Problem with return value in recursion.

Question

This program is to count sum of elements of an array that meets the target value, using recursion. The last count value won't get added to previous count value.
https://code.sololearn.com/cFkh5POV5O4S/?ref=app

Answer 1

# does this fullfill the challenge :

does_it = lambda x,y,target=11: True if x+y==11 else False

from itertools import product
from random import sample

#lili = sample(range(20),20)
lili = [3,4,7,7]
lolo = list(product(lili,lili))
lala = [(i,j) for i,j in lolo if does_it(i,j,11)]
result = len(lala)
print(lala)
print(result)

# i am not using recursion
# as this is more obvious

Answer 2

Thankyou Ervis Meta - SoloHelper for your code.
But my approach is with recursion....!

Answer 3

# what about this ?

#Possible combination---> (4,7)(4,7)(7,4)(7,4)

def summ(arr,target,count=0):
if target==0:
return True
if target<0:
return False
for i in range(len(arr)):
rem=target-arr[i]
if summ(arr[:i]+arr[i+1:],rem,count)==True:
# di[target]=count
count+=1
return rem

print(summ([3,4,7,7],11))

Answer 4

No, it won't work for other conditions..
I can't return rem value bcse..
I was just trying to count possible combinations

Answer 5

# maybe this ?

def summ(lili,target,count=0,res=[]):
if target == 0:
return True
if target<0:
return False
for i in lili:
for j in lili:
if i+j == (target-count):
# recursive part (•‿•)
summ(lili,target-1,count+1,res=res.append((i,j)))
return res

lolo = summ([3,4,7,7],11)
print(lolo)

Answer 6

Good one,
but it checks only for 2 pair of combinations
for example : [3,4,4,8] target 11
it fails in this case (3,4,4) or for more no. of combinations

Is two for loop required...? The Time complexity of your code is o(n^2 * n^2) (in worst case) it can be solved in more efficient way.

and you haven't used terminate conditions. (starting 2 if conditions)

Answer 7

# let's low down that damn time !

def subset_sum_iter(array, target):
sign = 1
array = sorted(array)
if target < 0:
array = reversed(array)
sign = -1
last_index = {0: [-1]}
for i in range(len(array)):
for s in list(last_index.keys()):
new_s = s + array[i]
if 0 < (new_s - target) * sign:
pass # Cannot lead to target
elif new_s in last_index:
last_index[new_s].append(i)
else:
last_index[new_s] = [i]
# Now yield up the answers.
def recur(new_target, max_i):
for i in last_index[new_target]:
if i == -1:
yield [] # Empty sum.
elif max_i <= i:
break # Not our solution.
else:
for answer in recur(new_target - array[i], i):
answer.append(array[i])
yield answer
for answer in recur(target, len(array)):
yield answer

Answer 8

Nice 😊👍.
The time complex has been reduced 🙃...
but the problem with my recursion code is still pending..🥲

Answer 9

Could you please show me some more examples of your problem ?

Answer 10

Arr=[3,4,5]
Traget=8
O/p: 2 (count val)#--->[(3,5),(5,3)].
#--->No repetition [(4,4)]❌

Arr=[3,4]
Traget=8
O/p: 0 (count val)#--->[].

Arr=[3,4,7,7]
Traget=11
O/p: 4 (count val)#--->[(4,7),(4,7),(7,4),(7,4)].

Answer 11

Problem with my code is, the last return value of count won't get added..
Its just returning new count value