Editorial for Singularity Cup P2 - Reverse Substring Partitioning


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Author: Mystical

If S_1 \ne S_N, we cannot perform any RSPs at all so the answer is just N.

Otherwise, the key observation is that groups of consecutive letters can all be merged into one letter, and we can treat S as circular because the letters at the very start of the string will end up being adjacent to the letters at the very end of the string once an RSP is performed.

Thus, the problem reduces to counting the number of groups of consecutive letters while treating S as a circle.

Time complexity: \mathcal{O}(N)


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