Solution Explanation:

The problem asks to find the number of rectangles that can form the largest possible square. The largest square from a rectangle [l, w] has a side length of min(l, w). The solution involves finding the maximum of these minimum side lengths across all rectangles and then counting how many rectangles can produce a square of that maximum side length.

Approach:

1. Initialization: Initialize two variables:

   • maxLen: To store the maximum side length of a square found so far (initialized to 0).
   • count: To store the number of rectangles that can form a square with maxLen (initialized to 0).

2. Iteration: Iterate through each rectangle [l, w] in the rectangles array:

   • Calculate sideLen = min(l, w). This represents the maximum side length of a square that can be formed from the current rectangle.

3. Comparison and Update:

   • If sideLen > maxLen, it means we've found a larger square. Update maxLen to sideLen and reset count to 1 (only one rectangle so far has this maximum side length).
   • If sideLen == maxLen, it means we've found another rectangle that can form a square of the current maximum size. Increment count by 1.

4. Return: After iterating through all rectangles, return count.

Time and Space Complexity:

• Time Complexity: O(n), where n is the number of rectangles. We iterate through the array of rectangles once.
• Space Complexity: O(1). We only use a few constant extra variables.

Code Implementation (Python):

def countGoodRectangles(rectangles):
    max_len = 0
    count = 0
    for l, w in rectangles:
        side_len = min(l, w)
        if side_len > max_len:
            max_len = side_len
            count = 1  # Reset count for the new max
        elif side_len == max_len:
            count += 1
    return count
 

Code Implementation (Java):

class Solution {
    public int countGoodRectangles(int[][] rectangles) {
        int maxLen = 0;
        int count = 0;
        for (int[] rect : rectangles) {
            int sideLen = Math.min(rect[0], rect[1]);
            if (sideLen > maxLen) {
                maxLen = sideLen;
                count = 1;
            } else if (sideLen == maxLen) {
                count++;
            }
        }
        return count;
    }
}

Code Implementation (C++):

class Solution {
public:
    int countGoodRectangles(vector<vector<int>>& rectangles) {
        int maxLen = 0;
        int count = 0;
        for (const auto& rect : rectangles) {
            int sideLen = min(rect[0], rect[1]);
            if (sideLen > maxLen) {
                maxLen = sideLen;
                count = 1;
            } else if (sideLen == maxLen) {
                count++;
            }
        }
        return count;
    }
};

The other languages (Go, TypeScript, etc.) would follow a very similar structure, maintaining the same time and space complexity. The core logic remains consistent across all implementations.