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

Solution 1: Mathematics

Assume the three consecutive integers are \(x-1\), \(x\), and \(x+1\). Their sum is \(3x\), so \(\textit{num}\) must be a multiple of \(3\). If \(\textit{num}\) 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 \(x = \frac{\textit{num}}{3}\), then \(x-1\), \(x\), and \(x+1\) are the three consecutive integers whose sum is \(\textit{num}\).

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

Python3JavaC++GoTypeScriptRustJavaScript

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

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};
    }
};

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];
}

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]
    }
}

/**
 * @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];
};

Comments