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.
\n
Example 1:
\n\n\nInput: nums = [18,43,36,13,7]\nOutput: 54\nExplanation: The pairs (i, j) that satisfy the conditions are:\n- (0, 2), both numbers have a sum of digits equal to 9, and their sum is 18 + 36 = 54.\n- (1, 4), both numbers have a sum of digits equal to 7, and their sum is 43 + 7 = 50.\nSo the maximum sum that we can obtain is 54.\n\n\n
Example 2:
\n\n\nInput: nums = [10,12,19,14]\nOutput: -1\nExplanation: There are no two numbers that satisfy the conditions, so we return -1.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 109给你一个下标从 0 开始的数组 nums ，数组中的元素都是 正 整数。请你选出两个下标 i 和 j（i != j），且 nums[i] 的数位和 与 nums[j] 的数位和相等。
请你找出所有满足条件的下标 i 和 j ，找出并返回 nums[i] + nums[j] 可以得到的 最大值 。
\n\n
示例 1：
\n\n\n输入：nums = [18,43,36,13,7]\n输出：54\n解释：满足条件的数对 (i, j) 为：\n- (0, 2) ，两个数字的数位和都是 9 ，相加得到 18 + 36 = 54 。\n- (1, 4) ，两个数字的数位和都是 7 ，相加得到 43 + 7 = 50 。\n所以可以获得的最大和是 54 。\n\n
示例 2：
\n\n\n输入：nums = [10,12,19,14]\n输出：-1\n解释：不存在满足条件的数对，返回 -1 。\n\n\n
\n\n
提示：
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