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feat: add solutions to lc problem: No.0719
No.0719.Find K-th Smallest Pair Distance
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‎solution/0700-0799/0719.Find K-th Smallest Pair Distance/README.md

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@@ -53,22 +53,71 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 二分查找**
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先对 $nums$ 数组进行排序，然后在 $[0, nums[n-1]-nums[0]]$ 范围内二分枚举数对距离 $dist$，若 $nums$ 中数对距离小于等于 $dist$ 的数量 $cnt$ 大于等于 $k$，则尝试缩小 $dist$，否则尝试扩大 $dist$。
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时间复杂度 $O(nlogn×logm)$，其中 $n$ 表示 $nums$ 的长度，$m$ 表示 $nums$ 中两个数的最大差值。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def smallestDistancePair(self, nums: List[int], k: int) -> int:
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def count(dist):
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cnt = 0
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for i, b in enumerate(nums):
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a = b - dist
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j = bisect_left(nums, a, 0, i)
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cnt += i - j
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return cnt
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nums.sort()
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return bisect_left(range(nums[-1] - nums[0]), k, key=count)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int smallestDistancePair(int[] nums, int k) {
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Arrays.sort(nums);
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int left = 0, right = nums[nums.length - 1] - nums[0];
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while (left < right) {
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int mid = (left + right) >> 1;
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if (count(mid, nums) >= k) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private int count(int dist, int[] nums) {
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int cnt = 0;
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for (int i = 0; i < nums.length; ++i) {
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int left = 0, right = i;
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while (left < right) {
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int mid = (left + right) >> 1;
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int target = nums[i] - dist;
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if (nums[mid] >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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cnt += i - left;
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}
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return cnt;
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}
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}
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```
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### **TypeScript**
@@ -100,6 +149,72 @@ function smallestDistancePair(nums: number[], k: number): number {
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int smallestDistancePair(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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int left = 0, right = nums.back() - nums.front();
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while (left < right)
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{
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int mid = (left + right) >> 1;
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if (count(mid, k, nums) >= k) right = mid;
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else left = mid + 1;
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}
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return left;
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}
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int count(int dist, int k, vector<int>& nums) {
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int cnt = 0;
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for (int i = 0; i < nums.size(); ++i)
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{
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int target = nums[i] - dist;
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int j = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
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cnt += i - j;
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}
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return cnt;
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}
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};
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```
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### **Go**
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```go
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func smallestDistancePair(nums []int, k int) int {
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sort.Ints(nums)
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n := len(nums)
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left, right := 0, nums[n-1]-nums[0]
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count := func(dist int) int {
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cnt := 0
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for i, v := range nums {
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target := v - dist
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left, right := 0, i
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for left < right {
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mid := (left + right) >> 1
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if nums[mid] >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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cnt += i - left
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}
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return cnt
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}
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for left < right {
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mid := (left + right) >> 1
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if count(mid) >= k {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}
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```
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103218
### **...**
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```

