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Sliding Window Maximum – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3 Output: [3,3,5,5,6,7] Explanation: Window position Max --------------- ----- [1 3 -1] -3 5 3 6 7 3 1 [3 -1 -3] 5 3 6 7 3 1 3 [-1 -3 5] 3 6 7 5 1 3 -1 [-3 5 3] 6 7 5 1 3 -1 -3 [5 3 6] 7 6 1 3 -1 -3 5 [3 6 7] 7
Example 2:
Input: nums = [1], k = 1 Output: [1]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 1041 <= k <= nums.length
C++ Sliding Window Maximum LeetCode Solution
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
deque<int> dq;
vector<int> ans;
for (int i=0; i<nums.size(); i++) {
if (!dq.empty() && dq.front() == i-k) dq.pop_front();
while (!dq.empty() && nums[dq.back()] < nums[i])
dq.pop_back();
dq.push_back(i);
if (i>=k-1) ans.push_back(nums[dq.front()]);
}
return ans;
}
};
Java Sliding Window Maximum LeetCode Solution
public static int[] slidingWindowMax(final int[] in, final int w) {
final int[] max_left = new int[in.length];
final int[] max_right = new int[in.length];
max_left[0] = in[0];
max_right[in.length - 1] = in[in.length - 1];
for (int i = 1; i < in.length; i++) {
max_left[i] = (i % w == 0) ? in[i] : Math.max(max_left[i - 1], in[i]);
final int j = in.length - i - 1;
max_right[j] = (j % w == 0) ? in[j] : Math.max(max_right[j + 1], in[j]);
}
final int[] sliding_max = new int[in.length - w + 1];
for (int i = 0, j = 0; i + w <= in.length; i++) {
sliding_max[j++] = Math.max(max_right[i], max_left[i + w - 1]);
}
return sliding_max;
}
Python 3 Sliding Window Maximum LeetCode Solution
import collections
class Solution(object):
def maxSlidingWindow(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
d = collections.deque()
out = []
for i, n in enumerate(nums):
print("i = {}, curr element = {}, d = {} and out = {}".format(i, n, d, out))
while d and nums[d[-1]] < n:
d.pop()
print("\t Popped from d because d has elements and nums[d.top] < curr element")
d.append(i)
print("\t Added i to d")
if d[0] == i - k:
d.popleft()
print("\t Popped left from d because it's outside the window's leftmost (i-k)")
if i>=k-1:
out.append(nums[d[0]])
print("\t Append nums[d[0]] = {} to out".format(nums[d[0]]))
return out
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
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