Container With Most Water

Try every combination of two lines and compute the water they can hold. Track the maximum area found.

1. 1For each pair (i, j) with i < j, compute area = min(height[i], height[j]) × (j - i).
2. 2Update maxArea if this area is larger.
3. 3Return maxArea.

Why it's not optimal: O(n²) time — too slow for n up to 10⁵. The two-pointer approach eliminates impossible pairs greedily and runs in O(n).

def maxArea(height):
    maxA = 0
    for i in range(len(height)):
        for j in range(i + 1, len(height)):
            area = min(height[i], height[j]) * (j - i)
            maxA = max(maxA, area)
    return maxA

The area between two lines is limited by the shorter one. If we move the taller wall inward, we can only decrease or maintain width — with no chance of increasing height. But if we move the shorter wall, we might find a taller wall and gain area.

1. 1Start with L=0 and R=n-1 (widest container).
2. 2Compute area = min(height[L], height[R]) × (R-L), update max.
3. 3Move the pointer at the shorter wall inward — the taller wall is already contributing its full value, only a taller replacement can help.
4. 4Repeat until pointers meet.