0 - 1 Knapsack Problem
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 3
- Core for
- 9 roles
The problem
You are given weights and values of n items, and a knapsack with capacity W. Each item can be included at most once. Find the maximum total value that can be put in the knapsack.
Example 1
- Input
- values = [60,100,120], weights = [10,20,30], W = 50
- Output
- 220
- Why
- Items with value 100 and 120 fit within capacity 50, giving total value 220.
Example 2
- Input
- values = [10,20,30], weights = [5,10,15], W = 20
- Output
- 50
- Why
- Items with value 20 and 30 fit within capacity 20, giving total value 50.
Example 3
- Input
- values = [1], weights = [5], W = 3
- Output
- 0
- Why
- The item is too heavy for the knapsack, so no value can be obtained.
Constraints
- 1 <= n <= 1000
- 1 <= W <= 1000
- 1 <= weights[i] <= W
- 1 <= values[i] <= 1000
How to think about it
Updated 2026-09-29Every item presents an all-or-nothing choice: pack it or leave it. Packing it consumes capacity and yields value, reducing remaining capacity for earlier items. Walking capacities backward prevents an item from being packed multiple times into the same knapsack.
Approaches, worst first
Subsets recursion
time O(2^n) · space O(n)
Branch on taking or skipping each item. Explores 2^n possibilities, generating exponential duplicate subproblems for overlapping remaining capacities.
2D dynamic programming grid
time O(n * W) · space O(n * W)
Build table dp[i][w] representing max value using a subset of the first i items with capacity w. If weight <= w, dp[i][w] = max(dp[i-1][w], dp[i-1][w - weight] + value).
1D backward scanWrite this one
time O(n * W) · space O(W)
Collapse the table into a single array dp of size W + 1. For each item, iterate w from W down to weight. Descending order guarantees dp[w - weight] holds values from the previous item step without self-overwriting.
Where people lose marks · 3
- Iterating the capacity forward (from weight up to W) in 1D space optimization. Forward iteration turns 0/1 knapsack into unbounded knapsack, allowing an item to be picked repeatedly.
- Overlooking cases where an item's weight exceeds the current capacity w. When weight > w, the state must inherit the previous value dp[w] unchanged.
- Allocating an array of size W instead of W + 1, causing index errors at capacity W.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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