Koko Eating Bananas
Koko loves to eat bananas. There are n
piles of bananas, the ith
pile has piles[i]
bananas. The guards have gone and will come back in h
hours.
Koko can decide her bananas-per-hour eating speed of k
. Each hour, she chooses some pile of bananas and eats k
bananas from that pile. If the pile has less than k
bananas, she eats all of them instead and will not eat any more bananas during this hour.
Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.
Return the minimum integer k
such that she can eat all the bananas within h
hours.
Example 1:
Input: piles = [3,6,7,11], h = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], h = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], h = 6
Output: 23
Constraints:
1 <= piles.length <= 104
piles.length <= h <= 109
1 <= piles[i] <= 109
Solution
We want to apply the binary search template, and implement the feasible
function.
Here, the feasible function is whether Koko can_finish_eating
all piles within h
hours while eating at speed k
per hour.
Since Koko eats at only one pile during each hour, ceil(float(p)/k)
is the time Koko takes to finish pile p
.
Note that p/k
does not work here because we want a whole number of hours so we needed to round up p/k
.
Therefore, the feasiblity is determined by whether Koko's hours_used
is within h
hours, where hours_used
is the total hours to finish all piles.
Recall example 1 with Input: piles = [3,6,7,11], h = 8
, and Output: 4
, the can_finish_eating
function returns the following results.
Implementation
1def can_finish_eating(piles, h, k):
2 hours_used = 0
3 for p in piles:
4 hours_used += ceil(float(p)/k)
5 return hours_used <= h
6
7def minEatingSpeed(piles, h):
8 left, right = 1, 1000000000 # 10^9 max length of the piles
9 ans = -1
10 while left <= right:
11 mid = (left + right) // 2
12 if can_finish_eating(piles, h, mid):
13 ans = mid
14 right = mid - 1
15 else:
16 left = mid + 1
17 return ans
How does merge sort divide the problem into subproblems?
What's the output of running the following function using input [30, 20, 10, 100, 33, 12]
?
1def fun(arr: List[int]) -> List[int]:
2 import heapq
3 heapq.heapify(arr)
4 res = []
5 for i in range(3):
6 res.append(heapq.heappop(arr))
7 return res
8
1public static int[] fun(int[] arr) {
2 int[] res = new int[3];
3 PriorityQueue<Integer> heap = new PriorityQueue<>();
4 for (int i = 0; i < arr.length; i++) {
5 heap.add(arr[i]);
6 }
7 for (int i = 0; i < 3; i++) {
8 res[i] = heap.poll();
9 }
10 return res;
11}
12
1class HeapItem {
2 constructor(item, priority = item) {
3 this.item = item;
4 this.priority = priority;
5 }
6}
7
8class MinHeap {
9 constructor() {
10 this.heap = [];
11 }
12
13 push(node) {
14 // insert the new node at the end of the heap array
15 this.heap.push(node);
16 // find the correct position for the new node
17 this.bubble_up();
18 }
19
20 bubble_up() {
21 let index = this.heap.length - 1;
22
23 while (index > 0) {
24 const element = this.heap[index];
25 const parentIndex = Math.floor((index - 1) / 2);
26 const parent = this.heap[parentIndex];
27
28 if (parent.priority <= element.priority) break;
29 // if the parent is bigger than the child then swap the parent and child
30 this.heap[index] = parent;
31 this.heap[parentIndex] = element;
32 index = parentIndex;
33 }
34 }
35
36 pop() {
37 const min = this.heap[0];
38 this.heap[0] = this.heap[this.size() - 1];
39 this.heap.pop();
40 this.bubble_down();
41 return min;
42 }
43
44 bubble_down() {
45 let index = 0;
46 let min = index;
47 const n = this.heap.length;
48
49 while (index < n) {
50 const left = 2 * index + 1;
51 const right = left + 1;
52
53 if (left < n && this.heap[left].priority < this.heap[min].priority) {
54 min = left;
55 }
56 if (right < n && this.heap[right].priority < this.heap[min].priority) {
57 min = right;
58 }
59 if (min === index) break;
60 [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61 index = min;
62 }
63 }
64
65 peek() {
66 return this.heap[0];
67 }
68
69 size() {
70 return this.heap.length;
71 }
72}
73
74function fun(arr) {
75 const heap = new MinHeap();
76 for (const x of arr) {
77 heap.push(new HeapItem(x));
78 }
79 const res = [];
80 for (let i = 0; i < 3; i++) {
81 res.push(heap.pop().item);
82 }
83 return res;
84}
85
Solution Implementation
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Which of the following method should we use to solve this problem?
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