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2177. Find Three Consecutive Integers That Sum to a Given Number
Description
Given an integer num, return three consecutive integers (as a sorted array) that sum to num. If num cannot be expressed as the sum of three consecutive integers, return an empty array.
Example 1:
Input: num = 33 Output: [10,11,12] Explanation: 33 can be expressed as 10 + 11 + 12 = 33. 10, 11, 12 are 3 consecutive integers, so we return [10, 11, 12].
Example 2:
Input: num = 4 Output: [] Explanation: There is no way to express 4 as the sum of 3 consecutive integers.
Constraints:
0 <= num <= 1015
Solutions
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class Solution { public long[] sumOfThree(long num) { if (num % 3 != 0) { return new long[] {}; } long x = num / 3; return new long[] {x - 1, x, x + 1}; } } -
class Solution { public: vector<long long> sumOfThree(long long num) { if (num % 3) { return {}; } long long x = num / 3; return {x - 1, x, x + 1}; } }; -
class Solution: def sumOfThree(self, num: int) -> List[int]: x, mod = divmod(num, 3) return [] if mod else [x - 1, x, x + 1] -
func sumOfThree(num int64) []int64 { if num%3 != 0 { return []int64{} } x := num / 3 return []int64{x - 1, x, x + 1} } -
function sumOfThree(num: number): number[] { if (num % 3) { return []; } const x = Math.floor(num / 3); return [x - 1, x, x + 1]; } -
/** * @param {number} num * @return {number[]} */ var sumOfThree = function (num) { if (num % 3) { return []; } const x = Math.floor(num / 3); return [x - 1, x, x + 1]; }; -
impl Solution { pub fn sum_of_three(num: i64) -> Vec<i64> { if num % 3 != 0 { return Vec::new(); } let x = num / 3; vec![x - 1, x, x + 1] } }