LeetCode #2177 — MEDIUM

Find Three Consecutive Integers That Sum to a Given Number

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Notice that if a solution exists, we can represent them as x-1, x, x+1. What does this tell us about the number?
Notice the sum of the numbers will be 3x. Can you solve for x?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2177: Find Three Consecutive Integers That Sum to a Given Number
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};
    }
}

// Accepted solution for LeetCode #2177: Find Three Consecutive Integers That Sum to a Given Number
func sumOfThree(num int64) []int64 {
	if num%3 != 0 {
		return []int64{}
	}
	x := num / 3
	return []int64{x - 1, x, x + 1}
}

# Accepted solution for LeetCode #2177: Find Three Consecutive Integers That Sum to a Given Number
class Solution:
    def sumOfThree(self, num: int) -> List[int]:
        x, mod = divmod(num, 3)
        return [] if mod else [x - 1, x, x + 1]

// Accepted solution for LeetCode #2177: Find Three Consecutive Integers That Sum to a Given Number
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]
    }
}

// Accepted solution for LeetCode #2177: Find Three Consecutive Integers That Sum to a Given Number
function sumOfThree(num: number): number[] {
    if (num % 3) {
        return [];
    }
    const x = Math.floor(num / 3);
    return [x - 1, x, x + 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.