701240. Sort by Absolute Difference
Sort by Absolute Difference
Summary
Given an array of integers, sort the elements based on their absolute differences with respect to a given target value. This problem involves arrays and sorting concepts.
Detailed Explanation
To solve this problem, we can use a simple sorting approach. First, calculate the absolute difference between each element in the array and the target value. Then, sort the array based on these calculated differences.
Here's a step-by-step breakdown of the solution:
- Initialize an empty array
resultto store the sorted elements. - Iterate through the input array
arr. - For each element
xinarr, calculate its absolute difference with respect to the target valuetarget. You can do this using the formulaabs(x - target). - Store the calculated difference along with the original element
xin a temporary array or list. - Sort the temporary array or list based on the calculated differences.
- Finally, construct the result array by iterating through the sorted temporary array and appending each element to the result.
Here's an ASCII art diagram illustrating this approach:
+---------------+
| Calculate |
| absolute |
| differences |
+---------------+
|
v
+---------------+
| Sort by |
| differences |
+---------------+
|
v
+---------------+
| Construct |
| result array|
+---------------+
Time complexity: O(n log n), where n is the size of the input array. This is because we are sorting the array based on calculated differences, which takes O(n log n) time.
Space complexity: O(1), as we only use a constant amount of space to store temporary variables and arrays.
Optimized Solutions
Java
public class SortByAbsoluteDifference {
public static int[] sort(int[] arr, int target) {
int[] result = new int[arr.length];
int[] differences = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
differences[i] = Math.abs(arr[i] - target);
}
Arrays.sort(differences);
int j = 0;
for (int difference : differences) {
while (j < arr.length && Math.abs(arr[j] - target) != difference) {
j++;
}
result[j++] = arr[j - 1];
}
return result;
}
}