knapsack problem using greedy method in python
# Python3 program to solve fractional
# Knapsack Problem
class ItemValue:
"""Item Value DataClass"""
def __init__(self, wt, val, ind):
self.wt = wt
self.val = val
self.ind = ind
self.cost = val // wt
def __lt__(self, other):
return self.cost < other.cost
# Greedy Approach
class FractionalKnapSack:
"""Time Complexity O(n log n)"""
@staticmethod
def getMaxValue(wt, val, capacity):
"""function to get maximum value """
iVal = []
for i in range(len(wt)):
iVal.append(ItemValue(wt[i], val[i], i))
# sorting items by value
iVal.sort(reverse = True)
totalValue = 0
for i in iVal:
curWt = int(i.wt)
curVal = int(i.val)
if capacity - curWt >= 0:
capacity -= curWt
totalValue += curVal
else:
fraction = capacity / curWt
totalValue += curVal * fraction
capacity = int(capacity - (curWt * fraction))
break
return totalValue
# Driver Code
if __name__ == "__main__":
wt = [10, 40, 20, 30]
val = [60, 40, 100, 120]
capacity = 50
maxValue = FractionalKnapSack.getMaxValue(wt, val, capacity)
print("Maximum value in Knapsack =", maxValue)
# This code is contributed by vibhu4agarwal
# a dynamic approach
# Returns the maximum value that can be stored by the bag
def knapSack(W, wt, val, n):
K = [[0 for x in range(W + 1)] for x in range(n + 1)]
#Table 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]
#Main
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print(knapSack(W, wt, val, n))
