K-diff Pairs in an Array

/*
LeetCode 532:
Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:
Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
Example 2:
Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
Example 3:
Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
Note:
The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7].

Solution:
Language: C++
Tag: sort binary search hashmap
*/

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        if (k < 0 || nums.size() <= 1) {
            return 0;
        }
        
        // 1. sort + binary search-- O(nlogn) time + O(1) space
        sort(nums.begin(), nums.end(), less<int>());
        //quicksort(nums, 0, nums.size() - 1);
        //heapsort(nums);
        //int *temp = new int[nums.size()];
        //mergesort(nums, 0, nums.size() - 1, temp);
        //free(temp);
        int count = 0;
        for (int i = 0; i < nums.size() - 1; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int start = i + 1, end = nums.size() - 1;
            int target = nums[i] + k;
            while (start + 1 < end) {
                int mid = start + (end - start) / 2;
                if (nums[mid] < target) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
            if (nums[start] == target || nums[end] == target) {
                ++count;
            }
        }
        return count;
        
        
        /*
        // 2. hashmap -- O(n) time + O(n) space
        unordered_map<int, int> hash;
        for (int num : nums) {
            if (hash.find(num) == hash.end()) {
                hash[num] = 1;
            } else {
                ++hash[num];
            }
        }
        int count = 0;
        for (pair<int, int> item : hash) {
            if (k == 0 && item.second >= 2) {
                ++count;
            } else if (k > 0) {
                if (hash.find(item.first + k) != hash.end()) {
                    ++count;
                }
            }
        }
        return count;
        */
    }
    
    void quicksort(vector<int> &nums, int start, int end) {
        if (start >= end) {
            return;
        }
        int left = start, right = end, pivot = nums[(start + end) / 2];
        while (left <= right) {
            while (left <= right && nums[left] < pivot) {
                ++left;
            }
            while (left <= right && nums[right] > pivot) {
                --right;
            }
            if (left <= right) {
                swap(nums[left++], nums[right--]);
            }
        }
        quicksort(nums, start, right);
        quicksort(nums, left, end);
    }
    
    void mergesort(vector<int> &nums, int start, int end, int *temp) {
        if (start >= end) {
            return;
        }
        int mid = start + (end - start) / 2;
        mergesort(nums, start, mid, temp);
        mergesort(nums, mid + 1, end, temp);
        merge(nums, start, end, temp);
    }
    
    void merge(vector<int> &nums, int start, int end, int *temp) {
        int mid = start + (end - start) / 2;
        int left = start, right = mid + 1, idx = start; 
        while (left <= mid && right <= end) {
            if (nums[left] <= nums[right]) {
                temp[idx++] = nums[left++];
            } else {
                temp[idx++] = nums[right++];
            }
        }
        while (left <= mid) {
            temp[idx++] = nums[left++];
        }
        while (right <= end) {
            temp[idx++] = nums[right++];
        }
        for (int i = start; i <= end; ++i) {
            nums[i] = temp[i];
        }
    }
    
    void heapsort(vector<int> &nums) {
        int size = nums.size();
        for (int i = (size - 1) / 2; i >= 0; --i) {
            siftdown(nums, i, size);
        }
        while (size > 0) {
            swap(nums[0], nums[--size]);
            siftdown(nums, 0, size);
        }
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            swap(nums[left++], nums[right--]);
        }
    }
    
    void siftdown(vector<int> &nums, int father, int size) {
        while (father < size) {
            int min_idx = father;
            int left = father * 2 + 1, right = father * 2 + 2;
            if (left < size && nums[left] < nums[min_idx]) {
                min_idx = left;
            }
            if (right < size && nums[right] < nums[min_idx]) {
                min_idx = right;
            }
            if (father == min_idx) {
                break;
            }
            swap(nums[father], nums[min_idx]);
            father = min_idx;
        }
    }
};