What is the brute force approach and complexity for Trapping Rain Water?

For each cell, scan left for the tallest bar and right for the tallest bar with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Trapping Rain Water?

Settle the cell on the side whose running max is smaller with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Trapping Rain Water?

Look for key signals in the prompt: "water trapped between bars" indicates Each cell's water = min(tallest on left, tallest on right) − its height → a per-cell formula that needs info from both sides.; "tallest on the left / right" indicates Running maxes from both ends. Two pointers walking inward can carry one running max each.; "limited by the lower wall" indicates The side with the smaller max is the one whose cells can be decided now → move that pointer..

Trapping Rain Water

Hard·LeetCode #42 Pro OnlyTricky / Edge Case

LeetCode

Problem Statement

You are given n bars standing side by side, each 1 unit wide. height[i] is the height of bar i.

Rain falls on them. Water collects in the dips between taller bars, and anything that isn't held in by a bar on both sides flows away.

Return the total amount of water trapped.

Input

An integer array height of length n.

Output

One integer: the total units of trapped water.

Example 1

Input: height = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]

Output: 6

Water per cell is [0, 0, 1, 0, 1, 2, 1, 0, 0, 1, 0, 0], which adds up to 6.

Example 2

Input: height = [4, 2, 0, 3, 2, 5]

Output: 9

The walls 4 and 5 hold the water. Per cell: [0, 2, 4, 1, 2, 0] = 9.

Constraints

• 1 ≤ n ≤ 2 · 10^4
• 0 ≤ height[i] ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“water trapped between bars”

→ Each cell's water = min(tallest on left, tallest on right) − its height → a per-cell formula that needs info from both sides.

“tallest on the left / right”

→ Running maxes from both ends. Two pointers walking inward can carry one running max each.

“limited by the lower wall”

→ The side with the smaller max is the one whose cells can be decided now → move that pointer.

Approach Complexity Ladder

Brute Force — scan both sides for every cellO(N²)O(1)

For each cell, scan left for the tallest bar and right for the tallest bar

Better — prefix max and suffix max arraysO(N)O(N)

Precompute the tallest bar up to each index from both sides

Optimal — two pointers with leftMax and rightMaxO(N)O(1)

Settle the cell on the side whose running max is smaller

Brute Force Intuition

For each cell i, scan everything to its left to find the tallest bar, and everything to its right to find the tallest bar (both including i). The water stands at the lower of those two walls, so cell i holds min(leftMax, rightMax) − height[i]. Add it up.

Recognition rule for this sub-pattern:

Start at both ends of a sorted (or symmetric) structure. Each step, move the side that cannot do better.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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