How do you solve Find the Punishment Number of an Integer on LeetCode?

Find the Punishment Number of an Integer (LeetCode #2698) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m²) time complexity and O(n * m) space complexity. Key insight: Understanding how to partition strings is crucial for solving this problem.

What is the brute force solution for Find the Punishment Number of an Integer?

The brute force approach for Find the Punishment Number of an Integer is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each integer from 1 to n, calculating its square, and then determining if we can partition that square into substrings that sum up to the integer itself. This is straightforward but can be inefficient. The optimal Optimal Solution approach improves this to O(n * m²).

What algorithm or data structure is used to solve Find the Punishment Number of an Integer?

Find the Punishment Number of an Integer uses the following concepts: Math, Backtracking. The recommended patterns to study are: Backtracking, Dynamic Programming.

What are the interview tips for Find the Punishment Number of an Integer?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as the smallest and largest values of n. Explain your thought process while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Find the Punishment Number of an Integer?

Failing to consider all possible partitions of the square string. Not using memoization effectively, leading to excessive recursive calls.

#2698

Find the Punishment Number of an Integer

Medium

MathBacktrackingBacktrackingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m²)
Space	O(1)	O(n * m)

💡

Intuition

Time O(n * m²)Space O(n * m)

The optimal solution leverages memoization to avoid recalculating partitions for the same substrings. This significantly reduces the number of recursive calls and speeds up the process.

⚙️

Algorithm

7 steps

1Step 1: Initialize a variable to hold the punishment number sum.
2Step 2: Create a memoization dictionary to store results of previously computed partitions.
3Step 3: Loop through each integer i from 1 to n.
4Step 4: Calculate i * i and convert it to a string.
5Step 5: Use the memoization function to check if the square can be partitioned to sum to i.
6Step 6: If it can, add i * i to the punishment number sum.
7Step 7: Return the total punishment number sum.

solution.py20 lines

1def punishment_number(n):
2    memo = {}
3    def can_partition(s, target):
4        if (s, target) in memo:
5            return memo[(s, target)]
6        if not s:
7            return target == 0
8        for i in range(1, len(s) + 1):
9            if can_partition(s[i:], target - int(s[:i])):
10                memo[(s, target)] = True
11                return True
12        memo[(s, target)] = False
13        return False
14
15    punishment_sum = 0
16    for i in range(1, n + 1):
17        square_str = str(i * i)
18        if can_partition(square_str, i):
19            punishment_sum += i * i
20    return punishment_sum

ℹ

Complexity note: The time complexity is O(n * m²) where n is the number of integers and m is the maximum length of the square string representation. The space complexity is O(n * m) due to the memoization dictionary storing results for each substring and target combination.

1Understanding how to partition strings is crucial for solving this problem.
2Using memoization can drastically improve performance by avoiding redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.