Problem Description

The task here is to find the so-called "punishment number" for a given positive integer n. A "punishment number" is essentially the sum of the squares of particular integers ranging from 1 to n. An integer i (within this range) contributes to this sum if and only if the square of i (i * i) can be split into one or more contiguous substrings, and when these substrings are converted to integers and added together, the sum equals i.

For example, if n is 81, we look at each number from 1 to 81, square them, and see if the square can be partitioned into substrings that add up to the original number. In this case, if n is 81, 81 itself does not contribute because 8181 is 6561 which cannot be split into substrings that add to 81. But the number 9 does contribute because 99 is 81, and 8+1 is 9.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem does not involve nodes and edges as in graph problems.

Need to solve for kth smallest/largest?

• No: The problem is not about finding the kth smallest or largest element.

Involves Linked Lists?

• No: The problem does not involve linked lists.

Does the problem have small constraints?

• Yes: The problem often involves small constraints, as finding punishment numbers might require analyzing a manageable amount of data points or scenarios.

Brute force / Backtracking?

• Yes: Brute force or backtracking could be suitable to explore all possible combinations or operations in searching for the punishment number.

Conclusion: The flowchart suggests an approach that could involve attempting various combinations or procedures akin to backtracking to effectively discover the punishment number of an integer.

Intuition

The solution involves iterating through each number i from 1 to n, calculating i * i, and checking if this square can meet the condition described above.

The intuition behind the implementation is that we can use recursion to check all possible substrings. For every position in the string representation of i * i, we recursively check all ways we could split the remaining string. If we reach the end of the string and the sum of the substrings equals i, then i * i is valid and should be included in our "punishment number" sum.

The check function in the code takes a string s, which represents the square of a number, an index i which indicates where to start partitioning, and an integer x that we are trying to match using the sum of substrings. It tries all possible ways to partition the string from index i onwards and returns true if any of those partitions sum up to x.

The main solution function punishmentNumber goes from 1 to n, checking each number i by calling the check function. If check returns true for a given i, it means we can sum up the substrings of i * i to get i, hence, we add i * i to our cumulative ans which is our punishment number.

This approach efficiently checks all possible partition points in the sequence and uses memorization by the nature of recursion to avoid rechecking the same combination of splits, leading to an optimal solution.

Learn more about Math and Backtracking patterns.

Solution Approach

To solve the "punishment number" problem, the implemented solution leverages recursion and backtracking as its core algorithmic concepts.

The check Helper Function

• This is a recursive function designed to check if a given squared number can be split into substrings that sum up to the original number before squaring.
• It takes the string s which represents the square of i, an index i which represents the current position in s that we're examining, and an integer x which is the value of the original number that we're aiming to reconstruct from the substrings.
• It iterates through the string starting from the current index i and accumulates a value y. If at any point the accumulated y is greater than x, there's no need to continue along that path.
• If y equals x and we are at the end of the string, then we know this is a valid partitioning. If not, we call check recursively with a new starting position just past the current substring we're considering. This recursive call will return True if the subsequent partitioning works out, or keep trying until all possibilities have been exhausted.

The punishmentNumber Main Function

• It iterates from 1 to n. For each iteration:
  • Calculates i * i to get the square of the current number.
  • It turns the square into a string and calls the check function with that string, the number 0 for starting index, and the original number i we are trying to partition into.
• If the check function returns True, which indicates that the square of i can be split into substrings adding up to i, then i * i gets added to the cumulative sum ans.
• After iteration completes, ans holds the sum of all valid squares and is returned as the final answer.

Algorithm and Data Structures

• The main data structure used here is string manipulation, as we're constantly slicing and parsing substrings.
• The algorithm uses recursion extensively to check all possible partitions, which inherently also uses a form of memoization due to overlapping subproblems within the recursive calls.
• The overall approach is a backtracking problem, where we try out different substring splits and backtrack if we realize that the current path does not lead us to a solution.

This solution is efficient because it explores all possible partitions while avoiding unnecessary work, which is reduced by not exploring splits that would lead to values greater than the target sum x. As check progresses, it "cuts off" branches that cannot possibly result in a successful split, thereby streamlining the recursive search process.

Example Walkthrough

Let's walk through an example to illustrate how the implemented solution works. Imagine we are tasked with finding the punishment number for n = 10. We want to check each number from 1 to 10 to see if its square can be split into substrings that sum up to itself.

Step 1: We start with i = 1.

• We calculate its square: 1 * 1 = 1. There are no substrings to consider, so 1 is a valid partition by itself (1 = 1).
• The check function would receive ("1", 0, 1) and return True, so we add 1 * 1 = 1 to the cumulative sum ans.

Step 2: Next, i = 2.

• Square it: 2 * 2 = 4. Similar to 1, 4 does not have substrings to split, so it doesn't meet the condition.
• The check function would receive ("4", 0, 2) and return False.

And we continue this process...

Step 3: For i = 3.

• 3 * 3 = 9. There's no way to split 9 into substrings that add up to 3.
• The check function would receive ("9", 0, 3) and return False.

Step 4: For i = 9 (Skipping a few steps for brevity).

• 9 * 9 = 81. Here we can split 81 as 8 and 1, where 8 + 1 = 9.
• The check function would receive ("81", 0, 9) and it would check the substring "8" (recursively calling check with "1", 1, 1) which returns True.
• Therefore, we add 9 * 9 = 81 to the cumulative sum ans.

Continued Iteration: We would continue iterating over each number up to and including n, performing similar calculations.

Final Step:

• After iterating from 1 through n (which is 10 in our example), we end up with ans holding the sum of valid squares, which, based on the provided numbers, are just 1 and 81.
• The punishment number for n = 10 would therefore be 1 + 81 = 82.

This example demonstrates how the solution uses recursion to check possible splits for the square of each number and how the valid squares contribute to the punishment number.

Solution Implementation

Time and Space Complexity

The given Python code defines a method punishmentNumber which calculates the sum of squares of certain integers up to n that satisfy a particular condition. The condition is verified using a helper method check that implements a recursive approach to determine if the square of the number can be represented as a consecutive sum of integers ending with the number itself.

Time Complexity:

The main time complexity comes from two sources:

1. The outer for loop iterating from 1 to n.
2. The recursive check function called for each i.

For each i, the check function explores different partitions of the digits of i*i into sums that could add up to i. In the worst case, this might result in checking each partition which leads to a search space that is exponential in the number of digits of i*i. However, the recursive search is pruned whenever a partition sum exceeds i, which can significantly reduce the number of searched partitions.

The number of digits d in i*i is O(log(i)) which means the worst case complexity of the check function is O(2^d). Since d is O(log(n)), the recursive calls for each i may reach a complexity of O(2^(2 * log(n))) due to the squaring operation, simplifying the expression to O(n^2) for each i. Multiplying this by n for the for-loop, we get a total worst-case time complexity of O(n^3).

Space Complexity:

The space complexity of the code is primarily determined by the maximum depth of the recursive call stack used by the check function.

The depth of the recursion is limited by the number of digits in x (the square of i), which is O(log(x)) = O(log(i^2)) = O(2*log(i)) = O(log(n)). Therefore, the space complexity is O(log(n)).

Note, the space used to store integers and for iteration is only O(1), which does not contribute significantly compared to the recursive stack.

Learn more about how to find time and space complexity quickly using problem constraints.