Teaching Kids Programming: Videos on Data Structures and Algorithms
You are given two lists of integers weights and values which have the same length and an integer capacity. weights[i] and values[i] represent the weight and value of the ith item. Given that you can take at most capacity weights, and that you can only take at most one copy of each item, return the maximum amount of value you can get.
Constraints
n ≤ 250 where n is the length of weights and values
capacity ≤ 250Example 1
Input
weights = [1, 2, 3]
values = [1, 5, 3]
capacity = 5
Output
8
Similar to original knapsack, but how do you ensure the specific element is only included once?
Space-optimised Dynamic Programming Algorithm to Solve 0/1 Knapsack
Given items to pack in a knapsack with capacity . Each item has weight and value . We want to pack the items to gain the maximum value but the total weights from the chosen items should not exceed the knapsack capacity.
Let’s state this mathematically:
We want to maximize the ,
subject to ,
and
When means that we don’t pick the i-th item while means we pack i-th item in the knapsack.
The Dynamic Programming Equation is:
for .
The top down DP uses the Recursion with Memoization aka the magic @cache keyword which asks the computer to remember the intermediate values. The bottom up DP uses two dimensional array dp[i][c] to store the values proactively. For Top Down DP, we call the dp(n-1, c) which will be expanded top down, while the bottom up DP computes the dp[0][0..c] (first row) and then compute the second row dp[1], the third row dp[2] until we get the value dp[n-1][c].
However, dp[i] is based on dp[i-1] only, and thus we can compress the space. That will optimise the space usage from O(CN) to O(C). The inner loop j (capacity) should be inversely enumerated (downwards) to avoid dp[0..j] is overwritten.
1 2 3 4 5 6 7 8 | class Solution: def solve(self, weights, values, capacity): n = len(values) dp = [0 for _ in range(capacity + 1)] for i in range(n): for j in range(capacity, weights[i] - 1, -1): dp[j] = max(dp[j], dp[j - weights[i]] + values[i]) return dp[capacity] |
class Solution: def solve(self, weights, values, capacity): n = len(values) dp = [0 for _ in range(capacity + 1)] for i in range(n): for j in range(capacity, weights[i] - 1, -1): dp[j] = max(dp[j], dp[j - weights[i]] + values[i]) return dp[capacity]
The time complexity is still O(CN).
Knapsack Problems
- Teaching Kids Programming - 0/1 Knapsack: Length of the Longest Subsequence That Sums to Target (Recursion+Memoization=Top Down Dynamic Programming)
- Teaching Kids Programming - 0/1 Knapsack Space Optimised Dynamic Programming Algorithm
- Teaching Kids Programming - Using Bottom Up Dynamic Programming Algorithm to Solve 0/1 Knapsack Problem
- Teaching Kids Programming - 0/1 Knapsack Problem via Top Down Dynamic Programming Algorithm
- Teaching Kids Programming - Multiple Strange Coin Flips/Toss Bottom Up Dynamic Programming Algorithm (Knapsack Variant)
- Teaching Kids Programming - Multiple Strange Coin Flips/Toss Top Down Dynamic Programming Algorithm (Knapsack Variant)
- Teaching Kids Programming - Max Profit of Rod Cutting (Unbounded Knapsack) via Bottom Up Dynamic Programming Algorithm
- Teaching Kids Programming - Max Profit of Rod Cutting (Unbounded Knapsack) via Top Down Dynamic Programming Algorithm
- Teaching Kids Programming - Brick Layout (Unlimited Knapsack) via Bottom Up Dynamic Programming Algorithm
- Teaching Kids Programming - Brick Layout (Unlimited Knapsack) via Top Down Dynamic Programming Algorithm
- Count Multiset Sum (Knapsacks) by Dynamic Programming Algorithm
- Count Multiset Sum (Knapsacks) by Recursive BackTracking Algorithm
- Dynamic Programming Algorithm to Solve the Poly Knapsack (Ubounded) Problem
- Dynamic Programming Algorithm to Solve 0-1 Knapsack Problem
- Classic Unlimited Knapsack Problem Variant: Coin Change via Dynamic Programming and Depth First Search Algorithm
- Classic Knapsack Problem Variant: Coin Change via Dynamic Programming and Breadth First Search Algorithm
- Complete Knapsack Problem
- 0/1 Knapsack Problem
- Teaching Kids Programming - Combination Sum Up to Target (Unique Numbers) by Dynamic Programming Algorithms
- Algorithms Series: 0/1 BackPack - Dynamic Programming and BackTracking
- Using BackTracking Algorithm to Find the Combination Integer Sum
- Facing Heads Probabilties of Tossing Strange Coins using Dynamic Programming Algorithm
- Partition Equal Subset Sum Algorithms using DFS, Top-Down and Bottom-up DP
–EOF (The Ultimate Computing & Technology Blog) —
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