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feat: update solutions to lc problem: No.2007 #2595

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e817756
feat: update solutions to lc problem: No.2007
yanglbme Apr 16, 2024
8927523
fix: java code
yanglbme Apr 16, 2024
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100 changes: 40 additions & 60 deletions solution/2000-2099/2007.Find Original Array From Doubled Array/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -50,66 +50,59 @@

## 解法

### 方法一：排序 + 计数 + 遍历
### 方法一：排序

我们先判断数组 `changed` 的长度 $n$ 是否为奇数，若是，则直接返回空数组。
我们注意到，如果数组 `changed` 是一个双倍数组，那么数组 `changed` 中最小的元素也一定是原数组中的元素，因此，我们可以先对数组 `changed` 进行排序，然后以第一个元素作为起点，从小到大遍历数组 `changed`。

然后对数组 `changed` 进行排序，并且用哈希表或数组 `cnt` 统计数组 `changed` 中每个元素出现的次数。
我们使用一个哈希表或数组 $cnt$ 来统计数组 `changed` 中每个元素的出现次数。对于数组 `changed` 中的每个元素 $x$，我们首先检查 $x$ 是否存在于 $cnt$ 中。如果不存在，我们直接跳过这个元素。否则，我们将 $cnt[x]$ 减一，并检查 $x \times 2$ 是否存在于 $cnt$ 中。如果不存在，我们直接返回一个空数组。否则，我们将 $cnt[x \times 2]$ 减一，并将 $x$ 加入答案数组中。

接下来遍历数组 `changed`，对于数组 `changed` 中的每个元素 $x$，我们首先判断哈希表 `cnt` 中是否存在 $x$，若不存在，则直接跳过该元素。否则，我们判断 `cnt` 中是否存在 $x \times 2$，若不存在，则直接返回空数组。否则，我们将 $x$ 加入答案数组 `ans` 中，并且将 `cnt` 中 $x$ 和 $x \times 2$ 的出现次数分别减 $1$。
遍历结束后，返回答案数组即可。

遍历结束后，我们判断答案数组 `ans` 的长度是否为 $\frac{n}{2}$，若是，则返回 `ans`，否则返回空数组。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `changed` 的长度。
时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$，其中 $n$ 为数组 `changed` 的长度。

<!-- tabs:start -->

```python
class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n & 1:
return []
cnt = Counter(changed)
changed.sort()
cnt = Counter(changed)
ans = []
for x in changed:
if cnt[x] == 0:
continue
if cnt[x * 2] <= 0:
cnt[x] -= 1
if cnt[x << 1] <= 0:
return []
cnt[x << 1] -= 1
ans.append(x)
cnt[x] -= 1
cnt[x * 2] -= 1
return ans if len(ans) == n // 2 else []
return ans
```

```java
class Solution {
public int[] findOriginalArray(int[] changed) {
int n = changed.length;
if (n % 2 == 1) {
return new int[] {};
}
Arrays.sort(changed);
int[] cnt = new int[changed[n - 1] + 1];
for (int x : changed) {
++cnt[x];
}
int[] ans = new int[n / 2];
int[] ans = new int[n >> 1];
int i = 0;
for (int x : changed) {
if (cnt[x] == 0) {
continue;
}
if (x * 2 >= cnt.length || cnt[x * 2] <= 0) {
return new int[] {};
--cnt[x];
int y = x << 1;
if (y >= cnt.length || cnt[y] <= 0) {
return new int[0];
}
--cnt[y];
ans[i++] = x;
cnt[x]--;
cnt[x * 2]--;
}
return i == n / 2 ? ans : new int[] {};
return ans;
}
}
```
Expand All @@ -118,86 +111,73 @@ class Solution {
class Solution {
public:
vector<int> findOriginalArray(vector<int>& changed) {
int n = changed.size();
if (n & 1) {
return {};
}
sort(changed.begin(), changed.end());
vector<int> cnt(changed.back() + 1);
for (int& x : changed) {
for (int x : changed) {
++cnt[x];
}
vector<int> ans;
for (int& x : changed) {
for (int x : changed) {
if (cnt[x] == 0) {
continue;
}
if (x * 2 >= cnt.size() || cnt[x * 2] <= 0) {
--cnt[x];
int y = x << 1;
if (y >= cnt.size() || cnt[y] <= 0) {
return {};
}
--cnt[y];
ans.push_back(x);
--cnt[x];
--cnt[x * 2];
}
return ans.size() == n / 2 ? ans : vector<int>();
return ans;
}
};
```

