알고리즘 읽기 노트

Knapsack Problem

In 0/1 knapsack, each item has a weight and value and may be chosen at most once. The goal is maximum total value without exceeding capacity.

유형알고리즘카테고리Sets & Dynamic Programming깊이고급일반적인 복잡도O(nW) DP

짧은 면접 답변

이 멘탈 모델부터 시작하세요

In 0/1 knapsack, each item has a weight and value and may be chosen at most once. The goal is maximum total value without exceeding capacity.

간단한 설명

불필요한 전문용어 없이 동작 방식 이해하기

A DP state can represent the best value using the first i items with capacity c.

For each item, either skip it or take it if it fits.

실행

알고리즘을 연속된 의사결정으로 읽기

Define the state precisely.

Write the transition/choice from smaller states.

Set base cases, then decide memoization, tabulation or backtracking.

기억해 둘 작은 예제

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

읽기 쉬운 JavaScript 구현JavaScript

function knapsack(items, capacity) {
  const dp = Array(capacity + 1).fill(0);
  for (const { weight, value } of items) {
    for (let c = capacity; c >= weight; c--) {
      dp[c] = Math.max(dp[c], dp[c - weight] + value);
    }
  }
  return dp[capacity];
}

이 코드를 읽는 방법

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

복잡도

O(nW) DP

Do not quote this complexity mechanically.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 DP when subproblems overlap and optimal solutions can be composed from smaller optimal states.

흔한 실수

보통 어디에서 문제가 생기는가

Writing recurrence before defining state.
Using DP when state space is larger than the original problem.

면접에서 설명하는 방법

전문적인 답변 구조

Clarify the variant first—0/1, unbounded or fractional. Their correct algorithms differ substantially.

문제→핵심 아이디어→불변식 / 상태→Big-O→트레이드오프

빠른 면접 체크

What problem does Knapsack Problem solve?

In 0/1 knapsack, each item has a weight and value and may be chosen at most once. The goal is maximum total value without exceeding capacity.

복잡도에 대해 무엇을 언급해야 하나요?

Its typical cost here is O(nW) DP, but a complete answer should connect that cost to the representation, input constraints and the dominant operations.

가장 흔한 실수는 무엇인가요?

Writing recurrence before defining state.

계속 읽기MemoizationTabulationBacktracking

function knapsack(items, capacity) { const dp = Array(capacity + 1).fill(0); for (const { weight, value } of items) { for (let c = capacity; c >= weight; c--) { dp[c] = Math.max(dp[c], dp[c - weight] + value); } } return dp[capacity]; }