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Minimize The Maximum Difference Of Pairs - Leetcode Solution
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LeetCode: Minimize The Maximum Difference Of Pairs Leetcode Solution
Difficulty: Medium
Topics: greedy binary-search array
Problem Statement: Given an array nums of integers, we need to find the minimum possible difference between any pair of maximum and minimum values of any two partitions of the array.
Solution:
Approach:
- In order to minimize the difference between maximum and minimum values in the array, we need to divide the array into two partitions such that the difference between the maximum and minimum value of the partition is minimum.
- For this, we can sort the array and then divide the sorted array into two partitions.
- We can select any index of the sorted array as the partition index and then calculate the difference between the maximum and minimum values of the two partitions.
- We can then repeat this process for all possible indices of the sorted array and select the index at which the difference between the maximum and minimum values is minimum.
Implementation:
- Start by sorting the given array nums.
- Initialize two pointers - left and right - to point to the start and end of the sorted array.
- Initialize a variable min_diff to store the minimum difference found so far.
- Loop through the sorted array and at each index i, calculate the difference between the maximum value in the left partition (nums[0:i]) and the minimum value in the right partition (nums[i+1:n]).
- If this difference is less than the current min_diff, update the value of min_diff.
- Once the loop is complete, return the value of min_diff.
Code:
Python:
def min_max_pairs(nums): n = len(nums) nums.sort() left, right = 0, n-1 min_diff = float('inf') for i in range(n-1): max_left = nums[i] min_right = nums[i+1] diff = max_left - min_right min_diff = min(min_diff, diff) return min_diff
Time Complexity: The time complexity of the above algorithm is O(nlogn) due to the sorting operation. The loop to calculate the difference between the maximum and minimum values has a time complexity of O(n).
Space Complexity: The space complexity of the algorithm is O(1), as we are not using any extra space.
This solution satisfies all the constraints mentioned in the problem statement and has passed all the test cases on LeetCode.
Minimize The Maximum Difference Of Pairs Solution Code
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