Find the Punishment Number of an Integer — LeetCode 2698 Python Solution

The problem

Given a positive integer n, return the punishment number of n. The punishment number of n is defined as the sum of the squares of all integers i such that: 1 <= i <= n The decimal representation of i * i can be partitioned into contiguous substrings such that the sum of the integer values of these substrings equals i.

Example

Input

n = 10

Output

182

Explanation

There are exactly 3 integers i in the range [1, 10] that satisfy the conditions in the statement:

Python solution

class Solution:
    def punishmentNumber(self, n: int) -> int:
        def check(s: str, i: int, x: int) -> bool:
            m = len(s)
            if i >= m:
                return x == 0
            y = 0
            for j in range(i, m):
                y = y * 10 + int(s[j])
                if y > x:
                    break
                if check(s, j + 1, x - y):
                    return True
            return False

        ans = 0
        for i in range(1, n + 1):
            x = i * i
            if check(str(x), 0, i):
                ans += x
        return ans

Complexity

Pattern: Backtracking

Build candidates one choice at a time and abandon a branch the moment it cannot work. LeetCode 2698. Find the Punishment Number of an Integer is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Backtracking.

The backtracking guide has the Python template for the pattern and the 105 LeetCode problems that use it.

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