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 $num$ must be a multiple of $3$. If $num$ is not a multiple of $3$, it cannot be expressed as the sum of three consecutive integers, and we return an empty array. Otherwise, let $x = \frac{num}{3}$, 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 $O(1)$.

Python3JavaC++GoTypeScript

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];
}

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