Maximum Element After Decreasing and Rearranging

class Solution {
public:
    int maxElementAfterModifying(vector<int>& input) {
        int size = input.size();
        vector<int> frequency = vector(size + 1, 0);
        
        for (int element : input) {
            frequency[min(element, size)]++;
        }
        
        int result = 1;
        for (int index = 2; index <= size; index++) {
            result = min(result + frequency[index], index);
        }
        
        return result;
    }
};

The provided solution in C++ addresses the problem of finding the maximum element possible after rearranging and reducing a given array according to specific rules. The method maxElementAfterModifying is implemented to achieve this goal by first determining the frequency of elements up to the size of the input array. Then, it iteratively adjusts the maximum possible value considering how many times each value or lower can appear in a rearranged version of the array.

Steps involved in the solution:

1. Calculate the size of the given input vector and create a frequency vector of size size + 1, initialized with zeros.
2. Iterate over the input vector to populate the frequency vector. Update the frequency of each element but ensure that elements larger than size are adjusted to size.
3. Initialize a variable result to 1, which will store the maximum element in the rearranged array.
4. Iterate through the frequency vector starting from index 2 up to size, adjusting the result based on the frequency of indices and ensuring it never surpasses the current index.
5. Return the calculated result.

This implementation efficiently computes the maximum possible element by making use of a frequency-based approach, ensuring linear time complexity relative to the size of the input. The use of vector operations and standard library functions in C++ aids in handling the data compactly and performing operations efficiently.

class Solution {
    public int maxElementWithRearrange(int[] array) {
        int size = array.length;
        int[] frequency = new int[size + 1];
        
        for (int value : array) {
            frequency[Math.min(value, size)]++;
        }
        
        int maxPossible = 1;
        for (int i = 2; i <= size; i++) {
            maxPossible = Math.min(maxPossible + frequency[i], i);
        }
        
        return maxPossible;
    }
}

The provided Java solution seeks to find the maximum element possible after reducing the elements of an array and rearranging them.

• Begin by defining the array length size.
• Create an array frequency of size size+1 to store the frequency of occurrence of each element capped by size.
• Iterate over the given array and populate the frequency array, ensuring if values are greater than size, they are capped to size.
• Establish a variable maxPossible initialized to 1 to calculate the maximum potential value.
• Traverse from the second element to the end of the frequency array.
• Update maxPossible within each iteration to be the minimum of itself incremented by the current indexed frequency or the current index.
• The final value of maxPossible at the end of the loop will be the maximum possible value after reducing and rearranging the elements according to the prescribed conditions.
• The function returns this value as the result.

By following these outlined steps, this Java function effectively determines the highest element achievable under the specified operations on the input array.

class Solution:
    def maxElementPostModification(self, array: List[int]) -> int:
        length = len(array)
        frequencies = [0] * (length + 1)

        for value in array:
            frequencies[min(value, length)] += 1

        result = 1
        for value in range(2, length + 1):
            result = min(result + frequencies[value], value)

        return result

The Python program provided outlines a solution for finding the maximum possible element in an array after certain operations, where the array elements can be decreased and rearranged. The function maxElementPostModification calculates this maximum value through the following detailed steps:

1. Determine the length of the given array and initialize a frequency list of zeros that is one element larger than the array length.
2. Populate the frequency list to count occurrences of each integer value in the array, but capped at the array's length to address maximum cut-off.
3. Initialize a result variable starting at 1, as an array must have at least one element.
4. Iteratively calculate possible values, updating the result variable by leveraging a combined check of incremented result with the minimum between it and current range value, which considers rearranged positions.
5. The function returns result, representing the highest number attainable after enforcing the decrements and rearrangements as described by the problem constraints.

This method effectively helps in maximizing an element based on modified array conditions, using frequency counts and iterative checks for feasible maximum values.