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feat: add solutions to lc problem: No.3312 (#3606)
No.3312.Sorted GCD Pair Queries
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‎solution/3300-3399/3312.Sorted GCD Pair Queries/README.md‎

Lines changed: 118 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -84,32 +84,145 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3312.So
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<!-- solution:start -->
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### 方法一
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### 方法一：预处理 + 前缀和 + 二分查找
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我们可以预处理得到数组 $\textit{nums}$ 中的所有数对的最大公约数的出现次数，记录在数组 $\textit{cntG}$ 中。然后，我们计算数组 $\textit{cntG}$ 的前缀和。最后，对于每个查询，我们可以通过二分查找在数组 $\textit{cntG}$ 中找到第一个大于 $\textit{queries}[i]$ 的元素的下标，即为答案。
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我们用 $\textit{mx}$ 表示数组 $\textit{nums}$ 中的最大值，用 $\textit{cnt}$ 记录数组 $\textit{nums}$ 中每个数的出现次数。我们用 $\textit{cntG}[i]$ 表示数组 $\textit{nums}$ 中最大公约数等于 $i$ 的数对个数。为了计算 $\textit{cntG}[i]$，我们可以按照以下步骤进行：
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1. 计算数组 $\textit{nums}$ 中 $i$ 的倍数的出现次数 $v$，那么从这些元素中任选两个元素组成的数对一定满足最大公约数是 $i$ 的倍数，即 $\textit{cntG}[i]$ 需要增加 $v \times (v - 1) / 2$；
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1. 我们需要排除最大公约数是 $i$ 的倍数且大于 $i$ 的数对，因此，对于 $i$ 的倍数 $j$，我们需要减去 $\textit{cntG}[j]$。
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以上需要我们从大到小遍历 $i$，这样才能保证我们在计算 $\textit{cntG}[i]$ 时已经计算了所有的 $\textit{cntG}[j]$。
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最后，我们计算数组 $\textit{cntG}$ 的前缀和，然后对于每个查询，我们可以通过二分查找在数组 $\textit{cntG}$ 中找到第一个大于 $\textit{queries}[i]$ 的元素的下标，即为答案。
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时间复杂度 $O(n + (M + q) \times \log M)$，空间复杂度 $O(M)$。其中 $n$ 和 $M$ 分别是数组 $\textit{nums}$ 的长度和最大值，而 $q$ 是查询的数量。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def gcdValues(self, nums: List[int], queries: List[int]) -> List[int]:
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mx = max(nums)
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cnt = Counter(nums)
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cnt_g = [0] * (mx + 1)
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for i in range(mx, 0, -1):
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v = 0
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for j in range(i, mx + 1, i):
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v += cnt[j]
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cnt_g[i] -= cnt_g[j]
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cnt_g[i] += v * (v - 1) // 2
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s = list(accumulate(cnt_g))
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return [bisect_right(s, q) for q in queries]
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```
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#### Java
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```java
100-
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class Solution {
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public int[] gcdValues(int[] nums, long[] queries) {
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int mx = Arrays.stream(nums).max().getAsInt();
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int[] cnt = new int[mx + 1];
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long[] cntG = new long[mx + 1];
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for (int x : nums) {
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++cnt[x];
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}
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for (int i = mx; i > 0; --i) {
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int v = 0;
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for (int j = i; j <= mx; j += i) {
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v += cnt[j];
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cntG[i] -= cntG[j];
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}
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cntG[i] += 1L * v * (v - 1) / 2;
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}
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for (int i = 2; i <= mx; ++i) {
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cntG[i] += cntG[i - 1];
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}
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int m = queries.length;
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int[] ans = new int[m];
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for (int i = 0; i < m; ++i) {
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ans[i] = search(cntG, queries[i]);
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}
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return ans;
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}
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private int search(long[] nums, long x) {
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int n = nums.length;
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int l = 0, r = n;
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while (l < r) {
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int mid = l + r >> 1;
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if (nums[mid] > x) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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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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vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {
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int mx = ranges::max(nums);
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vector<int> cnt(mx + 1);
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vector<long long> cntG(mx + 1);
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for (int x : nums) {
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++cnt[x];
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}
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for (int i = mx; i; --i) {
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long long v = 0;
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for (int j = i; j <= mx; j += i) {
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v += cnt[j];
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cntG[i] -= cntG[j];
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}
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cntG[i] += 1LL * v * (v - 1) / 2;
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}
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for (int i = 2; i <= mx; ++i) {
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cntG[i] += cntG[i - 1];
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}
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vector<int> ans;
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for (auto&& q : queries) {
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ans.push_back(upper_bound(cntG.begin(), cntG.end(), q) - cntG.begin());
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}
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return ans;
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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 gcdValues(nums []int, queries []int64) (ans []int) {
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mx := slices.Max(nums)
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cnt := make([]int, mx+1)
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cntG := make([]int, mx+1)
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for _, x := range nums {
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cnt[x]++
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}
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for i := mx; i > 0; i-- {
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var v int
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for j := i; j <= mx; j += i {
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v += cnt[j]
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cntG[i] -= cntG[j]
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}
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cntG[i] += v * (v - 1) / 2
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}
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for i := 2; i <= mx; i++ {
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cntG[i] += cntG[i-1]
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}
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for _, q := range queries {
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ans = append(ans, sort.SearchInts(cntG, int(q)+1))
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}
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return
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}
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```
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<!-- tabs:end -->

‎solution/3300-3399/3312.Sorted GCD Pair Queries/README_EN.md‎

Lines changed: 118 additions & 5 deletions
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Preprocessing + Prefix Sum + Binary Search
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We can preprocess to obtain the occurrence count of the greatest common divisor (GCD) of all pairs in the array $\textit{nums}$, recorded in the array $\textit{cntG}$. Then, we calculate the prefix sum of the array $\textit{cntG}$. Finally, for each query, we can use binary search to find the index of the first element in the array $\textit{cntG}$ that is greater than $\textit{queries}[i]$, which is the answer.
