Python Program for 0-1 Knapsack Problem

Difficulty Level : Expert
Last Updated : 23 Oct, 2019

# A naive recursive implementation of 0-1 Knapsack Problem 
  
# Returns the maximum value that can be put in a knapsack of 
# capacity W 
def knapSack(W, wt, val, n): 
  
    # Base Case 
    if n == 0 or W == 0 : 
        return 0
  
    # If weight of the nth item is more than Knapsack of capacity 
    # W, then this item cannot be included in the optimal solution 
    if (wt[n-1] > W): 
        return knapSack(W, wt, val, n-1) 
  
    # return the maximum of two cases: 
    # (1) nth item included 
    # (2) not included 
    else: 
        return max(val[n-1] + knapSack(W-wt[n-1], wt, val, n-1), 
                   knapSack(W, wt, val, n-1)) 
  
# end of function knapSack 
  
# To test above function 
val = [60, 100, 120] 
wt = [10, 20, 30] 
W = 50
n = len(val) 
print knapSack(W, wt, val, n) 
  
# This code is contributed by Nikhil Kumar Singh 

Output:

# A Dynamic Programming based Python  
# Program for 0-1 Knapsack problem 
# Returns the maximum value that can  
# be put in a knapsack of capacity W 
def knapSack(W, wt, val, n): 
    K = [[0 for x in range(W + 1)] for x in range(n + 1)] 
  
    # Build table K[][] in bottom up manner 
    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] 
  
# Driver program to test above function 
val = [60, 100, 120] 
wt = [10, 20, 30] 
W = 50
n = len(val) 
print(knapSack(W, wt, val, n)) 
  
# This code is contributed by Bhavya Jain 

Output:

Please refer complete article on Dynamic Programming | Set 10 ( 0-1 Knapsack Problem) for more details!

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