Container With Most Water in DSA Python - Execution Trace

Concept Flow - Container With Most Water

Start with two pointers

↓

Calculate area between pointers

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Compare with max area

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Move pointer with smaller height inward

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Repeat until pointers meet

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Return max area found

We use two pointers at the ends of the array, calculate area, update max, then move the smaller pointer inward until they meet.

Execution Sample

DSA Python

def maxArea(height):
    left, right = 0, len(height) - 1
    max_area = 0
    while left < right:
        area = min(height[left], height[right]) * (right - left)
        max_area = max(max_area, area)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_area

This code finds the maximum water container area by moving two pointers inward and calculating areas.

Execution Table

Step	Operation	Left Pointer	Right Pointer	Height[left]	Height[right]	Area Calculated	Max Area	Pointer Moved	Visual State
1	Initialize pointers	0	8	1	7	min(1,7)*8=8	8	Move left (height[left]<height[right])	[1,8,6,2,5,4,8,3,7]
2	Calculate area	1	8	8	7	min(8,7)*7=49	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
3	Calculate area	1	7	8	3	min(8,3)*6=18	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
4	Calculate area	1	6	8	8	min(8,8)*5=40	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
5	Calculate area	1	5	8	4	min(8,4)*4=16	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
6	Calculate area	1	4	8	5	min(8,5)*3=15	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
7	Calculate area	1	3	8	2	min(8,2)*2=4	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
8	Calculate area	1	2	8	6	min(8,6)*1=6	49	Move right (height[left]>=height[right])	[1,8,6,2,5,4,8,3,7]
9	Pointers meet or cross	1	1	8	8	-	49	Stop	[1,8,6,2,5,4,8,3,7]

💡 Pointers meet at index 1, loop ends with max area 49

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	Final
left	0	1	1	1	1	1	1	1	1	1
right	8	8	7	6	5	4	3	2	1	1
max_area	0	8	49	49	49	49	49	49	49	49

Key Moments - 3 Insights

Why do we move the pointer with the smaller height?

Because the area is limited by the smaller height, moving the larger height pointer won't increase area. See steps 1 and 2 in execution_table where moving the smaller height pointer leads to larger area.

Why does the loop stop when left meets right?

When pointers meet, there is no width between them, so area is zero. The exit_note and step 9 in execution_table show this stopping condition.

Why do we calculate area as min(height[left], height[right]) * (right - left)?

Because water container height is limited by the shorter line, and width is the distance between pointers. This formula is used in every step in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the area calculated?

A 42

B 21

C 15

D 10

💡 Hint

Check the 'Area Calculated' column at step 6 in execution_table.

At which step does the max_area first become 49?

A Step 2

B Step 3

C Step 5

D Step 4

💡 Hint

Look at the 'Max Area' column in execution_table to find when it first reaches 49.

If height[left] was always greater than height[right], which pointer would move?

A Left pointer moves

B Right pointer moves

C Both pointers move

D Neither pointer moves

💡 Hint

Refer to the 'Pointer Moved' column in execution_table where height[left]>=height[right] leads to right pointer moving.

Concept Snapshot

Container With Most Water:
- Use two pointers at array ends
- Calculate area = min(height[left], height[right]) * distance
- Move pointer with smaller height inward
- Update max area each step
- Stop when pointers meet
- Returns max water container area

Full Transcript

This visualization shows how the Container With Most Water problem is solved using two pointers. We start with pointers at both ends of the height array. At each step, we calculate the area formed by the lines at these pointers and update the maximum area found. We then move the pointer pointing to the smaller height inward, hoping to find a taller line and a larger area. This process repeats until the pointers meet. The execution table tracks each step's pointers, heights, area calculation, and pointer movement. Key moments clarify why we move the smaller pointer and why the loop stops when pointers meet. The visual quiz tests understanding of area calculations and pointer movements. The concept snapshot summarizes the approach in simple steps.