I have to check if it is possible to get sum from given numbers >to be strictly greater than given sum
for example:
if given sum is 9 and the given numbers are [3,5,7] then the correct answer is 1*3 +1*7=10>9 so the out put should be [1,0,1]
example2:
if given sum is 13 and the numbers are [1,3,5] then correct answer is 0*1+3*3+1*5=14>13 but 9*1+0*3+5=14 is >wrong so the correct output is [0,3,1] how to approach for this problem
A solution can be obtained using a version of the unbounded Knapsack algorithm.
Here we modify code from Python Unbound Knapsack
Modify with two objectives:
Minimized the count of the values used from the array
Create the max sum less than the limit (ie given sum + 1)
def knapsack_unbounded_dp(arr, C):
C += 1 # actual limit
# Form index value pairs (to keep track of the original indexes after sorting)
items = [(i, v) for i, v in enumerate(arr)]
# Sort values in descending order
items = sorted(items, key=lambda item: item[1], reverse=True)
# Sack keeps track of:
# max value so i.e. sack[0] and
# count of how many each item are in the sack i.e. sack[1]
sack = [(0, [0 for i in items]) for i in range(0, C+1)] # value, [item counts]
for i,item in enumerate(items):
# For current item we check if a previous entry could have done better
# by adding this item to the sack
index, value = item
for c in range(value, C+1): # check all previous values
sackwithout = sack[c-value] # previous max sack to try adding this item to
trial = sackwithout[0] + value # adding value to max without using it
used = sackwithout[1][i] # count of i-them item
if sack[c][0] < trial:
# old max sack with this added item is better
sack[c] = (trial, sackwithout[1][:])
sack[c][1][i] +=1 # use one more
value, bagged = sack[C]
# index and count of each array value
new_bagged = [(i, v) for (i, _), v in zip(items, bagged)]
# Re-sort based upon original order of array indexes
new_bagged.sort(key=lambda t: t[0])
# counts based upon original array order
cnts = [v for i, v in new_bagged]
return sum(cnts), value, cnts
Test Code
for t in [([1, 3, 5], 8), ([3, 5, 7], 9), ([1, 3, 5], 13)]:
result = knapsack_unbounded_dp(t[0], t[1])
print(f'Test: {t[0]}, Given Sum {t[1]}')
print(f'Result counts {result[2]}, with max sum {result[1]}, Total Count {result[0]}\n')
Output
Test: [1, 3, 5], Given Sum 8
Result counts [0, 3, 0], with max sum 9, Total Count 3
Test: [3, 5, 7], Given Sum 9
Result counts [0, 2, 0], with max sum 10, Total Count 2
Test: [1, 3, 5], Given Sum 13
Result counts [0, 3, 1], with max sum 14, Total Count 4
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