index value of the tuples-CodePudding

write a function named sums that takes 3 arguments: 2 lists of integers nums1 and nums2, and an integer target; then return a tuple containing indices i and j of the two numbers in nums1 and nums2, respectively, where the numbers at that indices add up to target (i.e, nums1[i] num2[j] = target). If there is no solution, return “Not found”. The function sums must contain exactly only one loop (either a for-loop or a while-loop), otherwise you will get no point. A nested-loop is considered as more than one loop.

def sums(nums1, nums2, target):
nums1 = [11, 2, 15, 7, 8]
nums2 = [2, 3, 4, 5]
target_number = 9
for i in range(len(nums1)):
    for j in range(len(nums2)):
        total = nums1[i]   nums2[j]
        if total == target:
            a = (nums1[i], nums2[j])
            print(a)

This is the code that i have written and i am struggling to get the index values can anyone please help!

CodePudding user response：

You can solve it in O(n) with a dictionary and a single for loop;

nums1 = [11, 2, 15, 7, 8]
nums2 = [2, 3, 4, 5]
target_number = 9

def sums(nums1, nums2, target):
    d = dict(zip(nums2, range(len(nums2))))
    for i,n in enumerate(nums1):
        remainder = target-n
        if remainder in d:
            return (i, d[remainder])
    return 'Not found'

sums(nums1, nums2, target_number)

Output: (3, 0)

Be aware that while there is a single for loop, the code is looping to construct the dictionary, it's just not visible directly.

CodePudding user response：

Here's an alternative solution using basic index incrementing. It's not as efficient as using a dictionary and finding the difference, as this naively iterates over b as many times as there are items in a. In the worst case, with two equal-length lists, this solution is O(n^2).

Although it may actually be faster in smaller cases, if there happens to be a bottlneck around the creation of any of the intermediate data structures that @mozway uses. I'm only speculating on this, I don't actually know.

def target_sum_idxs(a, b, target):
    i = j = 0
    while i < len(a):
        if a[i]   b[j] == target:
            return i, j
        j  = 1
        if j == len(b):
            i  = 1
            j = 0
    return "Not found"

Demo:

>>> target_sum_idxs(nums1, nums2, target=9)
(3, 0)

>>> target_sum_idxs(nums2, nums1, target=9)
(0, 3)

>>> target_sum_idxs(nums1, nums2, target=999)
'Not found'

CodePudding user response：

just one little thing.

Check that the variable in the if statement is really your parameter, not the "target_number" variable. Also, in the if statement, you need to stop execution, maybe you can use a break or return the tuple.

I would use something like this:

def sums(nums1, nums2, target):
    for i in range(len(nums1)):
        for j in range(len(nums2)):
            total = nums1[i]   nums2[j]
            if total == target:
                a = (i, j)
                return a
                
    return "Not found"