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Code Block
language py 0-1 Knapsack Problem true true
```
##########################################
## 상향식(타뷸레이션): 미리 값을 다 찾아놓고 사용 ##
##########################################
# 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

####################################################
## 하향식(memoization): 재귀로 돌면서 필요한 값만 찾아 저장 ##
####################################################

# This is the memoization approach of
# 0 / 1 Knapsack in Python in simple
# we can say recursion + memoization = DP

# driver code
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)

# We initialize the matrix with -1 at first.
t = [[-1 for i in range(W + 1)] for j in range(n + 1)]

def knapsack(wt, val, W, n):

# base conditions
if n == 0 or W == 0:
return 0
if t[n][W] != -1:
return t[n][W]

# choice diagram code
if wt[n-1] <= W:
t[n][W] = max(
val[n-1] + knapsack(
wt, val, W-wt[n-1], n-1),
knapsack(wt, val, W, n-1))
return t[n][W]
elif wt[n-1] > W:
t[n][W] = knapsack(wt, val, W, n-1)
return t[n][W]

print(knapsack(wt, val, W, n))

# This code is contributed by Prosun Kumar Sarkar```

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