Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1085.html

1085. Sum of Digits in the Minimum Number

Level

Easy

Description

Given an array A of positive integers, let S be the sum of the digits of the minimal element of A.

Return 0 if S is odd, otherwise return 1.

Example 1:

Input: [34,23,1,24,75,33,54,8]

Output: 0

Explanation:

The minimal element is 1, and the sum of those digits is S = 1 which is odd, so the answer is 0.

Example 2:

Input: [99,77,33,66,55]

Output: 1

Explanation:

The minimal element is 33, and the sum of those digits is S = 3 + 3 = 6 which is even, so the answer is 1.

Note:

  1. 1 <= A.length <= 100
  2. 1 <= A[i].length <= 100

Solution

Loop over the array A to find the minimum number. Then calculate the sum of the digits of the minimum number. Finally check whether the sum is odd or even, and return 0 or 1 respectively.

  • class Solution {
    public int sumOfDigits(int[] A) {
        int min = Integer.MAX_VALUE;
        for (int num : A)
            min = Math.min(min, num);
        if (min == 0)
            return 1;
        int sum = 0;
        int temp = min;
        while (temp > 0) {
            sum += temp % 10;
            temp /= 10;
        }
        return sum % 2 == 0 ? 1 : 0;
    }
}

  • // OJ: https://leetcode.com/problems/sum-of-digits-in-the-minimum-number/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int sumOfDigits(vector<int>& A) {
        int n = *min_element(begin(A), end(A)), sum = 0;
        while (n) {
            sum += n;
            n /= 10;
        }
        return 1 - sum % 2;
    }
};

  • class Solution:
    def sumOfDigits(self, nums: List[int]) -> int:
        x = min(nums)
        s = 0
        while x:
            s += x % 10
            x //= 10
        return 0 if s % 2 else 1



  • func sumOfDigits(nums []int) int {
	x := 100
	for _, v := range nums {
		if v < x {
			x = v
		}
	}
	s := 0
	for ; x > 0; x /= 10 {
		s += x % 10
	}
	return s&1 ^ 1
}

All Problems

All Solutions