2342. Max Sum of a Pair With Equal Sum of Digits
Description
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 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.
Example 1:
Input: nums = [18,43,36,13,7] Output: 54 Explanation: The pairs (i, j) that satisfy the conditions are: - (0, 2), both numbers have a sum of digits equal to 9, and their sum is 18 + 36 = 54. - (1, 4), both numbers have a sum of digits equal to 7, and their sum is 43 + 7 = 50. So the maximum sum that we can obtain is 54.
Example 2:
Input: nums = [10,12,19,14] Output: -1 Explanation: There are no two numbers that satisfy the conditions, so we return -1.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
Solutions
Solution 1: Hash Table
We can use a hash table $d$ to record the maximum value corresponding to each digit sum, and initialize an answer variable $ans = -1$.
Next, we traverse the array $nums$. For each number $v$, we calculate its digit sum $x$. If $x$ exists in the hash table $d$, then we update the answer $ans = \max(ans, d[x] + v)$. Then update the hash table $d[x] = \max(d[x], v)$.
Finally, return the answer $ans$.
Since the maximum element in $nums$ is $10^9$, the maximum digit sum is $9 \times 9 = 81$. We can directly define an array $d$ of length $100$ to replace the hash table.
The time complexity is $O(n \times \log M)$, and the space complexity is $O(D)$. Here, $n$ is the length of the array $nums$, and $M$ and $D$ are the maximum value of the elements in the array $nums$ and the maximum value of the digit sum, respectively. In this problem, $M \leq 10^9$, $D \leq 81$.
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