Triangular Number Series - Problem

Given a positive integer n, generate the first n triangular numbers.

A triangular number is the sum of all natural numbers up to a given position. The sequence starts as: 1, 3, 6, 10, 15, 21, 28, ...

For example:

1st triangular number: 1 (sum of 1)
2nd triangular number: 3 (sum of 1+2)
3rd triangular number: 6 (sum of 1+2+3)
4th triangular number: 10 (sum of 1+2+3+4)

Return the result as an array of integers.

Input & Output

Example 1 — First 5 Triangular Numbers

$ Input: n = 5

› Output: [1, 3, 6, 10, 15]

💡 Note: T₁=1, T₂=1+2=3, T₃=1+2+3=6, T₄=1+2+3+4=10, T₅=1+2+3+4+5=15

Example 2 — First 3 Triangular Numbers

$ Input: n = 3

› Output: [1, 3, 6]

💡 Note: First triangular: 1, second: 1+2=3, third: 1+2+3=6

Example 3 — Edge Case Single Number

$ Input: n = 1

› Output: [1]

💡 Note: Only the first triangular number, which is 1

Constraints

1 ≤ n ≤ 1000
Result fits in 32-bit integer

Visualization

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Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is that triangular numbers follow the pattern T(k) = k×(k+1)/2. The optimal approach uses this mathematical formula for direct calculation instead of nested loops. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Nested Loops

⏱️ Time: O(n²) Space: O(n)

For each triangular number from 1 to n, use an inner loop to sum all numbers from 1 up to that position. This requires recalculating the sum completely for each position.

Mathematical Formula

⏱️ Time: O(n) Space: O(n)

Apply the well-known mathematical formula for triangular numbers: the kth triangular number equals k×(k+1)/2. This eliminates all loops and calculations by computing each value directly.

Running Sum Approach

⏱️ Time: O(n) Space: O(n)

Instead of recalculating each sum from scratch, maintain a running total and add the current position number to get the next triangular number. Each triangular number is the previous one plus the current position.

Brute Force - Nested Loops — Algorithm Steps

Outer loop from 1 to n
Inner loop sums 1+2+...+i for each i
Store result in array

Visualization

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Step-by-Step Walkthrough

Position 1

Sum 1 = 1

Position 2

Sum 1+2 = 3

Position 3

Sum 1+2+3 = 6

Code -

solution.c — C

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

void solution(int n, int* result) {
    for (int i = 1; i <= n; i++) {
        int triangularSum = 0;
        for (int j = 1; j <= i; j++) {
            triangularSum += j;
        }
        result[i - 1] = triangularSum;
    }
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop runs n times, inner loop averages n/2 iterations

✓ Linear Growth

Space Complexity

O(n)

Array to store n triangular numbers

⚡ Linearithmic Space

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Triangular Number Series - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Nested Loops — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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