1066MediumArrayTwo PointersGreedySorting

How Many Places Can Be Bettered

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Trains the technique from

LeetCode 2592Maximize Greatness of an Array

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A list of whole numbers reads values. Rearrange the same values into any order you like and lay the rearrangement alongside the original.

A position betters the original when the rearrangement's value there is strictly larger than the original's.

Return the most positions that can be made to better the original at once.

Examples

Example 1

Input: values = [2, 2, 2]
Output: 0

Every value is the same, so no rearrangement puts a larger value anywhere.

Example 2

Input: values = [1, 1, 2, 2]
Output: 2

The two twos can be laid over the two ones, bettering both of those positions, but nothing is left to better a two.

Example 3

Input: values = [5, 4, 3, 2, 1]
Output: 4

Shifting every value one place along leaves each position holding the next value up, which betters all but one.

Constraints

1 <= values.length <= 10^5
0 <= values[i] <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def maximize_greatness(values: list[int]) -> int:

Java

public int maximizeGreatness(int[] values)

The intended approach

Time O(n log n) for the sort, then one pass.Space O(n) for the sorted values.

Only the multiset of values matters, so sort them. Then the question becomes how many values can be matched to strictly smaller ones, each value used once on each side. Sort the values and walk them with two markers, one over the values being beaten and one over the values doing the beating. Step the beating marker forward; whenever the value it stands on is strictly larger than the value under the other marker, that is one more position bettered and the beaten marker steps on too. That greedy is right because the smallest unbeaten value is the easiest to beat, so beating something else instead can never help: any matching can be rewritten to match the smallest available value without losing a pairing.

Pythonentry point: maximize_greatness

def maximize_greatness(values):
    order = sorted(values)
    beaten = 0
    for value in order:
        if value > order[beaten]:
            beaten += 1
    return beaten

The obvious version

Pair the values off by hand. Sort them, then take each value in turn as a beater and hunt the unused values from the smallest upwards for the first one it strictly beats, marking both used. Count the pairings made. Costs a hunt per value rather than a single pass with two markers.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_maximize_greatness(values):
    order = sorted(values)
    n = len(order)
    used = [False] * n
    count = 0
    for at in range(n):
        for other in range(n):
            if used[other]:
                continue
            if order[at] > order[other]:
                used[other] = True
                count += 1
                break
    return count

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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