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Commit fdff3c3

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feat: add solutions to lc problem: No.1887 (#4051)
No.1887.Reduction Operations to Make the Array Elements Equal
1 parent 62334f8 commit fdff3c3

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‎solution/1800-1899/1887.Reduction Operations to Make the Array Elements Equal/README.md‎

Lines changed: 37 additions & 78 deletions
Original file line numberDiff line numberDiff line change
@@ -57,8 +57,8 @@ tags:
5757
<strong>输出:</strong>4
5858
<strong>解释:</strong>需要 4 次操作使 nums 中的所有元素相等:
5959
1. largest = 3 下标为 4 。nextLargest = 2 。将 nums[4] 减少到 2 。nums = [1,1,2,2,<strong>2</strong>] 。
60-
2. largest = 2 下标为 2 。nextLargest = 1 。将 nums[2] 减少到 1 。nums = [1,1,<strong>1</strong>,2,2] 。
61-
3. largest = 2 下标为 3 。nextLargest = 1 。将 nums[3] 减少到 1 。nums = [1,1,1,<strong>1</strong>,2] 。
60+
2. largest = 2 下标为 2 。nextLargest = 1 。将 nums[2] 减少到 1 。nums = [1,1,<strong>1</strong>,2,2] 。
61+
3. largest = 2 下标为 3 。nextLargest = 1 。将 nums[3] 减少到 1 。nums = [1,1,1,<strong>1</strong>,2] 。
6262
4. largest = 2 下标为 4 。nextLargest = 1 。将 nums[4] 减少到 1 。nums = [1,1,1,1,<strong>1</strong>] 。
6363
</pre>
6464

@@ -79,11 +79,9 @@ tags:
7979

8080
### 方法一:排序
8181

82-
对 $nums$ 进行排序,用 $cnt$ 表示元素所需的操作次数,初始时 $cnt=0$。
82+
我们首先对数组 $\textit{nums}$ 进行排序,然后从数组的第二个元素开始遍历,如果当前元素和前一个元素不相等,那么我们就将 $\textit{cnt}$ 加一,表示我们需要将当前元素减小到最小值的操作次数。然后我们将 $\textit{ans}$ 加上 $\textit{cnt},ドル继续遍历下一个元素。
8383

84-
遍历 $nums[1..n-1],ドル如果当前元素 $nums[i]$ 不等于 $nums[i-1],ドル则将 $cnt$ 加一。累加当前 $cnt$ 到答案 $ans$。
85-
86-
时间复杂度 $O(nlogn)$。
84+
时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(\log n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
8785

8886
<!-- tabs:start -->
8987

@@ -94,8 +92,8 @@ class Solution:
9492
def reductionOperations(self, nums: List[int]) -> int:
9593
nums.sort()
9694
ans = cnt = 0
97-
for i, v in enumerate(nums[1:]):
98-
if v != nums[i]:
95+
for a, b in pairwise(nums):
96+
if a != b:
9997
cnt += 1
10098
ans += cnt
10199
return ans
@@ -125,7 +123,7 @@ class Solution {
125123
class Solution {
126124
public:
127125
int reductionOperations(vector<int>& nums) {
128-
sort(nums.begin(), nums.end());
126+
ranges::sort(nums);
129127
int ans = 0, cnt = 0;
130128
for (int i = 1; i < nums.size(); ++i) {
131129
cnt += nums[i] != nums[i - 1];
@@ -139,16 +137,16 @@ public:
139137
#### Go
140138
141139
```go
142-
func reductionOperations(nums []int) int {
140+
func reductionOperations(nums []int) (ans int) {
143141
sort.Ints(nums)
144-
ans, cnt := 0, 0
145-
for i, v := range nums[1:] {
146-
if v != nums[i] {
142+
cnt := 0
143+
for i, x := range nums[1:] {
144+
if x != nums[i] {
147145
cnt++
148146
}
149147
ans += cnt
150148
}
151-
return ans
149+
return
152150
}
153151
```
154152

@@ -157,10 +155,9 @@ func reductionOperations(nums []int) int {
157155
```ts
158156
function reductionOperations(nums: number[]): number {
159157
nums.sort((a, b) => a - b);
160-
let ans = 0;
161-
let cnt = 0;
158+
let [ans, cnt] = [0, 0];
162159
for (let i = 1; i < nums.length; ++i) {
163-
if (nums[i] != nums[i - 1]) {
160+
if (nums[i] !== nums[i - 1]) {
164161
++cnt;
165162
}
166163
ans += cnt;
@@ -169,81 +166,43 @@ function reductionOperations(nums: number[]): number {
169166
}
170167
```
171168

