LeetCode 2177. Find Three Consecutive Integers That Sum to a Given Number Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Mathematics

Assume the three consecutive integers are x-1, x, and x+1. Their sum is 3x, so num must be a multiple of 3. If num is not a multiple of 3, it cannot be represented as the sum of three consecutive integers, and we return an empty array. Otherwise, let x = ^num⁄₃, then x-1, x, and x+1 are the three consecutive integers whose sum is num.

The time complexity is O(1), and the space complexity is O(1).

PythonJavaC++GoTypeScriptRustJavaScript

class Solution: def sumOfThree(self, num: int) -> List[int]: x, mod = divmod(num, 3) return [] if mod else [x - 1, x, x + 1](code-box)

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}; } }(code-box)

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}; } };(code-box)

func sumOfThree(num int64) []int64 { if num%3 != 0 { return []int64{} } x := num / 3 return []int64{x - 1, x, x + 1} }(code-box)

function sumOfThree(num: number): number[] { if (num % 3) { return []; } const x = Math.floor(num / 3); return [x - 1, x, x + 1]; }(code-box)

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] } }(code-box)

/** * @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]; };(code-box)