```go
func findOriginalArray(changed []int) []int {
n := len(changed)
ans := []int{}
if n&1 == 1 {
return ans
}
func findOriginalArray(changed []int) (ans []int) {
sort.Ints(changed)
cnt := make([]int, changed[n-1]+1)
cnt := make([]int, changed[len(changed)-1]+1)
for _, x := range changed {
cnt[x]++
}
for _, x := range changed {
if cnt[x] == 0 {
continue
}
if x*2 >= len(cnt) || cnt[x*2] <= 0 {
cnt[x]--
y := x << 1
if y >= len(cnt) || cnt[y] <= 0 {
return []int{}
}
cnt[y]--
ans = append(ans, x)
cnt[x]--
cnt[x*2]--
}
if len(ans) != n/2 {
return []int{}
}
return ans
return
}
```

```ts
function findOriginalArray(changed: number[]): number[] {
const n = changed.length;
if (n & 1) {
return [];
}
const cnt = new Map<number, number>();
changed.sort((a, b) => a - b);
const cnt: number[] = Array(changed.at(-1)! + 1).fill(0);
for (const x of changed) {
cnt.set(x, (cnt.get(x) || 0) + 1);
++cnt[x];
}
changed.sort((a, b) => a - b);
const ans: number[] = [];
for (const x of changed) {
if (cnt.get(x) == 0) {
if (cnt[x] === 0) {
continue;
}
if ((cnt.get(x * 2) || 0) <= 0) {
cnt[x]--;
const y = x << 1;
if (y >= cnt.length || cnt[y] <= 0) {
return [];
}
cnt[y]--;
ans.push(x);
cnt.set(x, (cnt.get(x) || 0) - 1);
cnt.set(x * 2, (cnt.get(x * 2) || 0) - 1);
}
return ans.length == n / 2 ? ans : [];
return ans;
}
```

Expand Down
100 changes: 40 additions & 60 deletions solution/2000-2099/2007.Find Original Array From Doubled Array/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -49,66 +49,59 @@ Other original arrays could be [4,3,1] or [3,1,4].

## Solutions

### Solution 1: Sorting + Counting + Traversal
### Solution 1: Sorting

First, we check if the length $n$ of the array `changed` is odd. If it is, we directly return an empty array.
We notice that if the array `changed` is a double array, then the smallest element in the array `changed` must also be an element in the original array. Therefore, we can first sort the array `changed`, and then start from the first element to traverse the array `changed` in ascending order.

Then, we sort the array `changed`, and use a hash table or array `cnt` to count the occurrence of each element in `changed`.
We use a hash table or array $cnt$ to count the occurrence of each element in the array `changed`. For each element $x$ in the array `changed`, we first check whether $x$ exists in $cnt$. If it does not exist, we skip this element. Otherwise, we subtract one from $cnt[x]$, and check whether $x \times 2$ exists in $cnt$. If it does not exist, we return an empty array directly. Otherwise, we subtract one from $cnt[x \times 2]$, and add $x$ to the answer array.

Next, we traverse the array `changed`. For each element $x$ in `changed`, we first check if $x$ exists in the hash table `cnt`. If it does not exist, we directly skip this element. Otherwise, we check if $x \times 2$ exists in `cnt`. If it does not exist, we directly return an empty array. Otherwise, we add $x$ to the answer array `ans`, and decrease the occurrence counts of $x$ and $x \times 2$ in `cnt` by $1$ each.
After the traversal, we return the answer array.

After the traversal, we check if the length of the answer array `ans` is $\frac{n}{2}$. If it is, we return `ans`, otherwise we return an empty array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array `changed`.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$, where $n$ is the length of the array `changed`.

<!-- tabs:start -->

