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Coral Pulse Count

Published at05 Jan 2026
Created by
Easy 10 views
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This problem is about tracking a simple growth pattern one layer at a time. In Coral Pulse Count, the coral starts with 1 pulse, and every new layer makes the current total grow in the same way.

The rule is that each layer creates three copies of the pulses you already have, then adds one more new pulse. So if the current total is p, the next total becomes p * 3 + 1.

For example, if the input is layers = 0, the answer is 1 because no growth has happened yet. Another example is layers = 1, which gives 4. Starting from 1 pulse, one layer makes it 1 * 3 + 1 = 4. If layers = 3, the totals go 1 -> 4 -> 13 -> 40, so the answer is 40.

So the job is to start from the initial pulse count of 1, apply this growth rule exactly layers times, and report the final number of pulses.

Example Input & Output

Example 1
Input
layers = 0
Output
1
Explanation

No layers are added yet, so the coral keeps its starting pulse count of 1.

Example 2
Input
layers = 1
Output
4
Explanation

Starting from 1 pulse, one layer gives 1 * 3 + 1 = 4.

Example 3
Input
layers = 3
Output
40
Explanation

Applying the rule three times gives the sequence 1, 4, 13, 40.

Algorithm Flow

Recommendation Algorithm Flow for Coral Pulse Count
Recommendation Algorithm Flow for Coral Pulse Count

Solution Approach

A good method for this problem is a simple iterative simulation. The idea is to start from the initial pulse count of 1 and apply the growth rule once for each layer.

The nice part about this approach is that it matches the story of the problem directly. We do not need any complicated data structure. We just keep one variable for the current number of pulses and update it layer by layer.

We start with the initial state:

let pulses = 1;

This means before any layers are added, the coral has exactly one pulse.

Next, we repeat the growth rule for each layer:

for (let i = 0; i < layers; i++) {
  pulses = pulses * 3 + 1;
}

Each loop step represents one new layer. The current total is tripled, and then one extra pulse is added.

After the loop finishes, pulses stores the final answer:

return pulses;

So the full idea is simple: begin at 1, then apply pulses = pulses * 3 + 1 exactly layers times. The time complexity is O(layers), and the extra space complexity is O(1).

Best Answers

java
class Solution {
    public int coral_pulse_count(Object layers) {
        int n = layers instanceof Integer ? (int) layers : 0;
        if (n == 0) return 1;
        int pulses = 1;
        for (int i = 0; i < n; i++) {
            pulses = 1 + 3 * pulses;
        }
        return pulses;
    }
}