fit into a string problem?+
The string only contains lowercase letters, so 26 bits suffice to represent which characters have odd counts. Flipping a bit is a single XOR operation. This compact state representation lets you hash it and count collisions efficiently, avoiding nested loops over character frequencies.
Will I have time to code this under pressure?+
If you know the bitmask pattern, yes. It's roughly 10-15 lines of code: initialize a bitmask, iterate the string, XOR each character's bit, hash the bitmask, and count. If you don't know the trick, you'll probably timeout or submit a brute force that fails. That's where StealthCoder hedges you during the live assessment.
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