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Jul 20, 2024
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feat: update solutions to lc problem: No.0945 #3293

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62 changes: 26 additions & 36 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -61,11 +61,13 @@ tags:

### 方法一：排序 + 贪心

我们首先对数组进行排序，然后从前往后遍历数组，对于每个元素 `nums[i]`，如果它小于等于前一个元素 `nums[i - 1]`，那么我们将它增加到 `nums[i - 1] + 1`，那么操作的次数就是 `nums[i - 1] - nums[i] + 1`，累加到结果中。
我们首先对数组 $\textit{nums}$ 进行排序，用一个变量 $\textit{y}$ 记录当前的最大值，初始时 $\textit{y} = -1$。

然后遍历数组 $\textit{nums}$，对于每个元素 $x$，我们将 $y$ 更新为 $\max(y + 1, x)$，并将操作次数 $y - x$ 累加到结果中。

遍历完成后，返回结果即可。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组 `nums` 的长度。
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 @@ -75,12 +77,10 @@ tags:
class Solution:
def minIncrementForUnique(self, nums: List[int]) -> int:
nums.sort()
ans = 0
for i in range(1, len(nums)):
if nums[i] <= nums[i - 1]:
d = nums[i - 1] - nums[i] + 1
nums[i] += d
ans += d
ans, y = 0, -1
for x in nums:
y = max(y + 1, x)
ans += y - x
return ans
```

Expand All @@ -90,13 +90,10 @@ class Solution:
class Solution {
public int minIncrementForUnique(int[] nums) {
Arrays.sort(nums);
int ans = 0;
for (int i = 1; i < nums.length; ++i) {
if (nums[i] <= nums[i - 1]) {
int d = nums[i - 1] - nums[i] + 1;
nums[i] += d;
ans += d;
}
int ans = 0, y = -1;
for (int x : nums) {
y = Math.max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand All @@ -110,13 +107,10 @@ class Solution {
public:
int minIncrementForUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
int ans = 0;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] <= nums[i - 1]) {
int d = nums[i - 1] - nums[i] + 1;
nums[i] += d;
ans += d;
}
int ans = 0, y = -1;
for (int x : nums) {
y = max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand All @@ -128,12 +122,10 @@ public:
```go
func minIncrementForUnique(nums []int) (ans int) {
sort.Ints(nums)
for i := 1; i < len(nums); i++ {
if nums[i] <= nums[i-1] {
d := nums[i-1] - nums[i] + 1
nums[i] += d
ans += d
}
y := -1
for _, x := range nums {
y = max(y+1, x)
ans += y - x
}
return
}
Expand All @@ -144,12 +136,10 @@ func minIncrementForUnique(nums []int) (ans int) {
```ts
function minIncrementForUnique(nums: number[]): number {
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 1; i < nums.length; ++i) {
if (nums[i] <= nums[i - 1]) {
ans += nums[i - 1] - nums[i] + 1;
nums[i] = nums[i - 1] + 1;
}
let [ans, y] = [0, -1];
for (const x of nums) {
y = Math.max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand All @@ -163,13 +153,13 @@ function minIncrementForUnique(nums: number[]): number {

### 方法二：计数 + 贪心

根据题目描述，结果数组的最大值 $m = \max(\text{nums}) + \text{len}(\text{nums})$，我们可以使用一个计数数组 `cnt` 来记录每个元素出现的次数。
根据题目描述，结果数组的最大值 $m = \max(\text{nums}) + \text{len}(\text{nums})$，我们可以使用一个计数数组 $\textit{cnt}$ 来记录每个元素出现的次数。

然后从 $0$ 到 $m - 1$ 遍历，对于每个元素 $i$，如果它出现的次数 $\text{cnt}[i]$ 大于 $1$，那么我们将 $\text{cnt}[i] - 1$ 个元素增加到 $i + 1$，并将操作次数累加到结果中。

遍历完成后，返回结果即可。

时间复杂度 $O(m)$，空间复杂度 $O(m)$。其中 $m$ 是数组 `nums` 的长度加上数组 `nums` 的最大值。
时间复杂度 $O(m)$，空间复杂度 $O(m)$。其中 $m$ 是数组 $\textit{nums}$ 的长度加上数组的最大值。

<!-- tabs:start -->

Expand Down
70 changes: 33 additions & 37 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -57,7 +57,15 @@ It can be shown with 5 or less moves that it is impossible for the array to have

<!-- solution:start -->

### Solution 1
### Solution 1: Sorting + Greedy

First, we sort the array $\textit{nums}$, and use a variable $\textit{y}$ to record the current maximum value, initially $\textit{y} = -1$.

Then, we iterate through the array $\textit{nums}$. For each element $x$, we update $y$ to $\max(y + 1, x)$, and accumulate the operation count $y - x$ into the result.

After completing the iteration, we return the result.

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

<!-- tabs:start -->

Expand All @@ -67,12 +75,10 @@ It can be shown with 5 or less moves that it is impossible for the array to have
class Solution:
def minIncrementForUnique(self, nums: List[int]) -> int:
nums.sort()
ans = 0
for i in range(1, len(nums)):
if nums[i] <= nums[i - 1]:
d = nums[i - 1] - nums[i] + 1
nums[i] += d
ans += d
ans, y = 0, -1
for x in nums:
y = max(y + 1, x)
ans += y - x
return ans
```

