Find N Unique Integers Sum up to Zero

Find N Unique Integers Sum up to Zero - Problem

Array Math Easy

Given an integer n, return any array containing n unique integers such that they add up to 0.

The array should contain exactly n distinct integers, and their sum must equal zero. You can return any valid combination that satisfies these constraints.

Note: There are multiple valid answers for each input, so any correct solution will be accepted.

Input & Output

Example 1 — Even Number

$ Input: n = 4

› Output: [1,-1,2,-2]

💡 Note: Create 2 pairs of opposite numbers: (1,-1) and (2,-2). Sum: 1 + (-1) + 2 + (-2) = 0

Example 2 — Odd Number

$ Input: n = 5

› Output: [0,1,-1,2,-2]

💡 Note: Since n is odd, add 0 first, then create 2 pairs: (1,-1) and (2,-2). Sum: 0 + 1 + (-1) + 2 + (-2) = 0

Example 3 — Minimum Case

$ Input: n = 1

› Output: [0]

💡 Note: Only one unique integer that sums to zero is 0 itself

Constraints

1 ≤ n ≤ 1000

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to use symmetric pairs of numbers that cancel each other out. Best approach uses mathematical pairing: for even n, create n/2 pairs like (1,-1), (2,-2). For odd n, add 0 first then create pairs. Time: O(n), Space: O(n).

Common Approaches

✓ Heap

⏱️ Time: N/A Space: N/A

Mathematical Symmetric Pairs

⏱️ Time: O(n) Space: O(n)

Create pairs of opposite numbers that cancel each other out. For even n, use n/2 pairs like (-1,1), (-2,2). For odd n, add 0 as the middle element.

Random Generation with Validation

⏱️ Time: O(n) Space: O(n)

Keep generating sets of n unique random integers, calculating their sum, and adjusting the last element to make the total zero. Repeat if duplicates are created.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>

void solution(int n, int result[], int* size) {
    *size = 0;
    
    // If n is odd, add 0
    if (n % 2 == 1) {
        result[(*size)++] = 0;
        n -= 1;
    }
    
    // Add symmetric pairs
    for (int i = 1; i <= n / 2; i++) {
        result[(*size)++] = i;
        result[(*size)++] = -i;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    
    int result[1000];
    int size;
    solution(n, result, &size);
    
    printf("[");
    for (int i = 0; i < size; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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