Leetcode 2592: Maximize Greatness of an Array
You are given an integer array
nums. You can permute (rearrange) the elements of nums into a new array perm. Define the greatness of nums as the number of indices 0 <= i < nums.length where perm[i] > nums[i]. Your task is to return the maximum possible greatness you can achieve after permuting nums.Problem
Approach
Steps
Complexity
Input: The input consists of a single integer array `nums` representing the array of integers.
Example: For example, `nums = [3, 1, 4, 2]`.
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^9
Output: The output should be a single integer representing the maximum possible greatness you can achieve after permuting `nums`.
Example: For `nums = [3, 1, 4, 2]`, the output is `3`.
Constraints:
• The output is an integer value representing the maximum greatness.
Goal: The goal is to permute the array `nums` to maximize the number of indices where `perm[i] > nums[i]`.
Steps:
• 1. Sort the array `nums` to create a sorted version.
• 2. Use a greedy approach to find the maximum number of indices where `perm[i] > nums[i]` by pairing elements from the sorted array with the original array.
Goal: The input array `nums` will have a length of at least 1 and at most 10^5. Each element in the array will be a non-negative integer less than or equal to 10^9.
Steps:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^9
Assumptions:
• The input array will always contain at least one integer, and no element will exceed 10^9 in value.
• Input: For `nums = [3, 1, 4, 2]`
• Explanation: By rearranging `nums` into `perm = [4, 2, 3, 1]`, the greatness is 3 because at indices `0, 1, 2`, `perm[i] > nums[i]`. This is the optimal rearrangement.
Approach: The approach involves sorting the array `nums` and then using a greedy approach to match the elements from the sorted array with the original array to maximize the number of times `perm[i] > nums[i]`.
Observations:
• To maximize greatness, we want to pair smaller values in `nums` with larger values from a sorted version of `nums`.
• Sorting the array and then using a greedy approach to match elements can help achieve the maximum greatness.
Steps:
• 1. Sort the array `nums` in non-decreasing order.
• 2. Initialize a pointer to traverse through the sorted array and another pointer to traverse through the original array.
• 3. Count how many times `perm[i] > nums[i]` can be achieved by advancing the pointers in a way that maximizes the number of such occurrences.
Empty Inputs:
• The input array is guaranteed to have at least one element, so there will be no empty inputs.
Large Inputs:
• The solution must handle large inputs efficiently, with the array size up to 10^5.
Special Values:
• If all elements in `nums` are the same, the result will be zero because no element can be greater than another.
Constraints:
• The input constraints ensure that the number of elements will be manageable within the time limits.
int maximizeGreatness(vector<int>& nums) {
// 1 1 1 2 3 3 5
sort(nums.begin(), nums.end());
int n = nums.size();
map<int, int> pos;
pos[-1] = -1;
for(int i = 0; i < n; i++) {
auto it = upper_bound(nums.begin() + pos[i - 1] + 1, nums.end(), nums[i]);
if(it == nums.end()) break;
int idx = it - nums.begin();
pos[i] = idx;
if(idx == n - 1) break;
}
return pos.size() - 1;
}
1 : Function Definition
int maximizeGreatness(vector<int>& nums) {
Defines the `maximizeGreatness` function that takes a vector of integers, `nums`, as an argument.
2 : Sort Array
sort(nums.begin(), nums.end());
Sorts the input array `nums` in non-decreasing order.
3 : Size Calculation
int n = nums.size();
Calculates the size of the sorted array and stores it in variable `n`.
4 : Map Initialization
map<int, int> pos;
Declares a map `pos` to track the positions of elements that are being considered for the result.
5 : Initial Value Assignment
pos[-1] = -1;
Initializes the map `pos` with a value for `-1` key set to `-1`, which serves as the starting point for comparisons.
6 : Loop
for(int i = 0; i < n; i++) {
Starts a loop that iterates over each element in the sorted array `nums`.
7 : Upper Bound Search
auto it = upper_bound(nums.begin() + pos[i - 1] + 1, nums.end(), nums[i]);
Uses the `upper_bound` function to find the first position in the array where the current element `nums[i]` can be inserted while maintaining sorted order.
8 : Break Condition
if(it == nums.end()) break;
Breaks the loop if no valid position is found for the current element.
9 : Index Calculation
int idx = it - nums.begin();
Calculates the index `idx` of the found position where the current element can be inserted.
10 : Map Update
pos[i] = idx;
Updates the `pos` map with the index `idx` for the current element at position `i`.
11 : End Condition Check
if(idx == n - 1) break;
Ends the loop early if the last element of the array is reached.
12 : Return Statement
return pos.size() - 1;
Returns the size of the `pos` map minus one, which represents the maximum greatness of the sequence.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is dominated by the sorting step, which is O(n log n).
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the additional space used for sorting the array.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus