Find the Smallest Divisor Given a Threshold

class Solution {
public:
    int sumDivisions(vector<int>& nums, int divisor) {
        int totalSum = 0;
        for (int num : nums) {
            totalSum += ceil(num / (double) divisor);
        }
        return totalSum;
    }

    int findSmallestDivisor(vector<int>& nums, int threshold) {
        int left = 1;
        int right = *max_element(nums.begin(), nums.end());
        int selectedDivisor = -1;

        while (left <= right) {
            int middle = (left + right) / 2;
            int tempResult = sumDivisions(nums, middle);
            if (tempResult <= threshold) {
                selectedDivisor = middle;
                right = middle - 1;
            } else {
                left = middle + 1;
            }
        }

        return selectedDivisor;
    }
};

Given the task of finding the smallest divisor such that the sum of the ceiling division of an array by that divisor stays within a specified threshold, the presented C++ solution employs a binary search strategy to efficiently pinpoint the optimal divisor. Here's a breakdown of how this code addresses the problem:

• The function sumDivisions computes the sum of the ceiling results of each element in the given array nums when divided by the proposed divisor. The use of ceil ensures that fractions are rounded up.

• The main function findSmallestDivisor initializes the search range for the divisor: left is set to 1 to start with the smallest possible divisor, while right is assigned the maximum value in the array, as no number larger than this could be a feasible divisor.

• In each iteration of the loop, the mid-point of the current range (left to right) is calculated and treated as the potential divisor. Its suitability is tested by the sumDivisions function; if the sum is within the threshold, the current mid-point could be a potential solution, and the search space is narrowed to the lower half (right = middle-1); otherwise, the search space shifts to the upper half (left = middle+1).

• This binary search continues until the possible range converges (left exceeds right). The value of the variable selectedDivisor at the end of the search will be the smallest divisor that satisfies the condition with respect to the given threshold.

This method maximizes efficiency by utilizing binary search, thus ensuring that the smallest possible divisor is found rapidly without needing to test every possible integer between 1 and the maximum of the array. The solution is directly applicable to scenarios where optimization based on division results is crucial, providing a robust approach to handling division-based threshold constraints.

class Solution {
    // Calculate the sum of elements divided by divisor 
    private int sumOfDivided(int[] elements, int divisor) {
        int total = 0;
        for (int element : elements) {
            total += Math.ceil(element / (double) divisor);
        }
        return total;
    }
    
    public int minimalDivisor(int[] elements, int maxThreshold) {
        int optimalResult = -1;
        
        int lower = 1;
        int upper = elements[0];
        for (int element : elements) {
            upper = Math.max(upper, element);
        }
        
        // Binary search to find the smallest divisor.
        while (lower <= upper) {
            int middle = (lower + upper) / 2;
            int computedResult = sumOfDivided(elements, middle);
            if (computedResult <= maxThreshold) {
                optimalResult = middle;
                upper = middle - 1;
            } else {
                lower = middle + 1;
            }
        }
        
        return optimalResult;
    }
}

To solve the problem of finding the smallest divisor given a threshold in Java, follow this approach:

• Define a helper function sumOfDivided which takes an array of elements and a divisor. This function calculates the sum of each element in the array divided by the divisor, rounded up using Math.ceil.

• Implement the minimalDivisor method which utilizes a binary search to efficiently determine the smallest divisor. Start by finding the range for the search. Set lower as 1 and upper as the maximum value found in the elements array.

• Within a while loop governed by lower <= upper, calculate the midpoint for the current range. Use this midpoint as a divisor in the sumOfDivided function.

• If the result of sumOfDivided is less than or equal to the threshold, update optimalResult to the current divisor (mid) and decrease upper to narrow down the search range.

• If the result is greater than the threshold, increase lower to search for a higher divisor.

• Continue this process until the optimal divisor is found and return it.

This binary search method ensures that the solution is efficient and handles large datasets effectively. By leveraging the properties of binary search, the algorithm significantly reduces the number of necessary calculations as compared to a linear search approach.

