876737. Sort an Array of 1 to n
Sort an Array of 1 to n
Summary
This problem involves sorting a given array of integers from 1 to n, where n is the length of the array. The task is to arrange the elements in ascending order using mathematical formulae.
Detailed Explanation
The key concept involved in this problem is the mathematical property that the sum of all numbers from 1 to n is equal to (n * (n + 1)) / 2. This property can be used to calculate the correct position of each number in the array based on its value.
Here's a step-by-step breakdown of the solution:
- Initialize an empty array
arrwith lengthn. - Calculate the sum of all numbers from 1 to n using the formula
(n * (n + 1)) / 2. Let this sum betotal_sum. - Iterate through the input array and for each number
i, calculate its correct position in the sorted array by subtracting 1 fromiand multiplying it with(total_sum - i) / (i * (i - 1)) / 2. This value represents the correct position ofiin the sorted array. - Place
iat the calculated position in thearrarray.
Here's a sample ASCII art diagram for better understanding:
Input Array: [3, 1, 4, 2]
Correct Positions:
- 3: (6 - 3) / (3 * (3 - 1)) = 2
- 1: (6 - 1) / (1 * (1 - 1)) = 0
- 4: (6 - 4) / (4 * (4 - 1)) = 4
- 2: (6 - 2) / (2 * (2 - 1)) = 3
Sorted Array: [1, 2, 3, 4]
Time complexity for this solution is O(n), as we are iterating through the input array once. Space complexity is also O(1), as we are only using a constant amount of space to store the arr and intermediate variables.
Optimized Solutions
Java
public static void sortArray(int[] arr) {
int n = arr.length;
int total_sum = (n * (n + 1)) / 2;
for (int i = 0; i < n; i++) {
int correct_pos = total_sum - i;
for (int j = 0; j < correct_pos; j++) {
arr[j] = arr[j + 1];
}
arr[correct_pos] = i + 1;
}
}