알고리즘 읽기 노트

Rain Terraces / Trapping Rain Water

Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.

유형알고리즘카테고리Classic Problems깊이초급일반적인 복잡도Varies by implementation / input constraints

짧은 면접 답변

이 멘탈 모델부터 시작하세요

Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.

간단한 설명

불필요한 전문용어 없이 동작 방식 이해하기

A two-pointer solution tracks leftMax and rightMax while moving the side with the smaller boundary.

This gives O(n) time and O(1) extra space.

실행

알고리즘을 연속된 의사결정으로 읽기

Restate constraints and edge cases.

Identify the pattern and invariant.

Walk through a small example before coding.

기억해 둘 작은 예제

Read a tiny input by hand and write down the state after every important operation. For Rain Terraces / Trapping Rain Water, the goal is to make the invariant visible rather than to memorize code.

읽기 쉬운 JavaScript 구현JavaScript

function trap(height) {
  let l = 0, r = height.length - 1;
  let leftMax = 0, rightMax = 0, water = 0;
  while (l < r) {
    if (height[l] <= height[r]) {
      leftMax = Math.max(leftMax, height[l]);
      water += leftMax - height[l++];
    } else {
      rightMax = Math.max(rightMax, height[r]);
      water += rightMax - height[r--];
    }
  }
  return water;
}

이 코드를 읽는 방법

문법을 외우지 마세요. 각 줄에서 바뀌는 상태를 따라가고 위에서 설명한 불변식과 연결하세요.

복잡도

Varies by implementation / input constraints

Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.

언제 사용하는가

적합한 경우

Use the recognized pattern when its invariant matches the problem constraints.

흔한 실수

보통 어디에서 문제가 생기는가

Jumping to code before defining the invariant.
Optimizing before producing a correct baseline.

면접에서 설명하는 방법

전문적인 답변 구조

Explain why moving the smaller boundary is safe: the smaller side already determines the maximum water level for that position.

문제→핵심 아이디어→불변식 / 상태→Big-O→트레이드오프

빠른 면접 체크

What problem does Rain Terraces / Trapping Rain Water solve?

Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.

복잡도에 대해 무엇을 언급해야 하나요?

Its typical cost here is Varies by implementation / input constraints, but a complete answer should connect that cost to the representation, input constraints and the dominant operations.

가장 흔한 실수는 무엇인가요?

Jumping to code before defining the invariant.

계속 읽기Problem patternsEdge casesComplexity trade-offs

function trap(height) { let l = 0, r = height.length - 1; let leftMax = 0, rightMax = 0, water = 0; while (l < r) { if (height[l] <= height[r]) { leftMax = Math.max(leftMax, height[l]); water += leftMax - height[l++]; } else { rightMax = Math.max(rightMax, height[r]); water += rightMax - height[r--]; } } return water; }