169+
#### JavaScript
170+
171+
```js
172+
/**
173+
* @param {number[]} nums
174+
* @return {number}
175+
*/
176+
var reductionOperations = function (nums) {
177+
nums.sort((a, b) => a - b);
178+
let [ans, cnt] = [0, 0];
179+
for (let i = 1; i < nums.length; ++i) {
180+
if (nums[i] !== nums[i - 1]) {
181+
++cnt;
182+
}
183+
ans += cnt;
184+
}
185+
return ans;
186+
};
187+
```
188+
172189
#### C#
173190

174191
```cs
175192
public class Solution {
176193
public int ReductionOperations(int[] nums) {
177194
Array.Sort(nums);
178-
int ans = 0, up = 0;
195+
int ans = 0, cnt = 0;
179196
for (int i = 1; i < nums.Length; i++) {
180197
if (nums[i] != nums[i - 1]) {
181-
up++;
198+
++cnt;
182199
}
183-
ans += up;
184-
}
185-
return ans;
186-
}
187-
}
188-
```
189-
190-
<!-- tabs:end -->
191-
192-
<!-- solution:end -->
193-
194-
<!-- solution:start -->
195-
196-
### 方法二
197-
198-
<!-- tabs:start -->
199-
200-
#### Python3
201-
202-
```python
203-
class Solution:
204-
def reductionOperations(self, nums: List[int]) -> int:
205-
ans = cnt = 0
206-
for _, v in sorted(Counter(nums).items()):
207-
ans += cnt * v
208-
cnt += 1
209-
return ans
210-
```
211-
212-
#### Java
213-
214-
```java
215-
class Solution {
216-
public int reductionOperations(int[] nums) {
217-
Map<Integer, Integer> tm = new TreeMap<>();
218-
for (int v : nums) {
219-
tm.put(v, tm.getOrDefault(v, 0) + 1);
220-
}
221-
int ans = 0, cnt = 0;
222-
for (int v : tm.values()) {
223-
ans += cnt * v;
224-
++cnt;
200+
ans += cnt;
225201
}
226202
return ans;
227203
}
228204
}
229-
```
230-
231-
#### C++
232205

233-
```cpp
234-
class Solution {
235-
public:
236-
int reductionOperations(vector<int>& nums) {
237-
map<int, int> m;
238-
for (int v : nums) ++m[v];
239-
int ans = 0, cnt = 0;
240-
for (auto [_, v] : m) {
241-
ans += cnt * v;
242-
++cnt;
243-
}
244-
return ans;
245-
}
246-
};
247206
```
248207

249208
<!-- tabs:end -->

‎solution/1800-1899/1887.Reduction Operations to Make the Array Elements Equal/README_EN.md‎

Lines changed: 38 additions & 74 deletions
Original file line numberDiff line numberDiff line change
@@ -75,7 +75,11 @@ tags:
7575

7676
<!-- solution:start -->
7777

78-
### Solution 1
78+
### Solution 1: Sorting
79+
80+
We first sort the array $\textit{nums},ドル then iterate from the second element of the array. If the current element is not equal to the previous element, we increment $\textit{cnt},ドル indicating the number of operations needed to reduce the current element to the minimum value. Then we add $\textit{cnt}$ to $\textit{ans}$ and continue to the next element.
81+
82+
The time complexity is $O(n \times \log n),ドル and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.
7983

8084
<!-- tabs:start -->
8185

@@ -86,8 +90,8 @@ class Solution:
8690
def reductionOperations(self, nums: List[int]) -> int:
8791
nums.sort()
8892
ans = cnt = 0
89-
for i, v in enumerate(nums[1:]):
90-
if v != nums[i]:
93+
for a, b in pairwise(nums):
94+
if a != b:
9195
cnt += 1
9296
ans += cnt
9397
return ans
@@ -117,7 +121,7 @@ class Solution {
117121
class Solution {
118122
public:
119123
int reductionOperations(vector<int>& nums) {
120-
sort(nums.begin(), nums.end());
124+
ranges::sort(nums);
121125
int ans = 0, cnt = 0;
122126
for (int i = 1; i < nums.size(); ++i) {
123127
cnt += nums[i] != nums[i - 1];
@@ -131,16 +135,16 @@ public:
131135
#### Go
132136
133137
```go
134-
func reductionOperations(nums []int) int {
138+
func reductionOperations(nums []int) (ans int) {
135139
sort.Ints(nums)
136-
ans, cnt := 0, 0
137-
for i, v := range nums[1:] {
138-
if v != nums[i] {
140+
cnt := 0
141+
for i, x := range nums[1:] {
142+
if x != nums[i] {
139143
cnt++
140144
}
141145
ans += cnt
142146
}
143-
return ans
147+
return
144148
}
145149
```
146150

