src.most_water_container

Classes

Solution()

class src.most_water_container.Solution

maxArea(height: list[int]) → int

Thought process

• Once again basic two point setup so l, r = 0, len(height) - 1 and while l < r

• The area is the base length times $ r - l $ the height of the smaller wall \(\text{min}(h[l], h[r])\)

• We can move the pointer with the smaller wall since moving the pointer with the larger wall will only decrease the area

• We repeat this process until the pointers cross over

Notes

• time complexity: \(O(n)\) just a single loop

• space complexity: \(O(1)\)

maxAreaBruteForce(height: list[int]) → int

Thought process

• We can brute force this by checking all possible areas

• We can do this by having two loops and checking all possible combinations of areas

• We can then return the maximum area

Notes

• time complexity: \(O(n^2)\) ⚠️ does not pass time limit

• space complexity: \(O(1)\)