2698. Find the Punishment Number of an Integer

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.
i * i 的十进制表示的字符串可以分割成若干连续子字符串，且这些子字符串对应的整数值之和等于 i。

https://leetcode.com/problems/find-the-punishment-number-of-an-integer/

Example 1:

Input: n = 10
Output: 182
Explanation: There are exactly 3 integers i in the range [1, 10] that satisfy the conditions in the statement:

1 since 1 * 1 = 1

9 since 9 * 9 = 81 and 81 can be partitioned into 8 and 1 with a sum equal to 8 + 1 == 9.

10 since 10 * 10 = 100 and 100 can be partitioned into 10 and 0 with a sum equal to 10 + 0 == 10.

Hence, the punishment number of 10 is 1 + 81 + 100 = 182

Example 2:

Input: n = 37
Output: 1478
Explanation: There are exactly 4 integers i in the range [1, 37] that satisfy the conditions in the statement:

1 since 1 * 1 = 1.

9 since 9 * 9 = 81 and 81 can be partitioned into 8 + 1.

10 since 10 * 10 = 100 and 100 can be partitioned into 10 + 0.

36 since 36 * 36 = 1296 and 1296 can be partitioned into 1 + 29 + 6.

Hence, the punishment number of 37 is 1 + 81 + 100 + 1296 = 1478

Constraints:

1 <= n <= 1000

Test Cases

1 2	class Solution: def punishmentNumber(self, n: int) -> int:

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('n, expected', [
    (10, 182),
    (37, 1478),

    (35, 182),
    (36, 1478),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, n, expected):
    assert sol.punishmentNumber(n) == expected

Thoughts

先找出 [1, n] 中所有符合条件的 i（即 i * i 的十进制表示的字符串可以分割成若干连续子字符串，且这些子字符串对应的整数值之和等于 i）。

对于指定的数字 i，判定是否符合上述条件。令 s = str(i * i)，用回溯穷举所有可能的组合，看是否存在一个组合，各部分数字之和等于 i。

对于 s，设起长度为 m，则依次取前 k 个数字（1 ≤ k ≤ m），递归地判定剩下的字符串（s[k:]）是否能够组合出 i - int(s[:k])。

这些可以放到预处理中，设 n 的上限为 N（题中 N = 1000）。对于数字 i，判断是否符合条件的时间复杂度大约为 O(2ᵐ) = O(i)（其中 m = O(log i)）。所以判断 [1, N] 所有数字是否符合条件的时间复杂度为 O(N²)。

用有序数组记录所有符合条件的数字，可以再用同样大小的数组，记录所有前缀子数组的平方和。这样对于指定的 n，可以用二分法快速查找到小于等于 n 的最大的符合条件的数字，然后直接利用第二个数组得到对应的 punishment number。

预处理的时间复杂度为 O(N²)。punishmentNumber 方法的时间复杂度为 O(log n)。

Code

solution.py

from bisect import bisect_right
from itertools import accumulate


def is_special(n: int) -> bool:
    s = str(n ** 2)
    m = len(s)

    def backtrack(pos: int, target: int) -> bool:
        if pos == m:
            return target == 0

        for i in range(pos, m):
            left = int(s[pos:i + 1])
            if left > target:
                break
            if backtrack(i + 1, target - left):
                return True

        return False

    return backtrack(0, n)


MAX = 1001
specials = [i for i in range(1, MAX) if is_special(i)]
square_sums = list(accumulate(i ** 2 for i in specials))


class Solution:
    def punishmentNumber(self, n: int) -> int:
        ip = bisect_right(specials, n)
        return square_sums[ip-1]

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。