Number of Equal Count Substrings
A medium-tier problem at 45% community acceptance, tagged with Hash Table, String, Sliding Window. Reported in interviews at Cisco and 0 others.
You're prepping for an online assessment and this problem shows up: count substrings where every character appears the same number of times. It looks straightforward until you hit the implementation and realize you need to track character frequencies efficiently across a sliding window. Cisco has asked this one. The acceptance rate sits at 44%, which tells you most candidates either nail it or miss the pattern entirely. If you blank on the sliding window setup during the live OA, StealthCoder surfaces a working solution invisibly so you don't lose the points.
Companies that ask "Number of Equal Count Substrings"
Number of Equal Count 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. Built by a senior engineer who knows the OA is theater. This is the script.
Get StealthCoderThe trick is recognizing this is a sliding window problem where you track character counts, not just check a fixed substring length. Many candidates start with a brute force approach, generate every substring and count characters, which works but tanks on time complexity. The real insight is that you only care about windows where all characters have the same frequency. You shrink and expand the window based on that constraint, using a hash table to maintain character counts. The common pitfall is overthinking: you don't need to store multiple frequency maps. Track counts in a single map and rebuild when needed. This combines hash table logic with string traversal in a way that feels mechanical until you see it once.
Pattern tags
You know the problem.
Make sure you actually pass it.
Number of Equal Count 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. Built by a senior engineer who knows the OA is theater. This is the script. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Number of Equal Count Substrings interview FAQ
Is this problem actually medium difficulty?+
The 44% acceptance rate suggests it's harder than the difficulty label. The sliding window pattern itself is medium, but the frequency-tracking condition trips up candidates who don't immediately see that you need to filter windows by a specific property, not just scan left-to-right.
Do I need a separate frequency tracker for each window?+
No. Use one hash table per attempt. Clear and recount when your window shrinks past a valid boundary. This cuts memory overhead and simplifies the logic. Many candidates waste time maintaining multiple maps.
What's the difference between this and a standard substring-counting problem?+
Standard substring problems iterate length or position. This one iterates character frequency constraints. You're filtering valid windows by a property that isn't length-based, so you can't use a simple two-pointer scan. You need to check the frequency condition at each step.
Will Cisco ask exactly this problem in their assessment?+
Unknown. Cisco has asked this type before, but they may ask a variation. The techniques here, hash table maintenance, sliding window filtering, character frequency tracking, are the real skills being tested. Master the pattern, not the exact problem.
How do I avoid timeout on large inputs?+
Don't rebuild the entire hash table on every iteration. Update counts incrementally as you slide. Use the fact that you only care about windows where all characters share one frequency value. Stop early if a window can't be fixed by further sliding.
Want the actual problem statement? View "Number of Equal Count Substrings" on LeetCode →