‎solution/0700-0799/0719.Find K-th Smallest Pair Distance/README_EN.md

Lines changed: 110 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -52,13 +52,56 @@ Then the 1<sup>st</sup> smallest distance pair is (1,1), and its distance is 0.
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### **Python3**
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```python
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class Solution:
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def smallestDistancePair(self, nums: List[int], k: int) -> int:
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def count(dist):
58+
cnt = 0
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for i, b in enumerate(nums):
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a = b - dist
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j = bisect_left(nums, a, 0, i)
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cnt += i - j
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return cnt
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nums.sort()
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return bisect_left(range(nums[-1] - nums[0]), k, key=count)
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```
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### **Java**
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```java
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class Solution {
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public int smallestDistancePair(int[] nums, int k) {
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Arrays.sort(nums);
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int left = 0, right = nums[nums.length - 1] - nums[0];
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while (left < right) {
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int mid = (left + right) >> 1;
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if (count(mid, nums) >= k) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private int count(int dist, int[] nums) {
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int cnt = 0;
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for (int i = 0; i < nums.length; ++i) {
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int left = 0, right = i;
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while (left < right) {
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int mid = (left + right) >> 1;
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int target = nums[i] - dist;
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if (nums[mid] >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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cnt += i - left;
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}
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return cnt;
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}
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}
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```
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### **TypeScript**
@@ -74,7 +117,6 @@ function smallestDistancePair(nums: number[], k: number): number {
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let count = 0,
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i = 0;
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for (let j = 0; j < n; j++) {
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// 索引[i, j]距离nums[j]的距离<=mid
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while (nums[j] - nums[i] > mid) {
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i++;
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}
@@ -90,6 +132,72 @@ function smallestDistancePair(nums: number[], k: number): number {
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int smallestDistancePair(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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int left = 0, right = nums.back() - nums.front();
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while (left < right)
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{
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int mid = (left + right) >> 1;
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if (count(mid, k, nums) >= k) right = mid;
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else left = mid + 1;
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}
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return left;
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}
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int count(int dist, int k, vector<int>& nums) {
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int cnt = 0;
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for (int i = 0; i < nums.size(); ++i)
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{
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int target = nums[i] - dist;
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int j = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
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cnt += i - j;
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}
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return cnt;
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}
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};
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```
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### **Go**
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```go
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func smallestDistancePair(nums []int, k int) int {
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sort.Ints(nums)
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n := len(nums)
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left, right := 0, nums[n-1]-nums[0]
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count := func(dist int) int {
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cnt := 0
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for i, v := range nums {
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target := v - dist
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left, right := 0, i
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for left < right {
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mid := (left + right) >> 1
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if nums[mid] >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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cnt += i - left
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}
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return cnt
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}
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for left < right {
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mid := (left + right) >> 1
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if count(mid) >= k {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}
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```
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### **...**
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```

‎solution/0700-0799/0719.Find K-th Smallest Pair Distance/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
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class Solution {
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public:
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int smallestDistancePair(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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int left = 0, right = nums.back() - nums.front();
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while (left < right)
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{
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int mid = (left + right) >> 1;
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if (count(mid, k, nums) >= k) right = mid;
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else left = mid + 1;
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}
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return left;
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}
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int count(int dist, int k, vector<int>& nums) {
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int cnt = 0;
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for (int i = 0; i < nums.size(); ++i)
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{
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int target = nums[i] - dist;
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int j = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
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cnt += i - j;
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}
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return cnt;
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}
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};

‎solution/0700-0799/0719.Find K-th Smallest Pair Distance/Solution.go

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,31 @@
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func smallestDistancePair(nums []int, k int) int {
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sort.Ints(nums)
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n := len(nums)
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left, right := 0, nums[n-1]-nums[0]
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count := func(dist int) int {
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cnt := 0
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for i, v := range nums {
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target := v - dist
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left, right := 0, i
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for left < right {
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mid := (left + right) >> 1
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if nums[mid] >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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cnt += i - left
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}
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return cnt
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}
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for left < right {
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mid := (left + right) >> 1
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if count(mid) >= k {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}

‎solution/0700-0799/0719.Find K-th Smallest Pair Distance/Solution.java

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,33 @@
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class Solution {
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public int smallestDistancePair(int[] nums, int k) {
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Arrays.sort(nums);
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int left = 0, right = nums[nums.length - 1] - nums[0];
5+
while (left < right) {
6+
int mid = (left + right) >> 1;
7+
if (count(mid, nums) >= k) {
8+
right = mid;
9+
} else {
10+
left = mid + 1;
11+
}
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}
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return left;
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}
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private int count(int dist, int[] nums) {
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int cnt = 0;
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for (int i = 0; i < nums.length; ++i) {
19+
int left = 0, right = i;
20+
while (left < right) {
21+
int mid = (left + right) >> 1;
22+
int target = nums[i] - dist;
23+
if (nums[mid] >= target) {
24+
right = mid;
25+
} else {
26+
left = mid + 1;
27+
}
28+
}
29+
cnt += i - left;
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}
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return cnt;
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}
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}

‎solution/0700-0799/0719.Find K-th Smallest Pair Distance/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
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class Solution:
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def smallestDistancePair(self, nums: List[int], k: int) -> int:
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def count(dist):
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cnt = 0
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for i, b in enumerate(nums):
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a = b - dist
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j = bisect_left(nums, a, 0, i)
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cnt += i - j
9+
return cnt
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nums.sort()
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return bisect_left(range(nums[-1] - nums[0]), k, key=count)

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