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Let $\textit{mx}$ denote the maximum value in the array $\textit{nums}$, and let $\textit{cnt}$ record the occurrence count of each number in the array $\textit{nums}$. Let $\textit{cntG}[i]$ denote the number of pairs in the array $\textit{nums}$ whose GCD is equal to $i$. To calculate $\textit{cntG}[i]$, we can follow these steps:
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1. Calculate the occurrence count $v$ of multiples of $i$ in the array $\textit{nums}$. Then, the number of pairs formed by any two elements from these multiples must have a GCD that is a multiple of $i$, i.e., $\textit{cntG}[i]$ needs to be increased by $v \times (v - 1) / 2$;
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2. We need to exclude pairs whose GCD is a multiple of $i$ and greater than $i$. Therefore, for multiples $j$ of $i$, we need to subtract $\textit{cntG}[j]$.
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The above steps require us to traverse $i$ from large to small so that when calculating $\textit{cntG}[i]$, we have already calculated all $\textit{cntG}[j]$.
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Finally, we calculate the prefix sum of the array $\textit{cntG}$, and for each query, we can use binary search to find the index of the first element in the array $\textit{cntG}$ that is greater than $\textit{queries}[i]$, which is the answer.
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The time complexity is $O(n + (M + q) \times \log M)$, and the space complexity is $O(M)$. Here, $n$ and $M$ represent the length and the maximum value of the array $\textit{nums}$, respectively, and $q$ represents the number of queries.
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<!-- tabs:start -->
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#### Python3
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```python
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105+
class Solution:
106+
def gcdValues(self, nums: List[int], queries: List[int]) -> List[int]:
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mx = max(nums)
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cnt = Counter(nums)
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cnt_g = [0] * (mx + 1)
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for i in range(mx, 0, -1):
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v = 0
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for j in range(i, mx + 1, i):
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v += cnt[j]
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cnt_g[i] -= cnt_g[j]
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cnt_g[i] += v * (v - 1) // 2
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s = list(accumulate(cnt_g))
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return [bisect_right(s, q) for q in queries]
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```
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#### Java
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```java
98-
123+
class Solution {
124+
public int[] gcdValues(int[] nums, long[] queries) {
125+
int mx = Arrays.stream(nums).max().getAsInt();
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int[] cnt = new int[mx + 1];
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long[] cntG = new long[mx + 1];
128+
for (int x : nums) {
129+
++cnt[x];
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}
131+
for (int i = mx; i > 0; --i) {
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int v = 0;
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for (int j = i; j <= mx; j += i) {
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v += cnt[j];
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cntG[i] -= cntG[j];
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}
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cntG[i] += 1L * v * (v - 1) / 2;
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}
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for (int i = 2; i <= mx; ++i) {
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cntG[i] += cntG[i - 1];
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}
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int m = queries.length;
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int[] ans = new int[m];
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for (int i = 0; i < m; ++i) {
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ans[i] = search(cntG, queries[i]);
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}
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return ans;
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}
149+
150+
private int search(long[] nums, long x) {
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int n = nums.length;
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int l = 0, r = n;
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while (l < r) {
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int mid = l + r >> 1;
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if (nums[mid] > x) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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}
99164
```
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#### C++
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103168
```cpp
104-
169+
class Solution {
170+
public:
171+
vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {
172+
int mx = ranges::max(nums);
173+
vector<int> cnt(mx + 1);
174+
vector<long long> cntG(mx + 1);
175+
for (int x : nums) {
176+
++cnt[x];
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}
178+
for (int i = mx; i; --i) {
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long long v = 0;
180+
for (int j = i; j <= mx; j += i) {
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v += cnt[j];
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cntG[i] -= cntG[j];
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}
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cntG[i] += 1LL * v * (v - 1) / 2;
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}
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for (int i = 2; i <= mx; ++i) {
187+
cntG[i] += cntG[i - 1];
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}
189+
vector<int> ans;
190+
for (auto&& q : queries) {
191+
ans.push_back(upper_bound(cntG.begin(), cntG.end(), q) - cntG.begin());
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}
