This one took me a minute to even parse what they were asking.
First, recognize that for each index i, C[i] can be any integer from 0 to A[i]-1, so the maximum distinct values is the size of the union of these ranges. Since the ranges are contiguous from 0, the union is simply [0, max(A)-1], so the answer is max(A). Then, justify why this is achievable by setting B[i] = i for i < max(A) and B[i] = 0 for the rest, ensuring all values 0..max(A)-1 appear.
Pro tip: Mention that the problem reduces to finding the maximum element, which is O(n) time and O(1) space, and emphasize that no sorting or complex data structures are needed. This shows you can simplify problems to their core.
For each i, C[i] = B[i] mod A[i] can be any integer from 0 to A[i]-1 because B[i] can be chosen freely. So the set of possible values for index i is exactly {0, 1, ..., A[i]-1}.
The overall set of possible distinct values is the union of these ranges. Since each range starts at 0, the union is simply {0, 1, ..., max(A)-1}, which has size max(A).
To achieve all values from 0 to max(A)-1, assign B[i] = i for indices where i < max(A), and B[i] = 0 for the rest. This ensures each value 0..max(A)-1 appears as C[i] for some i.
The maximum number of distinct values is simply the maximum element in A. Find it by a single pass through the array, which takes O(n) time and O(1) extra space.
Time complexity: O(n). Space complexity: O(1). Handle edge cases: if n=0 (though not possible for positive length), or if all elements are 1, then max=1 and only value 0 is possible.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Felt more approachable than the first problem.
First, clarify the problem: you can choose one letter from each string and delete any number of its occurrences. The goal is to determine if the resulting strings can be anagrams. The key insight is that after deletion, the only letters that can differ in frequency are the two chosen letters; all other letters must have identical frequencies in both strings. So, check if the strings differ in at most two letters, and if so, whether the frequency differences can be resolved by deleting occurrences of the chosen letters.
Pro tip: Mention that the chosen letters can be the same or different, and that you can delete zero occurrences. This shows attention to edge cases and can simplify the solution.
Restate the problem: pick one letter from s and one from t (possibly the same), and delete any number of occurrences of those letters from their respective strings. The resulting strings must be anagrams.
For the strings to become anagrams, all letters except the two chosen ones must have equal frequencies in s and t. Thus, the set of letters with differing frequencies must be a subset of the two chosen letters.
Compute frequency arrays for s and t. Find all letters where frequencies differ. If more than two letters differ, return false. If zero differ, return true (already anagrams). If one or two differ, proceed.
If exactly one letter differs, it must be chosen from both strings (or one string and the other chosen letter has zero difference). If two letters differ, say a and b, then the excess of a in one string must equal the deficit of b in the other, and vice versa, so that deleting occurrences of a and b can balance frequencies.
The algorithm runs in O(n + m) time and O(1) space (since alphabet size is 26), which is optimal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.