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

Math Simulation Medium

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.

Note: Consecutive integers are integers that follow each other in order without gaps. For example: 1, 2, 3 or -2, -1, 0.

Input & Output

Example 1 — Basic Case

$ Input: num = 33

› Output: [10,11,12]

💡 Note: Three consecutive integers 10, 11, 12 sum to 33: 10 + 11 + 12 = 33

Example 2 — Small Number

$ Input: num = 4

› Output: []

💡 Note: No three consecutive integers can sum to 4. The minimum sum is 0+1+2=3, next is 1+2+3=6

Example 3 — Negative Numbers

$ Input: num = 0

› Output: [-1,0,1]

💡 Note: Three consecutive integers -1, 0, 1 sum to 0: -1 + 0 + 1 = 0

Constraints

-10¹⁵ ≤ num ≤ 10¹⁵

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that three consecutive integers x, x+1, x+2 have sum 3x+3. So we need (num-3) to be divisible by 3. Best approach is the mathematical solution with direct calculation. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(√n) Space: O(1)

Start from a reasonable range and check every possible first integer to see if three consecutive integers sum to the target. This approach works but is inefficient for large numbers.

Mathematical Solution

⏱️ Time: O(1) Space: O(1)

If three consecutive integers x, x+1, x+2 sum to num, then 3x + 3 = num, so x = (num-3)/3. We just need to check if (num-3) is divisible by 3.

Brute Force — Algorithm Steps

Start from a reasonable lower bound
For each starting integer x, check if x + (x+1) + (x+2) = num
Return the triplet if found, empty array otherwise

Visualization

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Step-by-Step Walkthrough

Set Range

Calculate reasonable start and end bounds

Try Values

Test each x to see if x + (x+1) + (x+2) = target

Return Result

Return triplet when found or empty array

Code -

solution.c — C

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

long long abs_val(long long x) {
    return x < 0 ? -x : x;
}

void solution(long long num, long long* result, int* size) {
    // Try different starting points
    long long start = num / 3 - abs_val(num) / 3 - 1;
    long long end = num / 3 + abs_val(num) / 3 + 1;
    
    for (long long x = start; x <= end; x++) {
        if (x + (x + 1) + (x + 2) == num) {
            result[0] = x;
            result[1] = x + 1;
            result[2] = x + 2;
            *size = 3;
            return;
        }
    }
    
    *size = 0;
}

int main() {
    long long num;
    scanf("%lld", &num);
    
    long long result[3];
    int size;
    solution(num, result, &size);
    
    printf("[");
    for (int i = 0; i < size; i++) {
        printf("%lld", result[i]);
        if (i < size - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(√n)

We iterate through possible starting values in range [-num/3, num/3]

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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