Leetcode 2740: Find the Value of the Partition

Input: The input consists of a single array of positive integers `nums`.

Example: nums = [3, 7, 2, 8]

Constraints:

• 2 <= nums.length <= 100000

• 1 <= nums[i] <= 1000000000

Output: Return the integer denoting the value of the partition, i.e., the minimum possible absolute difference between the maximum of `nums1` and the minimum of `nums2`.

Example: Output: 1

Constraints:

Goal: To minimize the absolute difference between the maximum of `nums1` and the minimum of `nums2`.

Steps:

• Sort the array `nums` in ascending order.

• Find the smallest difference between two consecutive elements in the sorted array.

• Return this minimum difference as the result.

Goal: The array `nums` has at least two elements and each element is a positive integer within the given range.

Steps:

• 2 <= nums.length <= 100000

• 1 <= nums[i] <= 1000000000

Assumptions:

• All elements in `nums` are positive integers.

• The array is non-empty and contains at least two elements.

• Input: nums = [3, 7, 2, 8]

• Explanation: We can partition the array into `nums1 = [2, 7]` and `nums2 = [3, 8]`. The minimum absolute difference is `|7 - 3| = 1`.

• Input: nums = [15, 2, 5]

• Explanation: We can partition the array into `nums1 = [5]` and `nums2 = [15, 2]`. The minimum absolute difference is `|5 - 2| = 3`.

Approach: The approach involves sorting the array and calculating the smallest difference between two consecutive elements in the sorted array.

Observations:

• Sorting the array helps us easily find the smallest possible difference between two elements.

• The problem can be solved by comparing adjacent elements in the sorted array.

Steps:

• Sort the array in ascending order.

• Iterate through the array and find the minimum difference between consecutive elements.

• Return this minimum difference as the result.

Empty Inputs:

• The input will never be empty as per the constraints.

Large Inputs:

• For large arrays, sorting the array and finding the minimum difference should still be efficient.

Special Values:

• If the array contains two elements, the result is the absolute difference between those two elements.

Constraints:

• The solution needs to handle arrays with up to 100,000 elements efficiently.

int findValueOfPartition(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int res = nums[1] - nums[0];
    for(int i = 1; i < nums.size(); i++) {
        res = min(nums[i] - nums[i - 1], res);
    }
    return res;
}

1 : Function Definition

int findValueOfPartition(vector<int>& nums) {

The function `findValueOfPartition` is defined, which takes a vector of integers `nums` and returns the minimum difference between any two adjacent elements after sorting the array.

2 : Sorting

    sort(nums.begin(), nums.end());

The `nums` array is sorted in non-decreasing order using the `sort` function from the Standard Library.

3 : Variable Initialization

    int res = nums[1] - nums[0];

The variable `res` is initialized to the difference between the first two elements of the sorted array, which will hold the minimum difference found.

4 : Loop - Condition

    for(int i = 1; i < nums.size(); i++) {

A `for` loop is initiated, iterating over the sorted `nums` array starting from index 1.

5 : Min Calculation

        res = min(nums[i] - nums[i - 1], res);

Inside the loop, the difference between the current element `nums[i]` and the previous element `nums[i - 1]` is calculated. The result is compared with the current value of `res`, and the smaller value is assigned back to `res`.

6 : Return Statement

    return res;

The function returns the value of `res`, which contains the minimum difference between adjacent elements in the sorted array.

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(1)

Worst Case: O(n)

Description: The space complexity is O(n) for storing the sorted array, assuming the sorting algorithm uses extra space.

Leetcode 2740: Find the Value of the Partition

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