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APCS Recursion 2, Problem 3. We'll say that a "pair" in a string is two instances of a char separated by a char. So "AxA" the A's make a pair. Pair's can overlap, so "AxAxA" contains 3 pairs -- 2 for A and 1 for x. Recursively compute the number of pairs in the given string. R23_countPairs("axa") → 1 R23_countPairs("axax") → 2 R23_countPairs("axbx") → 1 ...Save, Compile, Run (ctrl-enter) |
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Difficulty: 300
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