<ul><li>The problem can be solved using two approaches:</li><li>Brute Force approach: Sort the array in ascending order and create a set for unique elements. Loop through the array and for each element, check if NUMS[i]-K and NUMS[i]+K are not present in the set. If not present, add them to the set. Return the size of the set for the maximum number of distinct elements.</li><li>Time complexity: O(n*log(n) + n*k)</li><li>Greedy approach: Sort the array in ascending order. Initialize a minimum value as INT_MIN. Loop through each element in the array and calculate the start value as max(minimum, NUMS[i]-K). If start ≤ NUMS[i]+K, increment the count. Update the minimum value to start + 1. Return the count for the maximum number of distinct elements.</li><li>Time complexity: O(n*log(n) + n)</li></ul>

3397. Maximum Number of Distinct Elements After Operations | Leetcode |Medium.

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