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Sliding Window Maximum

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C program to find the maximum element in every sliding window of size k in an array. Given an array of integers and a window size k, slide the window from left to right and return an array containing the maximum element in each window position.

Example 1
    
  
  • Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
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  • Output: [3,3,5,5,6,7]
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  • Explanation: 
    • Window [1,3,-1] -> maximum is 3. 
    • Window [3,-1,-3] -> maximum is 3. 
    • Window [-1,-3,5] -> maximum is 5. 
    • Continue sliding and finding maximum in each window.
    •
Example 2
    
  
  • Input: nums = [1], k = 1
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  • Output: [1]
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  • Explanation: 
    • Only one element in array and window size is 1. 
    • The maximum (and only) element in the window is 1. 
    • Result array contains single element [1].
    •
Constraints
    
  
  • 1 <= nums.length <= 10^5
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  • -10^4 <= nums[i] <= 10^4
    • 
  
  • 1 <= k <= nums.length
    • 
  
  • Time Complexity: O(n) using deque approach
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  • Space Complexity: O(k) for storing indices in deque
    •
ArraysDequeAdobeTutorix
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Hint
Solution

Solution Hints

    
  
  • Use a deque (implemented with array) to store indices of array elements
    • 
  
  • The deque will store indices in decreasing order of their corresponding values
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  • The front of deque always contains the index of maximum element in current window
    • 
  
  • Remove indices that are out of current window from front of deque
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  • Remove indices from rear of deque if their values are smaller than current element
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  • Add current element's index to rear of deque and store front element as maximum
    •

Steps to solve by this approach:

 Step 1: Initialize a deque to store indices of array elements in decreasing order of their values
 Step 2: Process the first window by adding indices while maintaining decreasing order in deque
 Step 3: The front of deque contains the index of maximum element in current window
 Step 4: For each subsequent element, remove indices that are outside current window from front
 Step 5: Remove indices from rear whose corresponding values are smaller than current element
 Step 6: Add current element's index to rear and record the maximum (front element) in result
 Step 7: Continue until all windows are processed, maintaining O(n) time complexity

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