Strings With No k Consecutive Identical Characters
Reported by candidates from Salesforce's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The mistake that sinks a first attempt on this Salesforce OA, reported February 2026, is brute-forcing or inclusion-exclusion on 26^n strings. n goes up to 10^5, so that dies fast. This is a counting DP in disguise. You track how long the current run of identical characters is, and you count valid strings by length. If you blank on the recurrence mid-assessment, StealthCoder runs invisibly on your screen and can hand you a working solution as a safety net. But the recurrence is short, and you can own it in ten minutes.
The problem
You are given two integers: n: length of string k: maximum allowed consecutive identical characters A string is invalid if it contains k or more consecutive identical characters. Return the total number of strings of length n formed using lowercase English letters such that no character appears k or more times consecutively. Return the answer modulo 10^9 + 7. Function stringsWithNoKConsecutiveIdenticalCharacters(n: int, k: int) → int Examples Example 1 n = 3 k = 2 return = 16250 Total strings of length 3 is 26^3 = 17576. Invalid strings are those with at least one adjacent equal pair. For this case, the valid count is 16250. Constraints 1 <= n <= 10^5 1 <= k < n
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is a DP over length where dp[i] is the number of valid strings of length i. Appending a character that differs from the previous one gives dp[i] = 25 * (dp[i-1] +... + dp[i-k+1]) for i past the first character, with the first character having 26 choices. Cleaner: let f[i] be the count ending in a run of exactly one character start. Use a sliding window sum so each step is O(1), giving O(n) total. The pitfall is the off-by-one on k. Invalid means k or more in a row, so runs are capped at k-1. Check the example: n=3, k=2 gives 26*25*25 = 16250. Also apply modulo 10^9+7 on every subtraction when you maintain the window, or you'll get negatives. If the recurrence slips under pressure, StealthCoder is the hedge on the live OA.
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Strings With No k Consecutive Identical Characters FAQ
How hard is this Salesforce OA question really?+
Medium. The idea is simple once you see it as counting by run length, but the off-by-one on k and the modulo handling on a sliding sum trip people up. Expect a clean O(n) DP, not anything exotic.
What's the trick to solving it?+
Build strings left to right and track the current run length, capped at k-1. A new run starts with one of 25 different letters, so dp[i] sums the previous k-1 states times 25. A running window sum keeps it linear.
How do I verify my answer fast?+
Use the given example. With n=3 and k=2, no two adjacent characters can match, so the count is 26*25*25 = 16250. Also test k=1 edge ideas and a tiny n by hand before submitting.
What time complexity do I need?+
n goes up to 10^5, so you need O(n) or O(n log n). A naive sum over the last k states each step is O(nk) and can time out. Maintain a prefix sum or running window to keep it linear.
How do I prepare in 48 hours?+
Practice two or three counting DPs with run-length limits and a sliding window sum. Write the recurrence on paper, then code it with modulo on every add and subtract. Focus on boundary cases, not on memorizing code.