Expand All @@ -82,13 +88,10 @@ class Solution:
class Solution {
public int minIncrementForUnique(int[] nums) {
Arrays.sort(nums);
int ans = 0;
for (int i = 1; i < nums.length; ++i) {
if (nums[i] <= nums[i - 1]) {
int d = nums[i - 1] - nums[i] + 1;
nums[i] += d;
ans += d;
}
int ans = 0, y = -1;
for (int x : nums) {
y = Math.max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand All @@ -102,13 +105,10 @@ class Solution {
public:
int minIncrementForUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
int ans = 0;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] <= nums[i - 1]) {
int d = nums[i - 1] - nums[i] + 1;
nums[i] += d;
ans += d;
}
int ans = 0, y = -1;
for (int x : nums) {
y = max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand All @@ -120,12 +120,10 @@ public:
```go
func minIncrementForUnique(nums []int) (ans int) {
sort.Ints(nums)
for i := 1; i < len(nums); i++ {
if nums[i] <= nums[i-1] {
d := nums[i-1] - nums[i] + 1
nums[i] += d
ans += d
}
y := -1
for _, x := range nums {
y = max(y+1, x)
ans += y - x
}
return
}
Expand All @@ -136,12 +134,10 @@ func minIncrementForUnique(nums []int) (ans int) {
```ts
function minIncrementForUnique(nums: number[]): number {
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 1; i < nums.length; ++i) {
if (nums[i] <= nums[i - 1]) {
ans += nums[i - 1] - nums[i] + 1;
nums[i] = nums[i - 1] + 1;
}
let [ans, y] = [0, -1];
for (const x of nums) {
y = Math.max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand All @@ -155,13 +151,13 @@ function minIncrementForUnique(nums: number[]): number {

### Solution 2: Counting + Greedy

According to the problem description, the maximum value of the result array $m = \max(\text{nums}) + \text{len}(\text{nums})$. We can use a counting array `cnt` to record the occurrence times of each element.
According to the problem description, the maximum value of the result array $m = \max(\text{nums}) + \text{len}(\text{nums})$. We can use a counting array $\textit{cnt}$ to record the occurrence count of each element.

Then, we iterate from $0$ to $m - 1$. For each element $i$, if its occurrence times $\text{cnt}[i]$ is greater than $1$, then we add $\text{cnt}[i] - 1$ elements to $i + 1$ and accumulate the operation times to the result.
Then, we iterate from $0$ to $m - 1$. For each element $i$, if its occurrence count $\textit{cnt}[i]$ is greater than $1$, then we add $\textit{cnt}[i] - 1$ elements to $i + 1$, and accumulate the operation count into the result.

After the iteration, we return the result.
After completing the iteration, we return the result.

The time complexity is $O(m)$, and the space complexity is $O(m)$. Here, $m$ is the length of the array `nums` plus the maximum value in the array `nums`.
The time complexity is $O(m)$, and the space complexity is $O(m)$. Here, $m$ is the length of the array $\textit{nums}$ plus the maximum value in the array.

<!-- tabs:start -->

Expand Down
11 changes: 4 additions & 7 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2,13 +2,10 @@ class Solution {
public:
int minIncrementForUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
int ans = 0;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] <= nums[i - 1]) {
int d = nums[i - 1] - nums[i] + 1;
nums[i] += d;
ans += d;
}
int ans = 0, y = -1;
for (int x : nums) {
y = max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand Down
15 changes: 6 additions & 9 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,12 +1,9 @@
func minIncrementForUnique(nums []int) int {
func minIncrementForUnique(nums []int) (ans int) {
sort.Ints(nums)
ans := 0
for i := 1; i < len(nums); i++ {
if nums[i] <= nums[i-1] {
d := nums[i-1] - nums[i] + 1
nums[i] += d
ans += d
}
y := -1
for _, x := range nums {
y = max(y+1, x)
ans += y - x
}
return ans
return
}
11 changes: 4 additions & 7 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/Solution.java
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,10 @@
class Solution {
public int minIncrementForUnique(int[] nums) {
Arrays.sort(nums);
int ans = 0;
for (int i = 1; i < nums.length; ++i) {
if (nums[i] <= nums[i - 1]) {
int d = nums[i - 1] - nums[i] + 1;
nums[i] += d;
ans += d;
}
int ans = 0, y = -1;
for (int x : nums) {
y = Math.max(y + 1, x);
ans += y - x;
}
return ans;
}
Expand Down
10 changes: 4 additions & 6 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/Solution.py
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,10 +1,8 @@
class Solution:
def minIncrementForUnique(self, nums: List[int]) -> int:
nums.sort()
ans = 0
for i in range(1, len(nums)):
if nums[i] <= nums[i - 1]:
d = nums[i - 1] - nums[i] + 1
nums[i] += d
ans += d
ans, y = 0, -1
for x in nums:
y = max(y + 1, x)
ans += y - x
return ans
10 changes: 4 additions & 6 deletions solution/0900-0999/0945.Minimum Increment to Make Array Unique/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,11 +1,9 @@
function minIncrementForUnique(nums: number[]): number {
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 1; i < nums.length; ++i) {
if (nums[i] <= nums[i - 1]) {
ans += nums[i - 1] - nums[i] + 1;
nums[i] = nums[i - 1] + 1;
}
let [ans, y] = [0, -1];
for (const x of nums) {
y = Math.max(y + 1, x);
ans += y - x;
}
return ans;
}
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