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feat: add solutions for lc No.2541 (#5192)
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‎solution/2500-2599/2540.Minimum Common Value/README.md‎

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@@ -214,39 +214,4 @@ int getCommon(int* nums1, int nums1Size, int* nums2, int nums2Size) {
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二
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<!-- tabs:start -->
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#### Rust
224-
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```rust
226-
impl Solution {
227-
pub fn get_common(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
228-
let mut iter1 = nums1.iter();
229-
let mut iter2 = nums2.iter();
230-
let mut num1 = iter1.next();
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let mut num2 = iter2.next();
232-
233-
while let (Some(n1), Some(n2)) = (num1, num2) {
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if n1 == n2 {
235-
return *n1;
236-
} else if n1 < n2 {
237-
num1 = iter1.next();
238-
} else {
239-
num2 = iter2.next();
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}
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}
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-1
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/2500-2599/2540.Minimum Common Value/README_EN.md‎

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@@ -214,39 +214,4 @@ int getCommon(int* nums1, int nums1Size, int* nums2, int nums2Size) {
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2
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<!-- tabs:start -->
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#### Rust
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```rust
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impl Solution {
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pub fn get_common(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
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let mut iter1 = nums1.iter();
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let mut iter2 = nums2.iter();
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let mut num1 = iter1.next();
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let mut num2 = iter2.next();
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while let (Some(n1), Some(n2)) = (num1, num2) {
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if n1 == n2 {
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return *n1;
236-
} else if n1 < n2 {
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num1 = iter1.next();
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} else {
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num2 = iter2.next();
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}
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}
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-1
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/2500-2599/2540.Minimum Common Value/Solution2.rs‎

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This file was deleted.

‎solution/2500-2599/2541.Minimum Operations to Make Array Equal II/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -67,13 +67,13 @@ tags:
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### 方法一：一次遍历
6969

70-
我们用变量 $x$ 记录加减次数的差值，用变量 $ans$ 记录操作次数。
70+
我们用两个变量 $a$ 和 $b$ 分别记录将 $\textit{nums1}$ 中的元素增加 $k$ 和减少 $k$ 的次数。
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72-
遍历数组，对于每个位置 $i$，如果存在 $k=0$ 并且 $a_i \neq b_i$，则无法使两个数组相等，返回 $-1$。否则，如果 $k \neq 0$，则 $a_i - b_i$ 必须是 $k$ 的倍数，否则无法使两个数组相等，返回 $-1$。接下来，我们更新 $x$ 和 $ans$。
72+
我们遍历两个数组，如果两个指针指向的元素相等，则继续；如果两个指针指向的元素不相等，则如果 $k$ 等于 $0$ 或者两个元素之差不能被 $k$ 整除，则返回 $-1$；否则将两个元素之差除以 $k$ 得到操作数 $t$，如果 $t$ 小于 $0$，则将 $-t$ 加到 $a$ 上，否则将 $t$ 加到 $b$ 上。
7373

74-
最后，如果 $x \neq 0$，则无法使两个数组相等，返回 $-1$。否则，返回 $\frac{ans}{2}$。
74+
最后如果 $a$ 和 $b$ 相等，则返回 $a$，否则返回 $-1$。
7575

76-
时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
76+
时间复杂度 $O(n)$，其中 $n$ 是两个数组的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -82,42 +82,42 @@ tags:
8282
```python
8383
class Solution:
8484
def minOperations(self, nums1: List[int], nums2: List[int], k: int) -> int:
85-
ans = x = 0
86-
for a, b in zip(nums1, nums2):
87-
if k == 0:
88-
if a != b:
89-
return -1
85+
a = b = 0
86+
for x, y in zip(nums1, nums2):
87+
if x == y:
9088
continue
91-
if (a - b) % k:
89+
if k == 0 or (x - y) % k:
9290
return -1
93-
y = (a - b) // k
94-
ans += abs(y)
95-
x += y
96-
return -1 if x else ans // 2
91+
t = (x - y) // k
92+
if t < 0:
93+
a += -t
94+
else:
95+
b += t
96+
return a if a == b else -1
9797
```
9898

