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feat: add solutions to lc problem: No.2040 (#3098)
No.2040.Kth Smallest Product of Two Sorted Arrays
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‎solution/2000-2099/2037.Minimum Number of Moves to Seat Everyone/README.md

+1-6
Original file line numberDiff line numberDiff line change
@@ -163,12 +163,7 @@ func abs(x int) int {
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function minMovesToSeat(seats: number[], students: number[]): number {
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seats.sort((a, b) => a - b);
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students.sort((a, b) => a - b);
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const n = seats.length;
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let ans = 0;
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for (let i = 0; i < n; i++) {
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ans += Math.abs(seats[i] - students[i]);
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}
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return ans;
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return seats.reduce((acc, seat, i) => acc + Math.abs(seat - students[i]), 0);
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}
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```
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‎solution/2000-2099/2037.Minimum Number of Moves to Seat Everyone/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -165,12 +165,7 @@ func abs(x int) int {
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function minMovesToSeat(seats: number[], students: number[]): number {
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seats.sort((a, b) => a - b);
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students.sort((a, b) => a - b);
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const n = seats.length;
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let ans = 0;
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for (let i = 0; i < n; i++) {
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ans += Math.abs(seats[i] - students[i]);
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}
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return ans;
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return seats.reduce((acc, seat, i) => acc + Math.abs(seat - students[i]), 0);
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}
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```
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‎solution/2000-2099/2037.Minimum Number of Moves to Seat Everyone/Solution.ts

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Original file line numberDiff line numberDiff line change
@@ -1,10 +1,5 @@
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function minMovesToSeat(seats: number[], students: number[]): number {
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seats.sort((a, b) => a - b);
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students.sort((a, b) => a - b);
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const n = seats.length;
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let ans = 0;
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for (let i = 0; i < n; i++) {
7-
ans += Math.abs(seats[i] - students[i]);
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}
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return ans;
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return seats.reduce((acc, seat, i) => acc + Math.abs(seat - students[i]), 0);
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}

‎solution/2000-2099/2040.Kth Smallest Product of Two Sorted Arrays/README.md

+195-5
Original file line numberDiff line numberDiff line change
@@ -75,32 +75,222 @@ tags:
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<!-- solution:start -->
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78-
### 方法一
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### 方法一：二分查找
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我们可以二分枚举乘积的值 $p$，定义二分的区间为 $[l, r]$，其中 $l = -\text{max}(|\text{nums1}[0]|, |\text{nums1}[n - 1]|) \times \text{max}(|\text{nums2}[0]|, |\text{nums2}[n - 1]|)$, $r = -l$。
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对于每个 $p$，我们计算出乘积小于等于 $p$ 的乘积的个数，如果这个个数大于等于 $k$，那么说明第 $k$ 小的乘积一定小于等于 $p$，我们就可以将区间右端点缩小到 $p$，否则我们将区间左端点增大到 $p + 1$。
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那么问题的关键就是如何计算乘积小于等于 $p$ 的乘积的个数。我们可以枚举 $\text{nums1}$ 中的每个数 $x$，分类讨论：
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- 如果 $x > 0$，那么 $x \times \text{nums2}[i]$ 随着 $i$ 的增大是单调递增的，我们可以使用二分查找找到最小的 $i$，使得 $x \times \text{nums2}[i] > p$，那么 $i$ 就是小于等于 $p$ 的乘积的个数，累加到个数 $\text{cnt}$ 中；
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- 如果 $x < 0$，那么 $x \times \text{nums2}[i]$ 随着 $i$ 的增大是单调递减的，我们可以使用二分查找找到最小的 $i$，使得 $x \times \text{nums2}[i] \leq p$，那么 $n - i$ 就是小于等于 $p$ 的乘积的个数，累加到个数 $\text{cnt}$ 中；
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- 如果 $x = 0$，那么 $x \times \text{nums2}[i] = 0$，如果 $p \geq 0$，那么 $n$ 就是小于等于 $p$ 的乘积的个数，累加到个数 $\text{cnt}$ 中。
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这样我们就可以通过二分查找找到第 $k$ 小的乘积。
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时间复杂度 $O(m \times \log n \times \log M)$，其中 $m$ 和 $n$ 分别为 $\text{nums1}$ 和 $\text{nums2}$ 的长度，而 $M$ 为 $\text{nums1}$ 和 $\text{nums2}$ 中的最大值的绝对值。
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<!-- tabs:start -->
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#### Python3
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8498
```python
85-
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class Solution:
100+
def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
101+
def count(p: int) -> int:
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cnt = 0
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n = len(nums2)
104+
for x in nums1:
105+
if x > 0:
106+
cnt += bisect_right(nums2, p / x)
107+
elif x < 0:
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cnt += n - bisect_left(nums2, p / x)
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else:
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cnt += n * int(p >= 0)
111+
return cnt
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mx = max(abs(nums1[0]), abs(nums1[-1])) * max(abs(nums2[0]), abs(nums2[-1]))
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return bisect_left(range(-mx, mx + 1), k, key=count) - mx
86115
```
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#### Java
89118

