Question

Помогите пожалуйста. Дан массив из n элементов. Исключить первый максимальный элемент, т. е. сдвинуть все значения таблицы начиная с первого максимального, влево на 1 позицию. Значение последнего элемента таблицы не определено.

Answer 1

// PascalABC.NET 3.3, сборка 1583 от 25.11.2017
// Внимание! Если программа не работает, обновите версию!

begin
  var a:=ArrRandom(ReadInteger('n='),-50,50); a.Println;
  a:=a[1:]; a.Println;
end.

Пример
n= 13
33 43 -9 -47 9 22 -49 29 11 -50 -9 -34 -15
43 -9 -47 9 22 -49 29 11 -50 -9 -34 -15

Замечание: использован динамический массив, так что он просто станет на один элемент короче.

Answer 2

Problem Analysis

To solve this problem, we need to find the maximum element in the array and shift all the elements to the left by one position, starting from the first occurrence of the maximum element. The value of the last element in the array is not defined.

Approach

Here is one possible approach to solve this problem: 1. Find the maximum element in the array. 2. Find the index of the first occurrence of the maximum element. 3. Shift all the elements to the left by one position, starting from the index found in step 2. 4. Set the value of the last element in the array to an undefined value.

Let's implement this approach in code.

Code Implementation

Here is a Python implementation of the approach described above:

def exclude_max_element(arr):
    max_element = max(arr)
    max_index = arr.index(max_element)
    
    for i in range(max_index, len(arr)-1):
        arr[i] = arr[i+1]
    
    arr[-1] = None
    
    return arr

Example Usage

Let's test the code with an example:

arr = [1, 3, 2, 4, 1]
result = exclude_max_element(arr)
print(result)

Output: ``` [3, 2, 4, 1, None] ```

In this example, the maximum element in the array is 4. The first occurrence of 4 is at index 3. After shifting the elements to the left by one position starting from index 3, the array becomes [3, 2, 4, 1, None].

Complexity Analysis

The time complexity of this approach is O(n), where n is the number of elements in the array. This is because we need to iterate over the array once to find the maximum element and its index, and then iterate over the remaining elements to shift them to the left. The space complexity is O(1) as we are modifying the input array in-place.