Problem statement

A neon number is a number whose square's digits add up to the number itself.

Given a non-negative integer n, return true if n is a neon number.

Example 1

Input: n = 9
Output: true
Explanation: 9² = 81 and 8 + 1 = 9.

Example 2

Input: n = 8
Output: false
Explanation: 8² = 64 and 6 + 4 = 10, not 8.

Example 3

Input: n = 1
Output: true
Explanation: 1² = 1.

Constraints

• 0 <= n <= 100000

Hints for Neon Number

1. Hint 1

   Square the number, then sum the digits of the square.

2. Hint 2

   Compare that digit sum with the original number.

3. Hint 3

   In the range of an int there are only four neon numbers — 0, 1, 9 — so most inputs answer false.

How to solve Neon Number

Square the number, sum the digits of the square, and compare with the original.

Approach

1. Compute sq = n * n.
2. Peel the digits of sq and add them up.
3. Return total == n.

Why it works

The definition is literal. The digit sum of a 2k-digit square is at most 18k, which grows much slower than n, so only 0, 1 and 9 ever satisfy it.

Complexity

• Time — O(log n)
• Space — O(1)

Pitfalls

• 0 is neon (0² = 0), and the loop handles it because it runs zero times and the total is 0.
• Watch the square's width in fixed-width languages.

Reference solution

Python

def isNeon(n: int) -> bool:
    sq = n * n
    total = 0
    while sq > 0:
        total += sq % 10
        sq //= 10
    return total == n

Java

public static boolean isNeon(int n) {
    long sq = (long) n * n;
    long total = 0;
    while (sq > 0) {
        total += sq % 10;
        sq /= 10;
    }
    return total == n;
}

C++

bool isNeon(int n) {
    long long sq = (long long) n * n;
    long long total = 0;
    while (sq > 0) {
        total += sq % 10;
        sq /= 10;
    }
    return total == (long long) n;
}

JavaScript

var isNeon = function(n) {
    let sq = n * n, total = 0;
    while (sq > 0) {
        total += sq % 10;
        sq = Math.floor(sq / 10);
    }
    return total === n;
};

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