@@ -149,10 +153,9 @@ func reductionOperations(nums []int) int {
149153
```ts
150154
function reductionOperations(nums: number[]): number {
151155
nums.sort((a, b) => a - b);
152-
let ans = 0;
153-
let cnt = 0;
156+
let [ans, cnt] = [0, 0];
154157
for (let i = 1; i < nums.length; ++i) {
155-
if (nums[i] != nums[i - 1]) {
158+
if (nums[i] !== nums[i - 1]) {
156159
++cnt;
157160
}
158161
ans += cnt;
@@ -161,83 +164,44 @@ function reductionOperations(nums: number[]): number {
161164
}
162165
```
163166

167+
#### JavaScript
168+
169+
```js
170+
/**
171+
* @param {number[]} nums
172+
* @return {number}
173+
*/
174+
var reductionOperations = function (nums) {
175+
nums.sort((a, b) => a - b);
176+
let [ans, cnt] = [0, 0];
177+
for (let i = 1; i < nums.length; ++i) {
178+
if (nums[i] !== nums[i - 1]) {
179+
++cnt;
180+
}
181+
ans += cnt;
182+
}
183+
return ans;
184+
};
185+
```
186+
164187
#### C#
165188

166189
```cs
167190
public class Solution {
168191
public int ReductionOperations(int[] nums) {
169192
Array.Sort(nums);
170-
int ans = 0, up = 0;
193+
int ans = 0, cnt = 0;
171194
for (int i = 1; i < nums.Length; i++) {
172195
if (nums[i] != nums[i - 1]) {
173-
up++;
196+
++cnt;
174197
}
175-
ans += up;
176-
}
177-
return ans;
178-
}
179-
}
180-
```
181-
182-
<!-- tabs:end -->
183-
184-
<!-- solution:end -->
185-
186-
<!-- solution:start -->
187-
188-
### Solution 2
189-
190-
<!-- tabs:start -->
191-
192-
#### Python3
193-
194-
```python
195-
class Solution:
196-
def reductionOperations(self, nums: List[int]) -> int:
197-
ans = cnt = 0
198-
for _, v in sorted(Counter(nums).items()):
199-
ans += cnt * v
200-
cnt += 1
201-
return ans
202-
```
203-
204-
#### Java
205-
206-
```java
207-
class Solution {
208-
public int reductionOperations(int[] nums) {
209-
Map<Integer, Integer> tm = new TreeMap<>();
210-
for (int v : nums) {
211-
tm.put(v, tm.getOrDefault(v, 0) + 1);
212-
}
213-
int ans = 0, cnt = 0;
214-
for (int v : tm.values()) {
215-
ans += cnt * v;
216-
++cnt;
198+
ans += cnt;
217199
}
218200
return ans;
219201
}
220202
}
221203
```
222204

223-
#### C++
224-
225-
```cpp
226-
class Solution {
227-
public:
228-
int reductionOperations(vector<int>& nums) {
229-
map<int, int> m;
230-
for (int v : nums) ++m[v];
231-
int ans = 0, cnt = 0;
232-
for (auto [_, v] : m) {
233-
ans += cnt * v;
234-
++cnt;
235-
}
236-
return ans;
237-
}
238-
};
239-
```
240-
241205
<!-- tabs:end -->
242206

243207
<!-- solution:end -->
Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,12 @@
11
class Solution {
22
public:
33
int reductionOperations(vector<int>& nums) {
4-
sort(nums.begin(), nums.end());
4+
ranges::sort(nums);
55
int ans = 0, cnt = 0;
66
for (int i = 1; i < nums.size(); ++i) {
77
cnt += nums[i] != nums[i - 1];
88
ans += cnt;
99
}
1010
return ans;
1111
}
12-
};
12+
};

‎solution/1800-1899/1887.Reduction Operations to Make the Array Elements Equal/Solution.cs‎

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,12 @@
11
public class Solution {
22
public int ReductionOperations(int[] nums) {
33
Array.Sort(nums);
4-
int ans = 0, up = 0;
4+
int ans = 0, cnt = 0;
55
for (int i = 1; i < nums.Length; i++) {
66
if (nums[i] != nums[i - 1]) {
7-
up++;
7+
++cnt;
88
}
9-
ans += up;
9+
ans += cnt;
1010
}
1111
return ans;
1212
}

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