```python
class Solution:
def findOriginalArray(self, changed: List[int]) -> List[int]:
n = len(changed)
if n & 1:
return []
cnt = Counter(changed)
changed.sort()
cnt = Counter(changed)
ans = []
for x in changed:
if cnt[x] == 0:
continue
if cnt[x * 2] <= 0:
cnt[x] -= 1
if cnt[x << 1] <= 0:
return []
cnt[x << 1] -= 1
ans.append(x)
cnt[x] -= 1
cnt[x * 2] -= 1
return ans if len(ans) == n // 2 else []
return ans
```

```java
class Solution {
public int[] findOriginalArray(int[] changed) {
int n = changed.length;
if (n % 2 == 1) {
return new int[] {};
}
Arrays.sort(changed);
int[] cnt = new int[changed[n - 1] + 1];
for (int x : changed) {
++cnt[x];
}
int[] ans = new int[n / 2];
int[] ans = new int[n >> 1];
int i = 0;
for (int x : changed) {
if (cnt[x] == 0) {
continue;
}
if (x * 2 >= cnt.length || cnt[x * 2] <= 0) {
return new int[] {};
--cnt[x];
int y = x << 1;
if (y >= cnt.length || cnt[y] <= 0) {
return new int[0];
}
--cnt[y];
ans[i++] = x;
cnt[x]--;
cnt[x * 2]--;
}
return i == n / 2 ? ans : new int[] {};
return ans;
}
}
```
Expand All @@ -117,86 +110,73 @@ class Solution {
class Solution {
public:
vector<int> findOriginalArray(vector<int>& changed) {
int n = changed.size();
if (n & 1) {
return {};
}
sort(changed.begin(), changed.end());
vector<int> cnt(changed.back() + 1);
for (int& x : changed) {
for (int x : changed) {
++cnt[x];
}
vector<int> ans;
for (int& x : changed) {
for (int x : changed) {
if (cnt[x] == 0) {
continue;
}
if (x * 2 >= cnt.size() || cnt[x * 2] <= 0) {
--cnt[x];
int y = x << 1;
if (y >= cnt.size() || cnt[y] <= 0) {
return {};
}
--cnt[y];
ans.push_back(x);
--cnt[x];
--cnt[x * 2];
}
return ans.size() == n / 2 ? ans : vector<int>();
return ans;
}
};
```

```go
func findOriginalArray(changed []int) []int {
n := len(changed)
ans := []int{}
if n&1 == 1 {
return ans
}
func findOriginalArray(changed []int) (ans []int) {
sort.Ints(changed)
cnt := make([]int, changed[n-1]+1)
cnt := make([]int, changed[len(changed)-1]+1)
for _, x := range changed {
cnt[x]++
}
for _, x := range changed {
if cnt[x] == 0 {
continue
}
if x*2 >= len(cnt) || cnt[x*2] <= 0 {
cnt[x]--
y := x << 1
if y >= len(cnt) || cnt[y] <= 0 {
return []int{}
}
cnt[y]--
ans = append(ans, x)
cnt[x]--
cnt[x*2]--
}
if len(ans) != n/2 {
return []int{}
}
return ans
return
}
```

```ts
function findOriginalArray(changed: number[]): number[] {
const n = changed.length;
if (n & 1) {
return [];
}
const cnt = new Map<number, number>();
changed.sort((a, b) => a - b);
const cnt: number[] = Array(changed.at(-1)! + 1).fill(0);
for (const x of changed) {
cnt.set(x, (cnt.get(x) || 0) + 1);
++cnt[x];
}
changed.sort((a, b) => a - b);
const ans: number[] = [];
for (const x of changed) {
if (cnt.get(x) == 0) {
if (cnt[x] === 0) {
continue;
}
if ((cnt.get(x * 2) || 0) <= 0) {
cnt[x]--;
const y = x << 1;
if (y >= cnt.length || cnt[y] <= 0) {
return [];
}
cnt[y]--;
ans.push(x);
cnt.set(x, (cnt.get(x) || 0) - 1);
cnt.set(x * 2, (cnt.get(x * 2) || 0) - 1);
}
return ans.length == n / 2 ? ans : [];
return ans;
}
```

Expand Down
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