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Feb 10, 2025
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feat: add solutions to lc problem: No.1887 #4051

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115 changes: 37 additions & 78 deletions .../1800-1899/1887.Reduction Operations to Make the Array Elements Equal/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -57,8 +57,8 @@ tags:
<strong>输出：</strong>4
<strong>解释：</strong>需要 4 次操作使 nums 中的所有元素相等：
1. largest = 3 下标为 4 。nextLargest = 2 。将 nums[4] 减少到 2 。nums = [1,1,2,2,<strong>2</strong>] 。
2. largest = 2 下标为 2 。nextLargest = 1 。将 nums[2] 减少到 1 。nums = [1,1,<strong>1</strong>,2,2] 。
3. largest = 2 下标为 3 。nextLargest = 1 。将 nums[3] 减少到 1 。nums = [1,1,1,<strong>1</strong>,2] 。
2. largest = 2 下标为 2 。nextLargest = 1 。将 nums[2] 减少到 1 。nums = [1,1,<strong>1</strong>,2,2] 。
3. largest = 2 下标为 3 。nextLargest = 1 。将 nums[3] 减少到 1 。nums = [1,1,1,<strong>1</strong>,2] 。
4. largest = 2 下标为 4 。nextLargest = 1 。将 nums[4] 减少到 1 。nums = [1,1,1,1,<strong>1</strong>] 。
</pre>

Expand All @@ -79,11 +79,9 @@ tags:

### 方法一：排序

对 $nums$ 进行排序，用 $cnt$ 表示元素所需的操作次数，初始时 $cnt=0$。
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 $\textit{cnt}$，继续遍历下一个元素。

遍历 $nums[1..n-1]$，如果当前元素 $nums[i]$ 不等于 $nums[i-1]$，则将 $cnt$ 加一。累加当前 $cnt$ 到答案 $ans$。

时间复杂度 $O(nlogn)$。
时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。

<!-- tabs:start -->

Expand All @@ -94,8 +92,8 @@ class Solution:
def reductionOperations(self, nums: List[int]) -> int:
nums.sort()
ans = cnt = 0
for i, v in enumerate(nums[1:]):
if v != nums[i]:
for a, b in pairwise(nums):
if a != b:
cnt += 1
ans += cnt
return ans
Expand Down Expand Up @@ -125,7 +123,7 @@ class Solution {
class Solution {
public:
int reductionOperations(vector<int>& nums) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
int ans = 0, cnt = 0;
for (int i = 1; i < nums.size(); ++i) {
cnt += nums[i] != nums[i - 1];
Expand All @@ -139,16 +137,16 @@ public:
#### Go

```go
func reductionOperations(nums []int) int {
func reductionOperations(nums []int) (ans int) {
sort.Ints(nums)
ans, cnt := 0, 0
for i, v := range nums[1:] {
if v != nums[i] {
cnt := 0
for i, x := range nums[1:] {
if x != nums[i] {
cnt++
}
ans += cnt
}
return ans
return
}
```

Expand All @@ -157,10 +155,9 @@ func reductionOperations(nums []int) int {
```ts
function reductionOperations(nums: number[]): number {
nums.sort((a, b) => a - b);
let ans = 0;
let cnt = 0;
let [ans, cnt] = [0, 0];
for (let i = 1; i < nums.length; ++i) {
if (nums[i] != nums[i - 1]) {
if (nums[i] !== nums[i - 1]) {
++cnt;
}
ans += cnt;
Expand All @@ -169,81 +166,43 @@ function reductionOperations(nums: number[]): number {
}
```

#### JavaScript

```js
/**
* @param {number[]} nums
* @return {number}
*/
var reductionOperations = function (nums) {
nums.sort((a, b) => a - b);
let [ans, cnt] = [0, 0];
for (let i = 1; i < nums.length; ++i) {
if (nums[i] !== nums[i - 1]) {
++cnt;
}
ans += cnt;
}
return ans;
};
```

#### C#

```cs
public class Solution {
public int ReductionOperations(int[] nums) {
Array.Sort(nums);
int ans = 0, up = 0;
int ans = 0, cnt = 0;
for (int i = 1; i < nums.Length; i++) {
if (nums[i] != nums[i - 1]) {
up++;
++cnt;
}
ans += up;
}
return ans;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### Python3

```python
class Solution:
def reductionOperations(self, nums: List[int]) -> int:
ans = cnt = 0
for _, v in sorted(Counter(nums).items()):
ans += cnt * v
cnt += 1
return ans
```

#### Java

