Minimum Substring Partition of Equal Character Frequency
A medium-tier problem at 39% community acceptance, tagged with Hash Table, String, Dynamic Programming. Reported in interviews at Mitsogo and 0 others.
You're handed a string and asked to partition it into substrings where each substring has a single character appearing at the maximum frequency. The acceptance rate sits at 39 percent, which means most candidates either misread the constraint or build out a greedy solution that hits an edge case. Mitsogo has asked this. If you haven't drilled the exact phrasing, you can blank on what "equal character frequency" actually means during the live OA. StealthCoder reads the problem statement and surfaces a working partition strategy in seconds, invisible to the proctor.
Companies that ask "Minimum Substring Partition of Equal Character Frequency"
Minimum Substring Partition of Equal Character Frequency 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 trap is thinking greedy works. You can't just scan left to right and cut whenever frequencies align. The actual solution requires dynamic programming: for each position, decide whether to cut or extend the current substring based on whether all remaining characters in that substring have the same frequency. Hash Table tracks character counts in your current window. Counting lets you detect when all non-zero frequencies are identical. Many candidates build the frequency map but fail to validate the stopping condition correctly. They cut too early or too late, producing substrings where one character has frequency 3 and another has 2. StealthCoder is the hedge if you hit this live and haven't internalized the DP transition.
Pattern tags
You know the problem.
Make sure you actually pass it.
Minimum Substring Partition of Equal Character Frequency 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.
Minimum Substring Partition of Equal Character Frequency interview FAQ
What does 'equal character frequency' actually mean?+
Every character present in the substring must appear the same number of times. If a substring is 'aabbb', it fails because 'a' appears twice and 'b' appears three times. Valid substrings have all non-zero frequencies equal to one another.
Why doesn't greedy left-to-right cutting work?+
Greedy cuts as soon as frequencies match, but that local choice blocks better partitions later. DP explores all cut positions and finds the minimum. A greedy cut might force you into a substring where frequencies are permanently unequal.
Is this really asked outside Mitsogo?+
Only one company explicitly reported it, so it's lower frequency in the big FAANG rotation. But the pattern (DP plus character counting) appears in many medium-tier problems, so drilling it builds broader skill.
How do you check if a substring has equal character frequencies?+
Count characters in a hash table. Check if all non-zero counts are identical. You can iterate the frequency map once and verify max equals min (ignoring zero-count chars). A set of values works too.
What's the state transition in the DP?+
State: dp[i] = minimum partitions for s[0:i]. For each i, try all j < i. If s[j:i] has equal frequencies, update dp[i] = min(dp[i], dp[j] + 1). Base case: dp[0] = 0.
Want the actual problem statement? View "Minimum Substring Partition of Equal Character Frequency" on LeetCode →