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Formatted question description: https://leetcode.ca/all/2535.html
2535. Difference Between Element Sum and Digit Sum of an Array
Description
You are given a positive integer array nums
.
- The element sum is the sum of all the elements in
nums
. - The digit sum is the sum of all the digits (not necessarily distinct) that appear in
nums
.
Return the absolute difference between the element sum and digit sum of nums
.
Note that the absolute difference between two integers x
and y
is defined as |x - y|
.
Example 1:
Input: nums = [1,15,6,3] Output: 9 Explanation: The element sum of nums is 1 + 15 + 6 + 3 = 25. The digit sum of nums is 1 + 1 + 5 + 6 + 3 = 16. The absolute difference between the element sum and digit sum is |25 - 16| = 9.
Example 2:
Input: nums = [1,2,3,4] Output: 0 Explanation: The element sum of nums is 1 + 2 + 3 + 4 = 10. The digit sum of nums is 1 + 2 + 3 + 4 = 10. The absolute difference between the element sum and digit sum is |10 - 10| = 0.
Constraints:
1 <= nums.length <= 2000
1 <= nums[i] <= 2000
Solutions
Solution 1: Simulation
We traverse the array $nums$, calculate the sum of elements $a$ and the sum of digits $b$, and finally return $ | a - b | $. |
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
-
class Solution { public int differenceOfSum(int[] nums) { int a = 0, b = 0; for (int x : nums) { a += x; for (; x > 0; x /= 10) { b += x % 10; } } return Math.abs(a - b); } }
-
class Solution { public: int differenceOfSum(vector<int>& nums) { int a = 0, b = 0; for (int x : nums) { a += x; for (; x; x /= 10) { b += x % 10; } } return abs(a - b); } };
-
class Solution: def differenceOfSum(self, nums: List[int]) -> int: a, b = sum(nums), 0 for x in nums: while x: b += x % 10 x //= 10 return abs(a - b)
-
func differenceOfSum(nums []int) int { a, b := 0, 0 for _, x := range nums { a += x for ; x > 0; x /= 10 { b += x % 10 } } return abs(a - b) } func abs(x int) int { if x < 0 { return -x } return x }
-
function differenceOfSum(nums: number[]): number { return nums.reduce((r, v) => { r += v; while (v !== 0) { r -= v % 10; v = Math.floor(v / 10); } return r; }, 0); }
-
impl Solution { pub fn difference_of_sum(nums: Vec<i32>) -> i32 { let mut ans = 0; for &num in nums.iter() { let mut num = num; ans += num; while num != 0 { ans -= num % 10; num /= 10; } } ans } }