```java
class Solution {
public int reductionOperations(int[] nums) {
Map<Integer, Integer> tm = new TreeMap<>();
for (int v : nums) {
tm.put(v, tm.getOrDefault(v, 0) + 1);
}
int ans = 0, cnt = 0;
for (int v : tm.values()) {
ans += cnt * v;
++cnt;
ans += cnt;
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int reductionOperations(vector<int>& nums) {
map<int, int> m;
for (int v : nums) ++m[v];
int ans = 0, cnt = 0;
for (auto [_, v] : m) {
ans += cnt * v;
++cnt;
}
return ans;
}
};
```

<!-- tabs:end -->
Expand Down
112 changes: 38 additions & 74 deletions ...00-1899/1887.Reduction Operations to Make the Array Elements Equal/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -75,7 +75,11 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Sorting

We first sort the array $\textit{nums}$, then iterate from the second element of the array. If the current element is not equal to the previous element, we increment $\textit{cnt}$, 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.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.

<!-- tabs:start -->

Expand All @@ -86,8 +90,8 @@ class Solution:
def reductionOperations(self, nums: List[int]) -> int:
nums.sort()
ans = cnt = 0
for i, v in enumerate(nums[1:]):
if v != nums[i]:
for a, b in pairwise(nums):
if a != b:
cnt += 1
ans += cnt
return ans
Expand Down Expand Up @@ -117,7 +121,7 @@ class Solution {
class Solution {
public:
int reductionOperations(vector<int>& nums) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
int ans = 0, cnt = 0;
for (int i = 1; i < nums.size(); ++i) {
cnt += nums[i] != nums[i - 1];
Expand All @@ -131,16 +135,16 @@ public:
#### Go

```go
func reductionOperations(nums []int) int {
func reductionOperations(nums []int) (ans int) {
sort.Ints(nums)
ans, cnt := 0, 0
for i, v := range nums[1:] {
if v != nums[i] {
cnt := 0
for i, x := range nums[1:] {
if x != nums[i] {
cnt++
}
ans += cnt
}
return ans
return
}
```

Expand All @@ -149,10 +153,9 @@ func reductionOperations(nums []int) int {
```ts
function reductionOperations(nums: number[]): number {
nums.sort((a, b) => a - b);
let ans = 0;
let cnt = 0;
let [ans, cnt] = [0, 0];
for (let i = 1; i < nums.length; ++i) {
if (nums[i] != nums[i - 1]) {
if (nums[i] !== nums[i - 1]) {
++cnt;
}
ans += cnt;
Expand All @@ -161,83 +164,44 @@ function reductionOperations(nums: number[]): number {
}
```

#### JavaScript

```js
/**
* @param {number[]} nums
* @return {number}
*/
var reductionOperations = function (nums) {
nums.sort((a, b) => a - b);
let [ans, cnt] = [0, 0];
for (let i = 1; i < nums.length; ++i) {
if (nums[i] !== nums[i - 1]) {
++cnt;
}
ans += cnt;
}
return ans;
};
```

#### C#

```cs
public class Solution {
public int ReductionOperations(int[] nums) {
Array.Sort(nums);
int ans = 0, up = 0;
int ans = 0, cnt = 0;
for (int i = 1; i < nums.Length; i++) {
if (nums[i] != nums[i - 1]) {
up++;
++cnt;
}
ans += up;
}
return ans;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### Python3

```python
class Solution:
def reductionOperations(self, nums: List[int]) -> int:
ans = cnt = 0
for _, v in sorted(Counter(nums).items()):
ans += cnt * v
cnt += 1
return ans
```

#### Java

```java
class Solution {
public int reductionOperations(int[] nums) {
Map<Integer, Integer> tm = new TreeMap<>();
for (int v : nums) {
tm.put(v, tm.getOrDefault(v, 0) + 1);
}
int ans = 0, cnt = 0;
for (int v : tm.values()) {
ans += cnt * v;
++cnt;
ans += cnt;
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int reductionOperations(vector<int>& nums) {
map<int, int> m;
for (int v : nums) ++m[v];
int ans = 0, cnt = 0;
for (auto [_, v] : m) {
ans += cnt * v;
++cnt;
}
return ans;
}
};
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
4 changes: 2 additions & 2 deletions solution/1800-1899/1887.Reduction Operations to Make the Array Elements Equal/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,12 +1,12 @@
class Solution {
public:
int reductionOperations(vector<int>& nums) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
int ans = 0, cnt = 0;
for (int i = 1; i < nums.size(); ++i) {
cnt += nums[i] != nums[i - 1];
ans += cnt;
}
return ans;
}
};
};
6 changes: 3 additions & 3 deletions solution/1800-1899/1887.Reduction Operations to Make the Array Elements Equal/Solution.cs
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,12 +1,12 @@
public class Solution {
public int ReductionOperations(int[] nums) {
Array.Sort(nums);
int ans = 0, up = 0;
int ans = 0, cnt = 0;
for (int i = 1; i < nums.Length; i++) {
if (nums[i] != nums[i - 1]) {
up++;
++cnt;
}
ans += up;
ans += cnt;
}
return ans;
}
Expand Down
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