710277. Maximum selections
Maximum selections
Slug: maximum-selections--170637 Difficulty: Medium Id: 710277 Topic Tags: Sorting, Algorithms Company Tags: None
Summary
The problem is to find the maximum number of pairs that can be formed from a given array of integers such that each pair has a difference of at most k. This problem involves sorting and optimization techniques.
Detailed Explanation
To solve this problem, we need to first sort the array in ascending order. Then, we initialize two pointers, i and j, to the start and end of the array respectively. We iterate through the array until i meets j. For each iteration, we check if the difference between the elements at indices i and j is less than or equal to k. If it is, we increment the count of pairs and move both pointers towards the center of the array. If not, we only move the pointer that is closer to the other. This ensures that we consider all possible pairs in the array while keeping track of the maximum number of pairs.
Here's a step-by-step breakdown of the solution:
- Sort the array in ascending order.
- Initialize
iandjto 0 andn-1respectively, wherenis the length of the array. - Iterate through the array until
imeetsj. - For each iteration, check if the difference between the elements at indices
iandjis less than or equal tok. If it is:- Increment the count of pairs.
- Move both pointers towards the center of the array (i.e., increment
iand decrementj).
- If not, move the pointer that is closer to the other.
- Repeat steps 4-5 until
imeetsj. - Return the count of pairs as the maximum number of selections.
Time complexity: O(n log n) due to sorting Space complexity: O(1)
Optimized Solutions
Java
public int maxSelections(int[] arr, int k) {
Arrays.sort(arr);
int i = 0, j = arr.length - 1;
int count = 0;
while (i < j) {
if (arr[j] - arr[i] <= k) {
count++;
i++;
j--;
} else if (arr[j] - arr[i] > k) {
j--;
} else {
i++;
}
}
return count;
}