Sorting is the key move here and I think I would've fumbled around with it if I hadn't seen similar interval/grouping problems before.
Sort the array first, then use a greedy algorithm to form groups by scanning from smallest to largest and starting a new group whenever the current element exceeds the smallest element in the current group by more than maxdiff. This yields the minimum number of groups because sorting ensures that any valid grouping must respect the sorted order, and the greedy choice is optimal.
Pro tip: After presenting the greedy solution, mention that the problem is equivalent to covering the sorted points with intervals of length maxdiff, and that the greedy algorithm is optimal. Also, discuss edge cases like empty array, maxdiff=0, and negative numbers.
Confirm understanding of the problem: partition array into minimum groups where within each group, max-min <= maxdiff. Ask about constraints (array size, value range) and edge cases (empty array, maxdiff negative).
Explain that sorting is the key first step because it allows a linear scan to form groups optimally. Sorting brings elements with close values together, simplifying the grouping condition.
Iterate through the sorted array, keeping track of the smallest element in the current group. If the current element minus that smallest exceeds maxdiff, start a new group and increment the count.
Argue that the greedy approach is optimal: any valid group must contain elements within a range of maxdiff, and by sorting, the greedy grouping minimizes the number of groups because it maximizes the size of each group.
State time complexity O(n log n) due to sorting, and space complexity O(1) if sorting in-place or O(n) if using extra space. Mention that the grouping scan is O(n).
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