CODEWITHSUNDEEP
python-3.xlistpython-itertools
StackGuru
StackGuru

Reputation: 503

How to find indices and combinations that adds upto given sum?

How to find the combinations and corresponding indices that adds upto given sum ? And also, can it be handled list of elements of size 500000 (higher size) ?

Input:

l1 = [9,1, 2, 7, 6, 1, 5] 
target = 8

**Constraints**
1<=(len(l1))<=500000
1<=each_list_element<=1000

Output:

Format : {index:element}
{1:1, 5:1, 4:6}   #Indices : 1,5,4   Elements : 1,1,6
{1:1, 2:2,  6:5}
{5:1, 2:2,  6:5}
{1:1,  3:7}
{5:1,  3:7}
{2:2,  4:6}

Tried:

from itertools import combinations

def test(l1, target):
    l2 = []
    l3 = []    
    if len(l1) > 0:
        for r in range(0,len(l1)+1):        
            l2 += list(combinations(l1, r))

        for i in l2:        
            if sum(i) == target:
                l3.append(i)

    return l3

l1 = [9,1, 2, 7, 6, 1, 5] 
target = 8
print(test(l1,target))

[(1, 7), (2, 6), (7, 1), (1, 2, 5), (1, 6, 1), (2, 1, 5)]

Can someone guide me ?

UPDATE

Apart from above, code fails to handle these scenarios
Input = [4,6,8,5,3]
target = 3

Outputs {} , need to output {4:3}

Input = [4,6,8,3,5,3]
target = 3

Outputs {} , need to output {5:3,3:3}   #corrected index

Input = [1,2,3,15]
target = 15

Outputs = {}, need to output {3:15}

Upvotes: 1

Views: 481

Answers (2)

Prathamesh
Prathamesh

Reputation: 1057

You can just use index function to get index and store them as key:value pair with help of dictionary in another list such as following,

from itertools import combinations

def test(l1, target):
    l2 = []
    l3 = []
    l4=[]
    dict1={}
    a=0
    if len(l1) > 0:
        for r in range(0,len(l1)+1):        
            l2 += list(combinations(l1, r))

        for i in l2:
            dict1={}
            if sum(i) == target:

                for j in i:
                    a=l1.index(j)
                    dict1[a]=j
                l4.append(dict1)
                l3.append(i)

    return l4

l1 = [4,6,8,5,3]
target = 3

print(test(l1,target))

output:

[{4: 3}]

as you can see the condition l1 = [4,6,8,5,3] target = 3 works which is previously not working.

Hope this helps!

Upvotes: 0

Chris Doyle
Chris Doyle

Reputation: 12199

Your code was close, i would use enumerate to get the index and value as tuple pairs. I am always dropping any of the index and value tuples where that value is greater than the target since that cannot possible be a match. this will generate less combinations. Then like you i just iterate through the permutations of tuples and sum the values in each permutation, if it sums to the target then yield that permutation. lastly in the loop to output the values i give the perm to dict to convert into the dict format you wanted

from itertools import combinations


def find_sum_with_index(l1, target):
    index_vals = [iv for iv in enumerate(l1) if iv[1] < target]
    for r in range(1, len(index_vals) + 1):
        for perm in combinations(index_vals, r):
            if sum([p[1] for p in perm]) == target:
                yield perm


l1 = [9, 1, 2, 7, 6, 1, 5]
target = 8
for match in find_sum_with_index(l1, target):
    print(dict(match))

OUTPUT

{1: 1, 3: 7}
{2: 2, 4: 6}
{3: 7, 5: 1}
{1: 1, 2: 2, 6: 5}
{1: 1, 4: 6, 5: 1}
{2: 2, 5: 1, 6: 5}

Upvotes: 2

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