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Use recursion to search an array for a 3 next to a 3. Returns true if nums has a 3 next to a 3 somewhere and false otherwise. nums will have at least one element has3nextTo3([1, 2, 3, 3], 0) → true has3nextTo3([1, 2, 2, 3], 0) → false has3nextTo3([3], 0) → false ...Save, Compile, Run (ctrl-enter) |
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