0 1 Knapsack

Given a list of items with weight w[i] and value v[i], what's the maximum value you can get given a knapsack with capacity C, i.e. it can hold items with at most weight C in total. You can pick each item at most once (i.e. either pick an item i 0 times or 1 time).

Algorithm

Let dp[i + 1][c] be the maximum value we can get using the first i + 1 items (index from 0 to i) and weight capped by c.

dp[i + 1][c] = max(
    dp[i][c],             // If we skip `i`-th item
    dp[i][c-w[i]] + v[i]  // If we pick `i`-th item
)

dp[0][c] = 0

Implementation

// Time: O(NC)
// Space: O(NC)
vector<vector<int>> dp(N + 1, vector<int>(C + 1));
for (int i = 0; i < N; ++i) {
    for (int c = w[i]; c <= C; ++c) {
        dp[i + 1][c] = max( dp[i][c], dp[i][c - w[i]] + v[i] );
    }
}
return dp[N][C];

Space Optimization

// Time: O(NC)
// Space: O(C)
vector<int> dp(C + 1);
for (int i = 0; i < N; ++i) {
    for (int c = C; c >= w[i]; --c) {
        dp[c] = max( dp[c], dp[c - w[i]] + v[i] );
    }
}
return dp[C];

A Constant Optimization

Since we only need dp[N][C], which only requires dp[N-1][C] and dp[N-1][C-w[N-1]]. So for i = N - 2 (the second from the last item), we only need to loop from max( w[N-2], C - w[N-1] ) to C.

dp[N-1][C-w[N-1]] requires dp[N-2][C] and dp[N-2][C-w[N-1]-w[N-2]], so for i = N - 3, we only need to loop from max( w[N-3], C - w[N-1] - w[N-2] ) to C.

Generalization: for i-th item, the inner loop should be from max( w[i], C - sum(w[j] | i < j < N ) ) to C.

// Time: O(NC)
// Space: O(C)
vector<int> dp(C + 1);
int sum = accumulate(w.begin(), w.end(), 0);
for (int i = 0; i < N; ++i) {
    sum -= w[i];
    for (int c = C; c >= max(w[i], C - sum); --c) {
        dp[c] = max( dp[c], dp[c - w[i]] + v[i] );
    }
}
return dp[C];

Problems

• 416. Partition Equal Subset Sum (Medium)
• 474. Ones and Zeroes (Medium)
• 494. Target Sum (Medium)