1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
|
# 2342. Max Sum of a Pair With Equal Sum of Digits
"""
you are given a 0-indexed array 'nums' consisting of positive integers. you
can choose two indices 'i' and 'j', such that 'i != j', and the sum of the
digits of the number 'nums[i]' is equal to that of 'nums[j]'. return the
maximum value of 'nums[i] + nums[j]' that you can obtain over all
possible indices 'i' and 'j' that satisfy the conditions.
"""
class Solution(object):
def maximumSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
map, ans = [-1] * 82, -1
for i in nums:
digit_sum = sum(int(j) for j in str(i))
if map[digit_sum] != -1:
ans = max(ans, i + map[digit_sum])
map[digit_sum] = max(map[digit_sum], i)
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.maximumSum(nums=[18, 43, 36, 13, 7]))
print(obj.maximumSum(nums=[10, 12, 19, 14]))
|