# Knapsack Problem
# Knapsack Problem Basics
The Problem (opens new window): Given a set of items where each item contains a weight and value, determine the number of each to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.
Pseudo code for Knapsack Problem
Given:
- Values(array v)
- Weights(array w)
- Number of distinct items(n)
- Capacity(W)
for j from 0 to W do:
m[0, j] := 0
for i from 1 to n do:
for j from 0 to W do:
if w[i] > j then:
m[i, j] := m[i-1, j]
else:
m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])
A simple implementation of the above pseudo code using Python:
def knapSack(W, wt, val, n):
K = [[0 for x in range(W+1)] for x in range(n+1)]
for i in range(n+1):
for w in range(W+1):
if i==0 or w==0:
K[i][w] = 0
elif wt[i-1] <= w:
K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w])
else:
K[i][w] = K[i-1][w]
return K[n][W]
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, n))
Running the code: Save this in a file named knapSack.py
$ python knapSack.py
220
Time Complexity of the above code: O(nW)
where n is the number of items and W is the capacity of knapsack.
# Solution Implemented in C#
public class KnapsackProblem
{
private static int Knapsack(int w, int[] weight, int[] value, int n)
{
int i;
int[,] k = new int[n + 1, w + 1];
for (i = 0; i <= n; i++)
{
int b;
for (b = 0; b <= w; b++)
{
if (i==0 || b==0)
{
k[i, b] = 0;
}
else if (weight[i - 1] <= b)
{
k[i, b] = Math.Max(value[i - 1] + k[i - 1, b - weight[i - 1]], k[i - 1, b]);
}
else
{
k[i, b] = k[i - 1, b];
}
}
}
return k[n, w];
}
public static int Main(int nItems, int[] weights, int[] values)
{
int n = values.Length;
return Knapsack(nItems, weights, values, n);
}
}
# Remarks
The Knapsack problem mostly arises in resources allocation mechanisms. The name "Knapsack" was first introduced by Tobias Dantzig (opens new window).
Auxiliary Space: O(nw)
Time Complexity O(nw)