Find the Punishment Number of an Integer

class Solution {
public:
    bool isPartitionPossible(int number, int sum) {
        if (sum < 0 || number < sum) {
            return false;
        }

        if (number == sum) {
            return true;
        }

        return isPartitionPossible(number / 10, sum - number % 10) ||
               isPartitionPossible(number / 100, sum - number % 100) ||
               isPartitionPossible(number / 1000, sum - number % 1000);
    }

    int calculateTotalPenalty(int limit) {
        int totalPenalty = 0;

        for (int i = 1; i <= limit; i++) {
            int square = i * i;

            if (isPartitionPossible(square, i)) {
                totalPenalty += square;
            }
        }
        return totalPenalty;
    }
};

This C++ solution describes how to calculate the "total penalty" of integers from 1 up to a given limit. The penalty is calculated based on whether the square of an integer can be partitioned in a way where the sum of the parts equals the integer itself. The essential parts of the implementation include:

• isPartitionPossible Function: This recursive function checks if the square of a number can be split into parts such that the sum of the parts equals the original number. This check is performed by recursively reducing the problem size - first by considering the last digit, then the last two digits, and so on, of the squared number.

• calculateTotalPenalty Function: This function iterates from 1 to the specified limit, squares each integer, and uses isPartitionPossible to check if a valid partition exists. If it does, the square of the number is added to the total penalty.

Key Methods Used:

• Recursion in isPartitionPossible to explore different ways to split the square of the number.
• Loop in calculateTotalPenalty to accumulate penalties for numbers from 1 to the given limit.

This approach is efficient for calculating the sum of all such penalties up to a given limit and highlights how recursion can be utilized in breaking down complex computational problems into manageable sub-problems.

class Solution {

    public boolean checkPartition(int number, int goal) {
        if (goal < 0 || number < goal) {
            return false;
        }

        if (number == goal) {
            return true;
        }

        return (
            checkPartition(number / 10, goal - (number % 10)) ||
            checkPartition(number / 100, goal - (number % 100)) ||
            checkPartition(number / 1000, goal - (number % 1000))
        );
    }

    public int totalPunishment(int max) {
        int total = 0;

        for (int i = 1; i <= max; i++) {
            int squared = i * i;

            if (checkPartition(squared, i)) {
                total += squared;
            }
        }
        return total;
    }
}

The provided Java solution involves calculating the "Punishment Number" of integers up to a given maximum. It primarily consists of two pivotal methods: checkPartition() and totalPunishment(). The approach allows you to determine a specific property related to the numeric partitions of square numbers conformant to their original base value.

• Understanding checkPartition(int number, int goal):

  • This method recursively verifies whether a number can be partitioned in a way that the sum of some combination of its digits equals the goal. A number like 49 (7 squared), for example, can be split into separate digits, 4 and 9, where their sum equals 7.
  • The method diminishes the number and the goal through successive divisions and modulus operations, effectively stripping the last one, two, or three digits of the number and seeing if the reduced forms can still satisfy the condition.

• Understanding totalPunishment(int max):

  • Calculates the cumulative total of all square numbers up to max whose square can be partitioned into digits that sum to form the original number i.
  • Iteratively, each number from 1 through max is squared, after which checkPartition() is called to assess if the squared value can be broken down into a sum of its digits adding up to the original number. If true, its square is added to a running total.

When using this program, you get a resultant number from totalPunishment() indicating the aggregate of those special numerical properties up to a specified maximum. This implementation effectively combines iterative and recursive logic to compute a specific class of numbers satisfying an interesting digit-based property.

class Solution:
    def canDivideIntoSubsetsWithSum(self, bigNumber, sumTarget):
        if sumTarget < 0 or bigNumber < sumTarget:
            return False

        if bigNumber == sumTarget:
            return True

        return (
            self.canDivideIntoSubsetsWithSum(bigNumber // 10, sumTarget - bigNumber % 10)
            or self.canDivideIntoSubsetsWithSum(bigNumber // 100, sumTarget - bigNumber % 100)
            or self.canDivideIntoSubsetsWithSum(bigNumber // 1000, sumTarget - bigNumber % 1000)
        )

    def computePenalty(self, limit: int) -> int:
        total_penalty = 0

        for i in range(1, limit + 1):
            squaredValue = i * i

            if self.canDivideIntoSubsetsWithSum(squaredValue, i):
                total_penalty += squaredValue

        return total_penalty

The program written in Python3 defines a class Solution that provides methods to determine if a squared integer can be divided into subsets that sum up to the integer itself and ultimately computes a "punishment" value based on these findings.

• The canDivideIntoSubsetsWithSum(bigNumber, sumTarget) method checks whether a number, represented by bigNumber, can be divided into subsets which sum up to sumTarget. * The implementation involves recursively checking if the last digit, last two digits, or last three digits can contribute to subsets that equate to sumTarget. * Edge cases are handled at the beginning of the method by returning False if the conditions aren't favorable for forming such subsets. * If bigNumber matches sumTarget, return True.

• The computePenalty(limit: int) -> int method computes the total "punishment" for all integers up to a given limit. * It iterates through each natural number up to the limit, calculates its square and then checks if this squared number can be split into subsets that sum up to the original number using the previously defined method. * The sum of valid squared values constitutes the total "punishment number," which the method returns after finishing the iteration.

The process of calculation involves substantial use of recursion and conditional statements to determine subset divisibility, making it computationally intensive for large numbers. Pay attention to edge cases where the given integer's square precisely matches the integer, as this will automatically satisfy the subset sum condition. This solution offers a mathematical intrigue by connecting number properties (like divisibility and modular arithmetic) with recursive programming techniques.