Sliding Window Maximum.cpp

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.


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]


Example 3:
Input: nums = [1,-1], k = 1
Output: [1,-1]


Example 4:
Input: nums = [9,11], k = 2
Output: [11]


Example 5:
Input: nums = [4,-2], k = 2
Output: [4]
 
Constraints:
1. 1 <= nums.length <= 105
2. -104 <= nums[i] <= 104
3. 1 <= k <= nums.length





SOLUTION:



class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        
        int i,n;
        n=nums.size();
        
        deque<int> dq;
        vector<int> ans;
        
        for(i=0;i<n;i++)
        {
            if(!dq.empty() && dq.front()==i-k)     //Maximum element index is out of sliding window range
                dq.pop_front();
            
            while(!dq.empty() && nums[dq.back()]<nums[i])   //Remove elements that are less than new element
                dq.pop_back();
            
            dq.push_back(i);
            
            if(i>=k-1)
                ans.push_back(nums[dq.front()]);    //Maintain maximum element index at dequeu front
        }
        
        return ans;
        
    }
};