Find Three Consecutive Integers That Sum to a Given Number

Find Three Consecutive Integers That Sum to a Given Number - Problem

Math Simulation Medium

Given an integer num, your task is to find three consecutive integers that sum to exactly num. Return these three integers as a sorted array.

If it's impossible to express num as the sum of three consecutive integers, return an empty array.

What are consecutive integers? They are integers that follow each other in sequence with a difference of 1. For example: 4, 5, 6 or -2, -1, 0.

Key insight: If three consecutive integers start at x, they are: x, x+1, x+2, and their sum is 3x + 3 = 3(x+1).

Input & Output

example_1.py — Basic Case

$ Input: num = 9

› Output: [2, 3, 4]

💡 Note: 2 + 3 + 4 = 9, and these are three consecutive integers

example_2.py — Negative Numbers

$ Input: num = 0

› Output: [-1, 0, 1]

💡 Note: -1 + 0 + 1 = 0, consecutive integers can include negative numbers

example_3.py — Impossible Case

$ Input: num = 1

› Output: []

💡 Note: No three consecutive integers can sum to 1. The equation (1-3)/3 = -2/3 is not an integer

Constraints

-10⁹ ≤ num ≤ 10⁹
The three integers must be consecutive
Return empty array if no solution exists

Visualization

Tap to expand

Understanding the Visualization

Identify the pattern

Three consecutive integers: x, x+1, x+2

Set up equation

Sum = x + (x+1) + (x+2) = 3x + 3

Solve directly

If 3x + 3 = num, then x = (num-3)/3

Verify solution

Check if x is an integer (divisibility test)

Key Takeaway

🎯 Key Insight: The sum of three consecutive integers starting at x is always 3x+3, allowing us to solve directly with algebra instead of searching

Asked in

G Google 15 ⊞ Microsoft 12 a Amazon 8 f Meta 5

The optimal solution uses a mathematical approach: if three consecutive integers x, x+1, x+2 sum to num, then 3x + 3 = num, so x = (num-3)/3. We just check if this gives an integer solution in O(1) time.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Starting Points)	O(n)	O(1)	Try every possible starting integer within a reasonable range
Mathematical Formula (Optimal)	O(1)	O(1)	Use algebraic formula to directly calculate the answer

Brute Force (Try All Starting Points) — Algorithm Steps

Determine a reasonable range to search (from -num to num)
For each potential starting integer x in this range
Calculate sum = x + (x+1) + (x+2) = 3x + 3
If sum equals num, return [x, x+1, x+2]
If no valid starting point found, return empty array

Visualization

Tap to expand

Step-by-Step Walkthrough

Try x = -2

Check if -2 + (-1) + 0 = -3 equals target

Try x = -1

Check if -1 + 0 + 1 = 0 equals target

Try x = 0

Check if 0 + 1 + 2 = 3 equals target

Found!

When sum matches target, return the three integers

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int* solution(int num, int* returnSize) {
    // Try different starting points
    for (int x = -abs(num); x <= abs(num); x++) {
        if (x + (x + 1) + (x + 2) == num) {
            int* result = (int*)malloc(3 * sizeof(int));
            result[0] = x;
            result[1] = x + 1;
            result[2] = x + 2;
            *returnSize = 3;
            return result;
        }
    }
    *returnSize = 0;
    return NULL;
}

int main() {
    int num;
    scanf("%d", &num);
    int returnSize;
    int* result = solution(num, &returnSize);
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(", ");
    }
    printf("]\n");
    if (result) free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

In worst case, we might check up to |num| starting points

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

28.6K Views

Medium Frequency

~8 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Starting Points) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler