Question

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

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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, $M \leq 10^9$, $D \leq 81$.

class Solution:
  def maximumSum(self, nums: List[int]) -> int:
    d = defaultdict(int)
    ans = -1
    for v in nums:
      x, y = 0, v
      while y:
        x += y % 10
        y //= 10
      if x in d:
        ans = max(ans, d[x] + v)
      d[x] = max(d[x], v)
    return ans

class Solution {
  public int maximumSum(int[] nums) {
    int[] d = new int[100];
    int ans = -1;
    for (int v : nums) {
      int x = 0;
      for (int y = v; y > 0; y /= 10) {
        x += y % 10;
      }
      if (d[x] > 0) {
        ans = Math.max(ans, d[x] + v);
      }
      d[x] = Math.max(d[x], v);
    }
    return ans;
  }
}

class Solution {
public:
  int maximumSum(vector<int>& nums) {
    int d[100]{};
    int ans = -1;
    for (int v : nums) {
      int x = 0;
      for (int y = v; y; y /= 10) {
        x += y % 10;
      }
      if (d[x]) {
        ans = max(ans, d[x] + v);
      }
      d[x] = max(d[x], v);
    }
    return ans;
  }
};

func maximumSum(nums []int) int {
  d := [100]int{}
  ans := -1
  for _, v := range nums {
    x := 0
    for y := v; y > 0; y /= 10 {
      x += y % 10
    }
    if d[x] > 0 {
      ans = max(ans, d[x]+v)
    }
    d[x] = max(d[x], v)
  }
  return ans
}

function maximumSum(nums: number[]): number {
  const d: number[] = Array(100).fill(0);
  let ans = -1;
  for (const v of nums) {
    let x = 0;
    for (let y = v; y; y = (y / 10) | 0) {
      x += y % 10;
    }
    if (d[x]) {
      ans = Math.max(ans, d[x] + v);
    }
    d[x] = Math.max(d[x], v);
  }
  return ans;
}

impl Solution {
  pub fn maximum_sum(nums: Vec<i32>) -> i32 {
    let mut d = vec![0; 100];
    let mut ans = -1;

    for &v in nums.iter() {
      let mut x: usize = 0;
      let mut y = v;
      while y > 0 {
        x += (y % 10) as usize;
        y /= 10;
      }
      if d[x] > 0 {
        ans = ans.max(d[x] + v);
      }
      d[x] = d[x].max(v);
    }

    ans
  }
}