90119
```java
91-
120+
class Solution {
121+
private int[] nums1;
122+
private int[] nums2;
123+
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public long kthSmallestProduct(int[] nums1, int[] nums2, long k) {
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this.nums1 = nums1;
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this.nums2 = nums2;
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int m = nums1.length;
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int n = nums2.length;
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int a = Math.max(Math.abs(nums1[0]), Math.abs(nums1[m - 1]));
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int b = Math.max(Math.abs(nums2[0]), Math.abs(nums2[n - 1]));
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long r = (long) a * b;
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long l = (long) -a * b;
133+
while (l < r) {
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long mid = (l + r) >> 1;
135+
if (count(mid) >= k) {
136+
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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private long count(long p) {
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long cnt = 0;
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int n = nums2.length;
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for (int x : nums1) {
148+
if (x > 0) {
149+
int l = 0, r = n;
150+
while (l < r) {
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int mid = (l + r) >> 1;
152+
if ((long) x * nums2[mid] > p) {
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r = mid;
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} else {
155+
l = mid + 1;
156+
}
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}
158+
cnt += l;
159+
} else if (x < 0) {
160+
int l = 0, r = n;
161+
while (l < r) {
162+
int mid = (l + r) >> 1;
163+
if ((long) x * nums2[mid] <= p) {
164+
r = mid;
165+
} else {
166+
l = mid + 1;
167+
}
168+
}
169+
cnt += n - l;
170+
} else if (p >= 0) {
171+
cnt += n;
172+
}
173+
}
174+
return cnt;
175+
}
176+
}
92177
```
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94179
#### C++
95180

96181
```cpp
97-
182+
class Solution {
183+
public:
184+
long long kthSmallestProduct(vector<int>& nums1, vector<int>& nums2, long long k) {
185+
int m = nums1.size(), n = nums2.size();
186+
int a = max(abs(nums1[0]), abs(nums1[m - 1]));
187+
int b = max(abs(nums2[0]), abs(nums2[n - 1]));
188+
long long r = 1LL * a * b;
189+
long long l = -r;
190+
auto count = [&](long long p) {
191+
long long cnt = 0;
192+
for (int x : nums1) {
193+
if (x > 0) {
194+
int l = 0, r = n;
195+
while (l < r) {
196+
int mid = (l + r) >> 1;
197+
if (1LL * x * nums2[mid] > p) {
198+
r = mid;
199+
} else {
200+
l = mid + 1;
201+
}
202+
}
203+
cnt += l;
204+
} else if (x < 0) {
205+
int l = 0, r = n;
206+
while (l < r) {
207+
int mid = (l + r) >> 1;
208+
if (1LL * x * nums2[mid] <= p) {
209+
r = mid;
210+
} else {
211+
l = mid + 1;
212+
}
213+
}
214+
cnt += n - l;
215+
} else if (p >= 0) {
216+
cnt += n;
217+
}
218+
}
219+
return cnt;
220+
};
221+
while (l < r) {
222+
long long mid = (l + r) >> 1;
223+
if (count(mid) >= k) {
224+
r = mid;
225+
} else {
226+
l = mid + 1;
227+
}
228+
}
229+
return l;
230+
}
231+
};
98232
```
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100234
#### Go
101235
102236
```go
103-
237+
func kthSmallestProduct(nums1 []int, nums2 []int, k int64) int64 {
238+
m := len(nums1)
239+
n := len(nums2)
240+
a := max(abs(nums1[0]), abs(nums1[m-1]))
241+
b := max(abs(nums2[0]), abs(nums2[n-1]))
242+
r := int64(a) * int64(b)
243+
l := -r
244+
245+
count := func(p int64) int64 {
246+
var cnt int64
247+
for _, x := range nums1 {
248+
if x > 0 {
249+
l, r := 0, n
250+
for l < r {
251+
mid := (l + r) >> 1
252+
if int64(x)*int64(nums2[mid]) > p {
253+
r = mid
254+
} else {
255+
l = mid + 1
256+
}
257+
}
258+
cnt += int64(l)
259+
} else if x < 0 {
260+
l, r := 0, n
261+
for l < r {
262+
mid := (l + r) >> 1
263+
if int64(x)*int64(nums2[mid]) <= p {
264+
r = mid
265+
} else {
266+
l = mid + 1
267+
}
268+
}
269+
cnt += int64(n - l)
270+
} else if p >= 0 {
271+
cnt += int64(n)
272+
}
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}
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return cnt
275+
}
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277+
for l < r {
278+
mid := (l + r) >> 1
279+
if count(mid) >= k {
280+
r = mid
281+
} else {
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l = mid + 1
283+
}
284+
}
285+
return l
286+
}
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288+
func abs(x int) int {
289+
if x < 0 {
290+
return -x
291+
}
292+
return x
293+
}
104294
```
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<!-- tabs:end -->

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