Took me a while to even figure out what was being asked.
Recognize that the problem reduces to selecting a distinct hash value for each index, where the available values for index i are 0 to param[i]-1. Model this as a bipartite matching problem between indices and possible hash values, then find the maximum matching size. Alternatively, sort indices by param[i] and greedily assign the smallest available hash value to each index, which yields the optimal count.
Pro tip: Mention that the greedy approach works because the intervals [0, param[i]-1] are nested when sorted by param[i], so assigning the smallest available value preserves future options. This demonstrates insight into the problem's structure and avoids overcomplicating with full matching algorithms.
Restate the goal: we need to choose secretKey[i] such that hash[i] = secretKey[i] % param[i] are all distinct, and we want to maximize the number of distinct hash values. Note that secretKey[i] can be any integer, so hash[i] can be any integer from 0 to param[i]-1.
For each index i, the set of possible hash values is {0, 1, ..., param[i]-1}. We need to pick one value per index such that all picked values are distinct. This is a bipartite matching problem between indices and hash values.
Sort indices by param[i] in ascending order. For each index, assign the smallest non-negative integer not yet used that is less than param[i]. If no such integer exists, skip that index (it cannot be assigned a unique hash).
Argue that the greedy choice is safe: because intervals are nested, using the smallest available value leaves larger values for indices with larger param[i], maximizing the total number of assignments. This is equivalent to the standard interval scheduling/matching greedy proof.
Sorting takes O(n log n). Assigning values can be done in O(n) using a pointer or a set. Overall O(n log n) time and O(n) space. Mention that a full bipartite matching would be O(n^2) or worse, so greedy is more efficient.
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