LeetCode 3344: Maximum Sized Array Solution

Master LeetCode problem 3344 (Maximum Sized Array), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3344. Maximum Sized Array

Medium

Binary Search Bit Manipulation

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We can iterate through the array and keep track of the current window's size. We also need to maintain the maximum window size seen so far. We can achieve this by using two pointers: left and right to define the window.

Here are the detailed steps:

Initialize two pointers, left and right, both pointing to the beginning of the array.
Use a while loop to iterate through the array:
- Increment right pointer until the current window satisfies the condition.
- Update the maximum window size if needed.
- Increment left pointer until the current window no longer satisfies the condition.
Keep track of the maximum window size seen so far.

At the end of the iteration, the maximum window size seen will be the answer.

Time Complexity: O(n) where n is the number of elements in the input array. We only iterate through the array once.

Space Complexity: O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int maximumSizedArray(int[] nums) {
        int n = nums.length;
        int left = 0, right = 0;
        int maxWindowSize = 0;
        
        while (right < n) {
            // Expand the window to the right
            while (right < n && isValidWindow(nums, left, right)) {
                right++;
            }
            
            // Update maxWindowSize
            maxWindowSize = Math.max(maxWindowSize, right - left);
            
            // Shrink the window from the left
            while (left < right && !isValidWindow(nums, left, right)) {
                left++;
            }
        }
        
        return maxWindowSize;
    }
    
    private boolean isValidWindow(int[] nums, int left, int right) {
        // Add your condition here
        return true;
    }
}

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