LeetCode 3959: Check Good Integer ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2026-06-14

Submissions: Accepted (first try)

Approach
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squareSum - digitSum splits into a contribution of d² - d per digit. Since d is an integer 0–9, d² - d = d(d - 1) >= 0, so the gap only grows, and once it reaches >= 50 you can return True right away.

My Solution
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class Solution:
    def checkGoodInteger(self, n: int) -> bool:
        gap = 0
        while n:
            d = n % 10
            gap += d * d - d                 # each digit adds >= 0
            if gap >= 50:
                return True
            n //= 10
        return False

I originally tracked sum_ and sSum separately and compared the difference; a single gap is the same thing.

Complexity
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Time O(d), space O(1).

Mistakes to Avoid
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An early return needs the running value to be monotonic; if a digit could contribute a negative amount, you could not conclude midway.

Interview Explanation
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The gap is the sum of d² - d over the digits, which is never negative, so I accumulate as I go and return True at 50, or False after the last digit. Time O(d).