Simple without loops O(1) TC&SC solution||EASY SOLUTION✅
IntuitionSee they have asked the min num of decks that can be stored, so what you can do is without running a loop find n*n and maxWeigbt/w (to find the num of
# Intuition
See they have asked the min num of decks that can be stored, so what you can do is without running a loop find n*n and maxWeigbt/w (to find the num of decks which can be stored in the deck)

# Approach
<!-- Describe your approach to solving the problem. -->
declare the variables to calculate

# Complexity
- Time complexity:
- O(1) : No extra space used(no loops)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1): No extra space like vector, etc is used

# Code
```cpp []
class Solution {
public:
    int maxContainers(int n, int w, int maxWeight) {
        int totalDeck = n * n;
        int weight = maxWeight/w;

        return min(totalDeck, weight);
    }
};
```
['Math', 'C++']