9999
#### Java
100100

101101
```java
102102
class Solution {
103103
public long minOperations(int[] nums1, int[] nums2, int k) {
104-
long ans = 0, x = 0;
104+
long a = 0, b = 0;
105105
for (int i = 0; i < nums1.length; ++i) {
106-
int a = nums1[i], b = nums2[i];
107-
if (k == 0) {
108-
if (a != b) {
109-
return -1;
110-
}
106+
int x = nums1[i], y = nums2[i];
107+
if (x == y) {
111108
continue;
112109
}
113-
if ((a - b) % k != 0) {
110+
if (k == 0 || (x - y) % k != 0) {
114111
return -1;
115112
}
116-
int y = (a - b) / k;
117-
ans += Math.abs(y);
118-
x += y;
113+
int t = (x - y) / k;
114+
if (t < 0) {
115+
a += -t;
116+
} else {
117+
b += t;
118+
}
119119
}
120-
return x == 0 ? ans / 2 : -1;
120+
return a == b ? a : -1;
121121
}
122122
}
123123
```
@@ -128,23 +128,23 @@ class Solution {
128128
class Solution {
129129
public:
130130
long long minOperations(vector<int>& nums1, vector<int>& nums2, int k) {
131-
long long ans = 0, x = 0;
131+
long long a = 0, b = 0;
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for (int i = 0; i < nums1.size(); ++i) {
133-
int a = nums1[i], b = nums2[i];
134-
if (k == 0) {
135-
if (a != b) {
136-
return -1;
137-
}
133+
int x = nums1[i], y = nums2[i];
134+
if (x == y) {
138135
continue;
139136
}
140-
if ((a - b) % k != 0) {
137+
if (k == 0 || (x - y) % k != 0) {
141138
return -1;
142139
}
143-
int y = (a - b) / k;
144-
ans += abs(y);
145-
x += y;
140+
int t = (x - y) / k;
141+
if (t < 0) {
142+
a += -t;
143+
} else {
144+
b += t;
145+
}
146146
}
147-
return x == 0 ? ans / 2 : -1;
147+
return a == b ? a : -1;
148148
}
149149
};
150150
```
@@ -153,58 +153,50 @@ public:
153153
154154
```go
155155
func minOperations(nums1 []int, nums2 []int, k int) int64 {
156-
ans, x := 0, 0
157-
for i, a := range nums1 {
158-
b := nums2[i]
159-
if k == 0 {
160-
if a != b {
161-
return -1
162-
}
156+
var a, b int64
157+
for i, x := range nums1 {
158+
y := nums2[i]
159+
if x == y {
163160
continue
164161
}
165-
if (a-b)%k != 0 {
162+
if k == 0 || (x-y)%k != 0 {
166163
return -1
167164
}
168-
y := (a - b) / k
169-
ans += abs(y)
170-
x += y
171-
}
172-
if x != 0 {
173-
return -1
165+
t := (x - y) / k
166+
if t < 0 {
167+
a += int64(-t)
168+
} else {
169+
b += int64(t)
170+
}
174171
}
175-
return int64(ans / 2)
176-
}
177-
178-
func abs(x int) int {
179-
if x < 0 {
180-
return -x
172+
if a == b {
173+
return a
181174
}
182-
return x
175+
return -1
183176
}
184177
```
185178

186179
#### TypeScript
187180

188181
```ts
189182
function minOperations(nums1: number[], nums2: number[], k: number): number {
190-
const n = nums1.length;
191-
if (k === 0) {
192-
return nums1.every((v, i) => v === nums2[i]) ? 0 : -1;
193-
}
194-
let sum1 = 0;
195-
let sum2 = 0;
196-
for (let i = 0; i < n; i++) {
197-
const diff = nums1[i] - nums2[i];
198-
sum1 += diff;
199-
if (diff % k !== 0) {
183+
let [a, b] = [0, 0];
184+
for (let i = 0; i < nums1.length; ++i) {
185+
const [x, y] = [nums1[i], nums2[i]];
186+
if (x === y) {
187+
continue;
188+
}
189+
if (k === 0 || (x - y) % k !== 0) {
200190
return -1;
201191
}
202-
sum2 += Math.abs(diff);
203-
}
204-
if (sum1 !== 0) {
205-
return -1;
192+
const t = (x - y) / k;
193+
if (t < 0) {
194+
a += -t;
195+
} else {
196+
b += t;
197+
}
206198
}
207-
return sum2 / (k * 2);
199+
return a === b ? a : -1;
208200
}
209201
```
210202

@@ -213,29 +205,27 @@ function minOperations(nums1: number[], nums2: number[], k: number): number {
213205
```rust
214206
impl Solution {
215207
pub fn min_operations(nums1: Vec<i32>, nums2: Vec<i32>, k: i32) -> i64 {
216-
let k = k as i64;
217-
let n = nums1.len();
218-
if k == 0 {
219-
return if nums1.iter().enumerate().all(|(i, &v)| v == nums2[i]) {
220-
0
221-
} else {
222-
-1
223-
};
224-
}
225-
let mut sum1 = 0;
226-
let mut sum2 = 0;
227-
for i in 0..n {
228-
let diff = (nums1[i] - nums2[i]) as i64;
229-
sum1 += diff;
230-
if diff % k != 0 {
208+
let mut a: i64 = 0;
209+
let mut b: i64 = 0;
210+
for (&x, &y) in nums1.iter().zip(nums2.iter()) {
211+
if x == y {
212+
continue;
213+
}
214+
if k == 0 || (x - y) % k != 0 {
231215
return -1;
232216
}
233-
sum2 += diff.abs();
217+
let t = (x - y) / k;
218+
if t < 0 {
219+
a += (-t) as i64;
220+
} else {
221+
b += t as i64;
222+
}
234223
}
235-
if sum1 != 0 {
236-
return -1;
224+
if a == b {
225+
a
226+
} else {
227+
-1
237228
}
238-
sum2 / (k * 2)
239229
}
240230
}
241231
```
@@ -244,28 +234,23 @@ impl Solution {
244234

245235
```c
246236
long long minOperations(int* nums1, int nums1Size, int* nums2, int nums2Size, int k) {
247-
if (k == 0) {
248-
for (int i = 0; i < nums1Size; i++) {
249-
if (nums1[i] != nums2[i]) {
250-
return -1;
251-
}
237+
long long a = 0, b = 0;
238+
for (int i = 0; i < nums1Size; ++i) {
239+
int x = nums1[i], y = nums2[i];
240+
if (x == y) {
241+
continue;
252242
}
253-
return 0;
254-
}
255-
long long sum1 = 0;
256-
long long sum2 = 0;
257-
for (int i = 0; i < nums1Size; i++) {
258-
long long diff = nums1[i] - nums2[i];
259-
sum1 += diff;
260-
if (diff % k != 0) {
243+
if (k == 0 || (x - y) % k != 0) {
261244
return -1;
262245
}
263-
sum2 += llabs(diff);
264-
}
265-
if (sum1 != 0) {
266-
return -1;
246+
int t = (x - y) / k;
247+
if (t < 0) {
248+
a += -t;
249+
} else {
250+
b += t;
251+
}
267252
}
268-
return sum2 / (k * 2);
253+
return a == b ? a : -1;
269254
}
270255
```
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