Container With Most Water in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'Container With Most Water' problem ask you to find?

It asks to find two lines in an array that, together with the x-axis, form a container that holds the maximum amount of water.

intermediate

Why do we use two pointers in the 'Container With Most Water' problem?

Two pointers start at both ends and move inward to check all possible containers efficiently, reducing time from O(n²) to O(n).

beginner

How do you calculate the area of water between two lines at positions i and j?

Area = min(height[i], height[j]) * (j - i), where height[i] and height[j] are the heights of the lines.

intermediate

What is the rule for moving pointers in the 'Container With Most Water' algorithm?

Move the pointer at the shorter line inward, because moving the taller line won't increase the area.

beginner

What is the time complexity of the optimal solution for 'Container With Most Water'?

O(n), where n is the number of lines, because each pointer moves at most n times.

What does the 'Container With Most Water' problem primarily involve?

A Finding two lines that hold the maximum water

B Sorting an array of heights

C Finding the shortest line in the array

D Calculating the sum of all heights

The problem is about finding two lines that together hold the most water.

In the two-pointer approach, where do the pointers start?

A Both at the end of the array

B One at the start and one at the end

C Both at the start of the array

D Random positions in the array

Pointers start at the two ends to check the widest container first.

How do you decide which pointer to move inward?

A Move the pointer at the taller line

B Move the pointer at the middle

C Move both pointers simultaneously

D Move the pointer at the shorter line

Moving the shorter line pointer may find a taller line and increase area.

What is the formula to calculate the water area between two lines?

A max(height[i], height[j]) * (j - i)

B height[i] + height[j]

C min(height[i], height[j]) * (j - i)

D (height[i] + height[j]) / 2

Area depends on the shorter line and the distance between lines.

What is the time complexity of the two-pointer solution?

A O(n)

B O(n log n)

C O(n²)

D O(log n)

Two pointers move inward once each, making it O(n).

Explain how the two-pointer technique works in the 'Container With Most Water' problem.

Think about how moving the shorter line pointer can help find a bigger container.

Describe how to calculate the area between two lines and why the shorter line limits the water container.

Imagine water spilling over the shorter line if it was taller.