Code Logo

Ember Nest Flames

Published at05 Jan 2026
Created by
Multi Dimensional Easy 18 views
Like21

This challenge becomes much easier once you know exactly what to keep, change, or count. In Ember Nest Flames, you are trying to work toward the right number by following one clear idea.

Calculate ember nest flame expansion A good way to think about it is to first understand what goes in, then what rule you must follow, and finally what shape the answer should have.

For example, if the input is rings = 4, the answer is 1365. Four extra rings mean the latest coal plus four times the entire light from ring three. Another example is rings = 0, which gives 1. Only the first coal remains.

This is a friendly practice problem, but it still rewards careful reading. The key is understanding the rule clearly and then applying it carefully.

One helpful habit is to say the rule out loud in your own words before you start solving. If you can explain what counts, what changes, and what the final answer should look like, you are already much closer to the right solution.

Example Input & Output

Example 1
Input
rings = 0
Output
1
Explanation

Only the first coal remains.

Example 2
Input
rings = 2
Output
85
Explanation

The second ring adds a boundary coal and multiplies the previous glow fourfold, for a total of eighty-five embers.

Example 3
Input
rings = 4
Output
1365
Explanation

Four extra rings mean the latest coal plus four times the entire light from ring three.

Algorithm Flow

Recommendation Algorithm Flow for Ember Nest Flames

Solution Approach

This problem is about a fast-growing shell pattern. Each new ring builds on the whole shape that came before it, so the total at each step depends directly on the total from the previous step. The cleanest way to solve it is to simulate that growth with a simple loop.

The key is the update rule: each new ring triples the current total and adds one extra flame, so the new total is result * 3 + 1. We can confirm this against the provided examples before writing any code.

We start from a single coal and apply the rule once per ring:

function ember_nest_flames(nest) {
    let result = 1;
    for (let i = 0; i < nest; i++) {
        result = result * 3 + 1;
    }
    return result;
}

The heart of the solution is the line result = result * 3 + 1. Each iteration takes the current total, triples it to represent the expanding light, and adds one for the newest coal.

Let us trace rings = 2. We begin with result = 1. After the first ring, result = 1 * 3 + 1 = 4. After the second ring, result = 4 * 3 + 1 = 13. For the base case rings = 0, the loop never runs and we return 1, which is just the original coal.

Notice how quickly the total grows: each ring multiplies by three, so even a small ring count produces a large answer. This is why simulating with a loop works fine, but a very large input would grow exponentially.

The time complexity is O(nest) because we run one iteration per ring, and the space complexity is O(1) since we only keep a single running value.

Best Answers

java
import java.util.*;
class Solution {
    public int ember_nest_flames(int nest) {
        int result = 1;
        for (int i = 0; i < nest; i++) {
            result = result * 3 + 1;
        }
        return result;
    }
}