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return ans;
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}
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};
105196
```
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#### Go
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```go
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201+
func gcdValues(nums []int, queries []int64) (ans []int) {
202+
mx := slices.Max(nums)
203+
cnt := make([]int, mx+1)
204+
cntG := make([]int, mx+1)
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for _, x := range nums {
206+
cnt[x]++
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}
208+
for i := mx; i > 0; i-- {
209+
var v int
210+
for j := i; j <= mx; j += i {
211+
v += cnt[j]
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cntG[i] -= cntG[j]
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}
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cntG[i] += v * (v - 1) / 2
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}
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for i := 2; i <= mx; i++ {
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cntG[i] += cntG[i-1]
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}
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for _, q := range queries {
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ans = append(ans, sort.SearchInts(cntG, int(q)+1))
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}
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return
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}
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```
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<!-- tabs:end -->

‎solution/3300-3399/3312.Sorted GCD Pair Queries/Solution.cpp‎

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class Solution {
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public:
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vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {
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int mx = ranges::max(nums);
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vector<int> cnt(mx + 1);
6+
vector<long long> cntG(mx + 1);
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for (int x : nums) {
8+
++cnt[x];
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}
10+
for (int i = mx; i; --i) {
11+
long long v = 0;
12+
for (int j = i; j <= mx; j += i) {
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v += cnt[j];
14+
cntG[i] -= cntG[j];
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}
16+
cntG[i] += 1LL * v * (v - 1) / 2;
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}
18+
for (int i = 2; i <= mx; ++i) {
19+
cntG[i] += cntG[i - 1];
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}
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vector<int> ans;
22+
for (auto&& q : queries) {
23+
ans.push_back(upper_bound(cntG.begin(), cntG.end(), q) - cntG.begin());
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}
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return ans;
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}
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};

‎solution/3300-3399/3312.Sorted GCD Pair Queries/Solution.go‎

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func gcdValues(nums []int, queries []int64) (ans []int) {
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mx := slices.Max(nums)
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cnt := make([]int, mx+1)
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cntG := make([]int, mx+1)
5+
for _, x := range nums {
6+
cnt[x]++
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}
8+
for i := mx; i > 0; i-- {
9+
var v int
10+
for j := i; j <= mx; j += i {
11+
v += cnt[j]
12+
cntG[i] -= cntG[j]
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}
14+
cntG[i] += v * (v - 1) / 2
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}
16+
for i := 2; i <= mx; i++ {
17+
cntG[i] += cntG[i-1]
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}
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for _, q := range queries {
20+
ans = append(ans, sort.SearchInts(cntG, int(q)+1))
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}
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return
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}

‎solution/3300-3399/3312.Sorted GCD Pair Queries/Solution.java‎

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Original file line numberDiff line numberDiff line change
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class Solution {
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public int[] gcdValues(int[] nums, long[] queries) {
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int mx = Arrays.stream(nums).max().getAsInt();
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int[] cnt = new int[mx + 1];
5+
long[] cntG = new long[mx + 1];
6+
for (int x : nums) {
7+
++cnt[x];
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}
9+
for (int i = mx; i > 0; --i) {
10+
int v = 0;
11+
for (int j = i; j <= mx; j += i) {
12+
v += cnt[j];
13+
cntG[i] -= cntG[j];
14+
}
15+
cntG[i] += 1L * v * (v - 1) / 2;
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}
17+
for (int i = 2; i <= mx; ++i) {
18+
cntG[i] += cntG[i - 1];
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}
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int m = queries.length;
21+
int[] ans = new int[m];
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for (int i = 0; i < m; ++i) {
23+
ans[i] = search(cntG, queries[i]);
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}
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return ans;
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}
27+
28+
private int search(long[] nums, long x) {
29+
int n = nums.length;
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int l = 0, r = n;
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while (l < r) {
32+
int mid = l + r >> 1;
33+
if (nums[mid] > x) {
34+
r = mid;
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} else {
36+
l = mid + 1;
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}
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}
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return l;
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}
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}

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