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Knapsack with Duplicate Items

Medium Dynamic Programming Knapsack Original on GeeksforGeeks

Knapsack with Duplicate Items is a medium dynamic programming problem solved with the knapsack pattern. The best approach, optimal (1d array, capacity ascending), runs in O(n · W) time and O(W) space. Below are 2 approaches in Java, from recursion up.

Problem

Knapsack where each item can be taken unlimited times (unbounded knapsack). Return the maximum value within capacity W.

Examples

Example 1

Input
W = 8, val = [10, 40, 50, 70], wt = [1, 3, 4, 5]
Output
110
Why
The weight-5 item (70) plus the weight-3 item (40) fill all 8 units.

Example 2

Input
W = 3, val = [1, 1], wt = [2, 1]
Output
3

Constraints

  • 1 <= n, W <= 1000.
  • Each item can be taken any number of times.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
RecursionexponentialO(W)
Optimal (1D array, capacity ascending)O(n · W)O(W)

1Recursion

Timeexponential
SpaceO(W)

At item i, either take it and stay at i (it can be reused) or move on to i - 1.

  1. f(i, w) = max(f(i - 1, w), val[i] + f(i, w - wt[i])).
Java
class Solution {
    static int knapSack(int[] val, int[] wt, int capacity) {
        return f(val, wt, val.length - 1, capacity);
    }

    private static int f(int[] val, int[] wt, int i, int w) {
        if (i < 0) return 0;
        int skip = f(val, wt, i - 1, w);
        return wt[i] <= w ? Math.max(skip, val[i] + f(val, wt, i, w - wt[i])) : skip;
    }
}

2Optimal (1D array, capacity ascending)

TimeO(n · W)
SpaceO(W)

dp[w] = best value for capacity w. For each item, loop w upward: dp[w - wt] may already include this item, which is exactly what unlimited copies need.

  1. For each item, for w from wt to W: dp[w] = max(dp[w], val + dp[w - wt]).
Java
class Solution {
    static int knapSack(int[] val, int[] wt, int capacity) {
        int[] dp = new int[capacity + 1];
        for (int i = 0; i < val.length; i++)
            for (int w = wt[i]; w <= capacity; w++)
                dp[w] = Math.max(dp[w], val[i] + dp[w - wt[i]]);
        return dp[capacity];
    }
}

Edge cases to test

  • An item too heavy to ever fit

Hints

Hint 1

Same as 0/1 knapsack, but loop capacity upward so an item can be reused within the same pass.

FAQ

What is the best time complexity for Knapsack with Duplicate Items?

Optimal (1D array, capacity ascending) runs in O(n · W) time and O(W) extra space.

Which pattern does Knapsack with Duplicate Items use?

It is a dynamic programming problem that uses the knapsack pattern. Other problems with the same pattern: 0 - 1 Knapsack Problem, Fractional Knapsack, Coin Change.

Is there a brute force solution for Knapsack with Duplicate Items?

Yes. Recursion takes exponential time and O(W) space. At item i, either take it and stay at i (it can be reused) or move on to i - 1.

Which edge cases should I test for Knapsack with Duplicate Items?

An item too heavy to ever fit.