Sliding Window Maximum
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# LeetCode #239: Sliding Window Maximum (C/C++).

hard

source: https://leetcode.com/problems/sliding-window-maximum
C/C++ Solution to LeetCode problem 239. Sliding Window Maximum.

## Problem

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.

## Examples

### Example 1:

Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:

1 2 3 4 5 6 7 8 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] <= 104
• 1 <= k <= nums.length

## Solution

Two solutions:

1. Fast. Using deque. We will track the index of the max value for each window. - Iterate through the numbers
• If the first element (index) is outside the window, then we remove it (pop_front)
• We remove all the indexes that have nums < thant the one at the current index.
• We add the new index to the window/deque (push_back).
• Once the index is corresponds to the last element of the window
• We can add the num at the index of first element (front) of the deque to the results, as it holds the max value for the window.
2. Slow. Using multiset (using the greater function object) to sort the elements from greater to lower. - Iterate through the numbers.
• We add the number to the window.
• Once the multiset has the size of the window we take the first element, as it will be the greater one in the window.
• We remove the element that corresponds to the first element of the window.

### Solution 1:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { deque<int> dq; vector<int> result; 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) result.push_back(nums[dq.front()]); } return result; } }; 

### Solution 2:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> result; multiset<int> w; int i = 0; for (i=0; i<k-1; i+=1) w.insert(nums[i]); while (i<nums.size()) { w.insert(nums[i]); auto n = w.rbegin(); result.push_back(*n); w.erase(w.lower_bound(nums[i-k+1])); i += 1; } return result; } };