Take Gifts From the Richest Pileeasy

Problem Statement

You're presented with several piles of gifts, with each pile containing a certain number of gifts. Every second, you'll engage in the following activity:

• Pick the pile that contains the highest number of gifts. If multiple piles share this distinction, you can select any of them.
• Compute the square root of the number of gifts in the selected pile, and then leave behind that many gifts (rounded down). Take all the other gifts from this pile.
• You'll do this for "k" seconds. The objective is to find out how many gifts would still remain after these "k" seconds.

Examples

1. • Input: gifts = [4, 9, 16], k = 2
   • Expected Output: 11
   • Justification:
     1. Take from third pile (16 gifts): leave ( ) = 4 gifts, take 12. Remaining gifts = [4, 9, 4]
     2. Take from second pile (9 gifts): leave ( ) = 3 gifts, take 6. Remaining gifts = [4, 3, 4]
2. • Input: gifts = [1, 2, 3], k = 1
   • Expected Output: 4
   • Justification:
     1. Take from third pile (3 gifts): leave ( ) = 1 gift (rounded down), take 2. Remaining gifts = [1, 2, 1]
3. • Input: gifts = [25, 36, 49], k = 3
   • Expected Output: 18
   • Justification:
     1. Take from third pile (49 gifts): leave ( ) = 7 gifts, take 42. Remaining gifts = [25, 36, 7]
     2. Take from second pile (36 gifts): leave ( ) = 6 gifts, take 30. Remaining gifts = [25, 6, 7]
     3. Take from first pile (25 gifts): leave ( ) = 5 gifts, take 20. Remaining gifts = [5, 6, 7]

Constraints:

• 1 <= gifts.length <= 10³
• 1 <= gifts[i] <= 10⁹
• 1 <= k <= 10³

Try it yourself

Try solving this question here:

🎯 STRICT STANDOUT — Take Gifts From the Richest Pile

1. Why / judgment

Each second: replace richest pile with floor(sqrt(pile)). After k seconds, sum gifts left. Recognition: repeated touch current maximum -> max-heap. k log n per op. When-NOT: k huge and need formula — piles drop fast under sqrt, but still simulate with heap unless math jump-ahead required.

2. Big-O derivation (K11)

Each of k ops: heap pop/push O(log n) -> O(k log n).
Init heap O(n). Final sum O(n).
n,k up to 1e5 OK with log factors.

3. Pattern + when-NOT (K12)

Name: MAX-HEAP REPEATED REPLACE-WITH-f(max)

Recognition: always modify current maximum; aggregate after k steps.

When-NOT: Need min each time -> min-heap. Offline all queries -> different. Only sum after ops -> still heap.

4. Edge hand-run (K13)

k=0 -> sum original
all piles 1 -> sqrt stays 1
single pile -> iterate sqrt k times O(k)

5. Interviewer follow-ups (model answers)

Q1. Why heap not sort each time?
A: Resort O(n log n) per second -> O(kn log n) worse.

Q2. Integer sqrt?
A: floor(sqrt) — use int isqrt; floating error care for large ints.

Q3. When piles become 0?
A: sqrt(0)=0; still valid.

6. Short drills

Drill: piles=[25,64,9] k=4 simulate manually
Drill: prove heap top is always current max

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