Minimum Swaps to Group All 1's Together
Given two strings s
and t
of lengths m
and n
respectively, return the minimum window
substring
of s
such that every character in t
(including duplicates) is included in the window. If there is no such substring, return the empty string ""
.
The testcases will be generated such that the answer is unique.
Example 1:
Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
Example 2:
Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.
Example 3:
Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.
Constraints:
m == s.length
n == t.length
1 <= m, n <= 105
s
andt
consist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n)
time?
Solution
We will use a sliding window to find the minimum window.
We wish to use a charmap
to store the number of characters seen in s
, and we will use two indices to represent an interval (left inclusive, right exclusive) of the window
.
To ensure that all characters in t
is included, we can initialize charmap
to negative counts for each character in t
, so that when we iterate s
and increase the counter for each c
, we know that charmap[c] >= 0
means that s
contains enough c
to satisfy the condition.
So the window is valid when every value in charmap
is non-negative, thus we can find the smallest window.
Notice that our window interval is initialized with (-m, 0)
which has a length m
but will produce ""
if we were to obtain s[-m:0]
.
Implementation
1def minWindow(self, s: str, t: str) -> str:
2 m, n = len(s), len(t)
3 if m < n:
4 return ""
5 charmap = defaultdict(int)
6 for c in t:
7 charmap[c] -= 1
8 window, l = (-m, 0), 0
9
10 for r in range(m):
11 charmap[s[r]] += 1
12 r += 1
13 while min(charmap.values()) >= 0: # valid
14 if window[1] - window[0] >= r - l: # smaller than current
15 window = (l, r)
16 charmap[s[l]] -= 1
17 l += 1
18
19 return s[window[0] : window[1]]
What is an advantages of top-down dynamic programming vs bottom-up dynamic programming?
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
What is the space complexity of the following code?
1int sum(int n) {
2 if (n <= 0) {
3 return 0;
4 }
5 return n + sum(n - 1);
6}
What are the two properties the problem needs to have for dynamic programming to be applicable? (Select 2)
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