Leetcode 2616: Minimize the Maximum Difference of Pairs
You are given a 0-indexed integer array
nums and an integer p. Your task is to find p pairs of indices such that the maximum absolute difference between any of the pairs is minimized. Each index can be used in at most one pair. Return the minimized value of the maximum difference among all the pairs.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` and an integer `p`. The array `nums` contains integers, and `p` specifies the number of pairs to form.
Example: nums = [8, 3, 7, 1, 2, 9], p = 3
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^9
• 0 <= p <= (nums.length)/2
Output: Return the minimum possible value of the maximum absolute difference between any of the formed pairs.
Example: Output: 2
Constraints:
• The answer will be a non-negative integer representing the minimized maximum difference.
Goal: The goal is to select `p` pairs such that the maximum absolute difference between any pair is as small as possible.
Steps:
• Step 1: Sort the array to allow easier pairing of close values.
• Step 2: Use binary search to minimize the maximum difference by testing different possible differences and adjusting accordingly.
• Step 3: For each candidate difference, count the number of valid pairs that can be formed and determine if it's feasible to achieve that difference.
• Step 4: Return the smallest possible maximum difference after the binary search concludes.
Goal: The algorithm needs to efficiently handle input sizes up to 100,000 elements and values up to 10^9. The number of pairs `p` can be up to half of the array length.
Steps:
• The algorithm should run in a time complexity that allows handling arrays of size up to 100,000.
Assumptions:
• All elements in `nums` are non-negative integers.
• The array `nums` can contain duplicate values, and the goal is to pair distinct indices that minimize the maximum difference.
• Input: nums = [8, 3, 7, 1, 2, 9], p = 3
• Explanation: After sorting the array as [1, 2, 3, 7, 8, 9], the best way to form 3 pairs is by selecting (1, 2), (3, 7), and (8, 9). The maximum difference is 2 (from the pair (7, 3)), which is the minimized maximum difference.
• Input: nums = [10, 1, 2, 7, 1, 3], p = 2
• Explanation: The array is sorted as [1, 1, 2, 3, 7, 10]. The two best pairs are (1, 1) and (2, 3), with the maximum difference being 1, which is the minimized maximum.
Approach: The approach uses sorting and binary search to find the minimized maximum difference efficiently. The binary search narrows down the possible maximum differences, while sorting simplifies pairing close elements.
Observations:
• Brute force approaches could be too slow, so we need to optimize the solution.
• By sorting the array and applying binary search, we can efficiently minimize the maximum difference while ensuring the correct number of valid pairs.
Steps:
• Step 1: Sort the array `nums` to make pairing easier.
• Step 2: Initialize binary search bounds. The lower bound is 0, and the upper bound is the difference between the largest and smallest elements in `nums`.
• Step 3: Perform binary search. For each mid value (candidate for maximum difference), check how many valid pairs can be formed with differences less than or equal to `mid`.
• Step 4: If `p` valid pairs can be formed, adjust the binary search bounds to minimize the maximum difference.
• Step 5: Return the smallest maximum difference found.
Empty Inputs:
• If `nums` is empty, return 0.
Large Inputs:
• The solution should efficiently handle arrays up to 100,000 elements.
Special Values:
• If `p` is 0, the answer should be 0 since no pairs are formed.
Constraints:
• Ensure the algorithm runs in O(n log(maxDifference)) time where n is the size of the array and maxDifference is the range of values in `nums`.
int minimizeMax(vector<int>& nums, int p) {
sort(nums.begin(), nums.end());
int n = nums.size(), l = 0, r = nums[n - 1] - nums[0], ans = nums[n - 1] - nums[0];
while(l <= r) {
int mid = l + (r - l) / 2;
int k = p;
for(int i = 1; i < n && k > 0; i++) {
if(nums[i] - nums[i - 1] <= mid) {
k--;
i++;
}
}
if(k == 0) {
ans = mid;
r = mid - 1;
} else l = mid + 1;
}
return ans;
}
1 : Code Initialization
int minimizeMax(vector<int>& nums, int p) {
Initialize the function minimizeMax, taking a vector of integers and an integer p as input.
2 : Sorting
sort(nums.begin(), nums.end());
Sort the array nums to facilitate comparison of adjacent elements for minimizing the maximum difference.
3 : Variable Setup
int n = nums.size(), l = 0, r = nums[n - 1] - nums[0], ans = nums[n - 1] - nums[0];
Set the initial values for binary search range (l, r) and the answer variable (ans).
4 : Binary Search
while(l <= r) {
Start the binary search loop to narrow down the range for the minimum maximum difference.
5 : Middle Point Calculation
int mid = l + (r - l) / 2;
Calculate the midpoint (mid) for the current search range to test the possible maximum difference.
6 : Remaining Pairs Setup
int k = p;
Set the number of allowed pairs (k) to p, which is passed as an argument.
7 : Pair Evaluation
for(int i = 1; i < n && k > 0; i++) {
Loop through the array to evaluate possible pairs based on the current maximum allowed difference.
8 : Pair Selection
if(nums[i] - nums[i - 1] <= mid) {
Check if the current pair meets the condition of having a maximum difference less than or equal to mid.
9 : Decrement Allowed Pairs
k--;
Decrement the number of remaining pairs after successfully selecting a pair.
10 : Skip Next Element
i++;
Skip the next element in the array as it has been paired with the current element.
11 : Answer Update
if(k == 0) {
If no more pairs can be made, update the answer with the current mid value.
12 : Binary Search Adjustment
ans = mid;
Update the answer to the current mid value as it represents a potential minimum maximum difference.
13 : Search Range Update
r = mid - 1;
Narrow the binary search range by adjusting the right boundary to mid - 1.
14 : Search Range Update
} else l = mid + 1;
If no valid pairs are found, adjust the left boundary of the binary search range to mid + 1.
15 : Return Final Answer
return ans;
Return the result, which is the minimized maximum difference between pairs.
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 takes O(n log n), followed by a binary search for the maximum difference.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the space required to store the sorted array.
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