// Computes the total sum of divisions with 'divisor'.
const getTotalDivisionSum = (elements, divisor) => {
    let sum = 0;
    for (let i in elements) {
        const element = elements[i];
        sum += Math.ceil((1.0 * element) / divisor);
    }
    return sum;
}

const findOptimalDivisor = function(elements, threshold) {
    let solution = -1;

    // Setup the bounds for binary search with the maximum in elements
    let low = 1;
    let high = elements.reduce((max, n) => Math.max(max, n), elements[0]);

    // Binary search to find the smallest divisor
    while (low <= high) {
        const mid = Math.floor((low + high) / 2);
        const sumResult = getTotalDivisionSum(elements, mid);

        // If sumResult is within bounds, check for an even smaller divisor
        if (sumResult <= threshold) {
            solution = mid;
            high = mid - 1;
        } else {
            // Increase the divisor if threshold is not met
            low = mid + 1;
        }
    }

    return solution;
};

This JavaScript implementation solves the problem of finding the smallest divisor of a list of integers such that the sum of their quotients when divided by this divisor does not exceed a specified threshold. The task leverages binary search, optimizing the process of determining the right divisor.

The solution consists of two functions:

• getTotalDivisionSum(elements, divisor): Calculates the total sum of the elements of an array when each element is divided by a given divisor, and their results are rounded up to the nearest integer. The function iterates through each element, applies the division, rounds up using Math.ceil, and accumulates this in a sum.

• findOptimalDivisor(elements, threshold): Implements binary search to find the minimal divisor. This function starts by setting the initial bounds of the search between 1 and the maximum element of the array. During each iteration of the binary search loop, it calculates a mid-point, uses this as a divisor to get a total sum through getTotalDivisionSum, and adjusts the search bounds based on whether the calculated sum exceeds the threshold. If the sum is less than or equal to the threshold, it keeps a record of the divisor (as a potential smallest divisor) and narrows the search to try finding a smaller valid divisor.

The control flow in the binary search ensures that the smallest suitable divisor is efficiently found, effectively minimizing computational overhead by limiting unnecessary full searches through the input and directly targeting the range where the optimal divisor is likely to exist. At the end of the process, the function returns the divisor that meets the condition specified by the threshold constraint.

class Solution:
    def minDivisor(self, values: List[int], maxLimit: int) -> int:
        
        def compute_sum(div: int) -> int:
            total = 0
            for val in values:
                total += ceil((1.0 * val) / div)
            return total
        
        res = -1
        start = 1
        end = max(values)
        
        while start <= end:
            middle = (start + end) // 2
            calculated_sum = compute_sum(middle)
            if calculated_sum <= maxLimit:
                res = middle
                end = middle - 1
            else:
                start = middle + 1
        
        return res

Problem Summary:

The goal is to find the smallest divisor for an array of integers so that the sum of each integer divided by the divisor, and rounded up to the nearest whole number, is less than or equal to a specified threshold.

Solution Approach:

• Calculate the smallest divisor by implementing a binary search method from 1 to the maximum value in the list.
• Introduce a helper function to compute the sum of ceil values when the array elements are divided by the divisor.
• Start with the lowest possible divisor and increase it gradually using binary search principles until finding the smallest divisor that meets the threshold condition.

Key Variables:

• values: List of integers.
• maxLimit: The threshold sum that the calculation should not exceed.

Method Execution:

1. Define a helper function compute_sum which calculates the total sum required using the ceiling of division results, looping through the given values divided by a divisor.
2. Initialize variables for the binary search: start as 1, end as the maximum value in the list, and res for storing the result.
3. Use a while loop to perform the binary search:
   • Calculate the middle point divisor.
   • Use compute_sum to determine if the resulting sum with this divisor remains under or equal to the maxLimit.
4. Adjust search range based on comparison of the computed sum with maxLimit.
5. Return the smallest valid divisor once the condition is met.

This approach effectively reduces the number of checks needed by using binary search, optimizing the solution by narrowing down potential divisors from a calculated range and tailoring changes based on the sum calculated via the helper function. The algorithm guarantees the finding of a minimum divisor that adheres to the conditions set by the threshold.