Leetcode #2698: Find the Punishment Number of an Integer
In this guide, we solve Leetcode #2698 Find the Punishment Number of an Integer in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Math, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
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:
- 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
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
The time complexity is , and the space complexity is , where is the given positive integer. The space complexity is , where is the given positive integer.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.