_2342. Max Sum of a Pair With Equal Sum of Digits.md

2342. Max Sum of a Pair With Equal Sum of Digits

All prompts are owned by LeetCode. To view the prompt, click the title link above.

Back to top

---

First completed : February 12, 2025

Last updated : February 12, 2025

---

Related Topics : Array, Hash Table, Sorting, Heap (Priority Queue)

Acceptance Rate : 66.01 %

---

V1

My original response which was within expectations runtime wise and efficiency wise.

V2 and V3

V2 and V3 are identical, just with V3 being V2 expanded into a few lines for readability. When doing V1 I thought it would be fun to make a oneliner version and it was a good way for me to practice my itertools and function confidence.

To go line by line explaining:

# max of the list containing:
#   -1 default
#   for each digit sum that appears in nums, what's the highest sum of those values
return max(
    [-1] + # default
    [
        sum(heapq.nlargest(2, [-inf] + list(group[1])))  # heapq nlargest = largest 2 values
                                                         # [-inf] accounts for if a digit sum is unique
                                                         # since nlargest(2, ...) will return 1 value if there's
                                                         # only 1 value

        for group in groupby(                            # itertools.groupby(...)

            sorted(nums, key=lambda y: sum(map(int, str(y)))),  # groupby doesn't always work if the
                                                                # values aren't sorted in the intended
                                                                # use case order (order of digit sum)

            key=lambda x: sum(map(int, str(x)))                 # key for both: sum (int value of each digit)
                                                                # I found this to be much more efficient since
                                                                # it's an iternal method which I presume uses
                                                                # C instead of python. For a oneliner this was
                                                                # the best choice at least and saved me a
                                                                # few test cases of runtime.
        )
    ]
)

---

Solutions

• m2342 v1.py
• m2342 v2 ONELINER.py
• m2342 v3 ONELINER expanded.py

Python

m2342 v1.py

class Solution:
    def digit_sum(self, num: int) -> int :
        output = 0
        while num :
            output += num % 10
            num //= 10
        return output

    def maximumSum(self, nums: List[int]) -> int:
        digit_sums = [self.digit_sum(num) for num in nums]

        digit_sum_indxs = defaultdict(list)
        for i, x in enumerate(digit_sums) :
            digit_sum_indxs[x].append(i)

        output = -1
        for freq, indxs in digit_sum_indxs.items() :
            if len(indxs) < 2 :
                continue
            output = max(output, sum(sorted([nums[x] for x in indxs])[-2:]))
        return output

m2342 v2 ONELINER.py

class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        return max([-1] + [sum(heapq.nlargest(2, [-inf] + list(group[1]))) for group in groupby(sorted(nums, key=lambda y: sum(map(int, str(y)))), key=lambda x: sum(map(int, str(x))))])

m2342 v3 ONELINER expanded.py

class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        return max(
            [-1] + 
            [
                sum(heapq.nlargest(2, [-inf] + list(group[1]))) 
                for group in groupby(
                    sorted(nums, key=lambda y: sum(map(int, str(y)))),
                    key=lambda x: sum(map(int, str(x)))
                )
            ]
        )