Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
You are given an integer array gifts denoting the number of gifts in various piles. Every second, you do the following:
Return the number of gifts remaining after k seconds.
Example 1:
Input: gifts = [25,64,9,4,100], k = 4 Output: 29 Explanation: The gifts are taken in the following way: - In the first second, the last pile is chosen and 10 gifts are left behind. - Then the second pile is chosen and 8 gifts are left behind. - After that the first pile is chosen and 5 gifts are left behind. - Finally, the last pile is chosen again and 3 gifts are left behind. The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29.
Example 2:
Input: gifts = [1,1,1,1], k = 4 Output: 4 Explanation: In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile. That is, you can't take any pile with you. So, the total gifts remaining are 4.
Constraints:
1 <= gifts.length <= 1031 <= gifts[i] <= 1091 <= k <= 103Problem summary: You are given an integer array gifts denoting the number of gifts in various piles. Every second, you do the following: Choose the pile with the maximum number of gifts. If there is more than one pile with the maximum number of gifts, choose any. Reduce the number of gifts in the pile to the floor of the square root of the original number of gifts in the pile. Return the number of gifts remaining after k seconds.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[25,64,9,4,100] 4
[1,1,1,1] 4
remove-stones-to-minimize-the-total)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2558: Take Gifts From the Richest Pile
class Solution {
public long pickGifts(int[] gifts, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int v : gifts) {
pq.offer(v);
}
while (k-- > 0) {
pq.offer((int) Math.sqrt(pq.poll()));
}
long ans = 0;
for (int v : pq) {
ans += v;
}
return ans;
}
}
// Accepted solution for LeetCode #2558: Take Gifts From the Richest Pile
func pickGifts(gifts []int, k int) (ans int64) {
h := &hp{gifts}
heap.Init(h)
for ; k > 0; k-- {
gifts[0] = int(math.Sqrt(float64(gifts[0])))
heap.Fix(h, 0)
}
for _, x := range gifts {
ans += int64(x)
}
return
}
type hp struct{ sort.IntSlice }
func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (hp) Pop() (_ any) { return }
func (hp) Push(any) {}
# Accepted solution for LeetCode #2558: Take Gifts From the Richest Pile
class Solution:
def pickGifts(self, gifts: List[int], k: int) -> int:
h = [-v for v in gifts]
heapify(h)
for _ in range(k):
heapreplace(h, -int(sqrt(-h[0])))
return -sum(h)
// Accepted solution for LeetCode #2558: Take Gifts From the Richest Pile
impl Solution {
pub fn pick_gifts(gifts: Vec<i32>, k: i32) -> i64 {
let mut h = std::collections::BinaryHeap::from(gifts);
let mut ans = 0;
for _ in 0..k {
if let Some(mut max_gift) = h.pop() {
max_gift = (max_gift as f64).sqrt().floor() as i32;
h.push(max_gift);
}
}
for x in h {
ans += x as i64;
}
ans
}
}
// Accepted solution for LeetCode #2558: Take Gifts From the Richest Pile
function pickGifts(gifts: number[], k: number): number {
const pq = new MaxPriorityQueue<number>();
gifts.forEach(v => pq.enqueue(v));
while (k--) {
let v = pq.dequeue();
v = Math.floor(Math.sqrt(v));
pq.enqueue(v);
}
let ans = 0;
while (!pq.isEmpty()) {
ans += pq.dequeue();
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.