Maximum Number of Distinct Elements After Operations
MedPrompt
You are given an integer array nums and an integer k.
You are allowed to perform the following operation on each element of the array at most once:
- Add an integer in the range
[-k, k]to the element.
Return the maximum possible number of distinct elements in nums after performing the operations.
Example 1:
Input: nums = [1,2,2,3,3,4], k = 2
Output: 6
Explanation:
nums changes to [-1, 0, 1, 2, 3, 4] after performing operations on the first four elements.
Example 2:
Input: nums = [4,4,4,4], k = 1
Output: 3
Explanation:
By adding -1 to nums[0] and 1 to nums[1], nums changes to [3, 5, 4, 4].
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1090 <= k <= 109
Approaches
2 approaches with complexity analysis and trade-offs.
This approach attempts to solve the problem by exploring all possible ways to assign new values to the elements of nums. For each element, we can change it by adding any integer from -k to k. A recursive function is used to traverse this vast search space of choices to find the assignment that results in the maximum number of distinct elements.
Algorithm
- Define a recursive function, say
solve(index, used_values). indexis the current index in thenumsarray we are considering.used_valuesis a set containing the distinct values assigned to elements fromnums[0]tonums[index-1].- The base case for the recursion is when
index == nums.length. At this point, we have made assignments for all elements, and the number of distinct values isused_values.size(). We return this size. - In the recursive step for
nums[index], we explore all possible valid assignments. For each possible new valuevin the range[nums[index] - k, nums[index] + k], we can either add it to the set if it's new or reuse it if it's already present. - The function would look something like this:
function solve(index, used_values):- If
index == nums.length, returnused_values.size(). max_distinct = 0- For
dfrom-ktok:new_val = nums[index] + dmax_distinct = max(max_distinct, solve(index + 1, used_values U {new_val}))
- Return
max_distinct.
- If
- The initial call would be
solve(0, new HashSet<>()).
Walkthrough
We define a recursive function, say solve(index, used_values), where index is the current element nums[index] we are considering, and used_values is a set containing the distinct values assigned to elements from nums[0] to nums[index-1]. The base case for the recursion is when index reaches the end of the array. At this point, we have made assignments for all elements, and the number of distinct values is used_values.size(). We keep track of the maximum size found across all recursive paths.
In the recursive step for nums[index], we would try to assign it a new value. The new value v must be in the range [nums[index] - k, nums[index] + k]. We would have to iterate through all possible values v in this range, add v to a copy of the used_values set, and recurse with solve(index + 1, new_used_values). The final answer is the maximum size returned by any of these recursive explorations.
This method is a direct translation of the problem statement into a search algorithm. However, the number of choices at each step is 2k + 1, leading to a total number of states of roughly (2k+1)^N, which is computationally intractable for the given constraints.
Complexity
Time
O((2k+1)^N), where N is the length of `nums`. The search space is enormous as each of the N elements has up to `2k+1` choices. This is not a feasible solution.
Space
O(N), for the recursion stack depth and to store the set of used values in a single path.
Trade-offs
Pros
It is a correct formulation of the problem that theoretically explores all possibilities to find the true maximum.
Cons
Extremely inefficient and infeasible for the given constraints.
The range of
kcan be very large, making it impossible to iterate through all possible modificationsdfrom-ktok.
Solutions
Solution
class Solution {public int maxDistinctElements(int[] nums, int k) { Arrays.sort(nums); int n = nums.length; int ans = 0, pre = Integer.MIN_VALUE; for (int x : nums) { int cur = Math.min(x + k, Math.max(x - k, pre + 1)); if (cur > pre) { ++ans; pre = cur; } } return ans; }}Video walkthrough
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Same difficulty — related problems to reinforce the pattern.