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feat: add solutions to lc problem: No.2869 (#1731)
No.2869.Minimum Operations to Collect Elements
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‎solution/2800-2899/2869.Minimum Operations to Collect Elements/README.md

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Original file line numberDiff line numberDiff line change
@@ -53,14 +53,31 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：逆序遍历**
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我们可以逆序遍历数组，每次遍历到的元素如果小于等于 $k$，且没有被添加过，就将其添加到集合中，直到集合中包含了元素 $1$ 到 $k$ 为止。
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时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(k)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
63-
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class Solution:
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def minOperations(self, nums: List[int], k: int) -> int:
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is_added = [False] * k
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count = 0
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n = len(nums)
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for i in range(n - 1, -1, -1):
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if nums[i] > k or is_added[nums[i] - 1]:
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continue
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is_added[nums[i] - 1] = True
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count += 1
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if count == k:
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return n - i
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```
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### **Java**
@@ -73,7 +90,7 @@ class Solution {
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boolean[] isAdded = new boolean[k];
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int n = nums.size();
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int count = 0;
76-
for (int i = n - 1; i >= 0; i--) {
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for (int i = n - 1;; i--) {
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if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
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continue;
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}
@@ -83,28 +100,70 @@ class Solution {
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return n - i;
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}
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}
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return n;
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}
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}
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```
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### **C++**
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```cpp
95-
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class Solution {
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public:
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int minOperations(vector<int>& nums, int k) {
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int n = nums.size();
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vector<bool> isAdded(n);
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int count = 0;
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for (int i = n - 1;; --i) {
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if (nums[i] > k || isAdded[nums[i] - 1]) {
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continue;
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}
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isAdded[nums[i] - 1] = true;
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if (++count == k) {
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return n - i;
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}
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}
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}
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};
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```
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### **Go**
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```go
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func minOperations(nums []int, k int) int {
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isAdded := make([]bool, k)
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count := 0
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n := len(nums)
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for i := n - 1; ; i-- {
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if nums[i] > k || isAdded[nums[i]-1] {
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continue
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}
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isAdded[nums[i]-1] = true
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count++
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if count == k {
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return n - i
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}
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}
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}
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```
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### **TypeScript**
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```ts
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function minOperations(nums: number[], k: number): number {
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const n = nums.length;
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const isAdded = Array(k).fill(false);
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let count = 0;
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for (let i = n - 1; ; --i) {
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if (nums[i] > k || isAdded[nums[i] - 1]) {
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continue;
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}
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isAdded[nums[i] - 1] = true;
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++count;
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if (count === k) {
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return n - i;
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}
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}
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}
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```
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### **...**

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/README_EN.md

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@@ -47,12 +47,29 @@
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## Solutions
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**Solution 1: Traverse in Reverse Order**
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We can traverse the array in reverse order. For each element encountered during the traversal that is less than or equal to $k$ and has not been added to the set yet, we add it to the set until the set contains elements from $1$ to $k$.
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The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(k)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def minOperations(self, nums: List[int], k: int) -> int:
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is_added = [False] * k
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count = 0
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n = len(nums)
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for i in range(n - 1, -1, -1):
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if nums[i] > k or is_added[nums[i] - 1]:
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continue
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is_added[nums[i] - 1] = True
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count += 1
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if count == k:
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return n - i
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```
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### **Java**
@@ -63,7 +80,7 @@ class Solution {
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boolean[] isAdded = new boolean[k];
6481
int n = nums.size();
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int count = 0;
66-
for (int i = n - 1; i >= 0; i--) {
83+
for (int i = n - 1;; i--) {
6784
if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
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continue;
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}
@@ -73,28 +90,70 @@ class Solution {
7390
return n - i;
7491
}
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}
76-
return n;
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}
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}
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```
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### **C++**
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```cpp
85-
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class Solution {
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public:
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int minOperations(vector<int>& nums, int k) {
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int n = nums.size();
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vector<bool> isAdded(n);
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int count = 0;
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for (int i = n - 1;; --i) {
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if (nums[i] > k || isAdded[nums[i] - 1]) {
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continue;
109+
}
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isAdded[nums[i] - 1] = true;
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if (++count == k) {
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return n - i;
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}
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}
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}
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};
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```
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### **Go**
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```go
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func minOperations(nums []int, k int) int {
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isAdded := make([]bool, k)
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count := 0
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n := len(nums)
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for i := n - 1; ; i-- {
127+
if nums[i] > k || isAdded[nums[i]-1] {
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continue
129+
}
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isAdded[nums[i]-1] = true
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count++
132+
if count == k {
133+
return n - i
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}
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}
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}
92137
```
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### **TypeScript**
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```ts
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function minOperations(nums: number[], k: number): number {
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const n = nums.length;
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const isAdded = Array(k).fill(false);
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let count = 0;
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for (let i = n - 1; ; --i) {
147+
if (nums[i] > k || isAdded[nums[i] - 1]) {
148+
continue;
149+
}
150+
isAdded[nums[i] - 1] = true;
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++count;
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if (count === k) {
153+
return n - i;
154+
}
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}
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}
98157
```
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100159
### **...**

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
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class Solution {
2+
public:
3+
int minOperations(vector<int>& nums, int k) {
4+
int n = nums.size();
5+
vector<bool> isAdded(n);
6+
int count = 0;
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for (int i = n - 1;; --i) {
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if (nums[i] > k || isAdded[nums[i] - 1]) {
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continue;
10+
}
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isAdded[nums[i] - 1] = true;
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if (++count == k) {
13+
return n - i;
14+
}
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}
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}
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};

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/Solution.go

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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func minOperations(nums []int, k int) int {
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isAdded := make([]bool, k)
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count := 0
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n := len(nums)
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for i := n - 1; ; i-- {
6+
if nums[i] > k || isAdded[nums[i]-1] {
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continue
8+
}
9+
isAdded[nums[i]-1] = true
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count++
11+
if count == k {
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return n - i
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}
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}
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}

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/Solution.java

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@@ -3,7 +3,7 @@ public int minOperations(List<Integer> nums, int k) {
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boolean[] isAdded = new boolean[k];
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int n = nums.size();
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int count = 0;
6-
for (int i = n - 1; i >= 0; i--) {
6+
for (int i = n - 1;; i--) {
77
if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
88
continue;
99
}
@@ -13,6 +13,5 @@ public int minOperations(List<Integer> nums, int k) {
1313
return n - i;
1414
}
1515
}
16-
return n;
1716
}
1817
}

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/Solution.py

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Original file line numberDiff line numberDiff line change
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1+
class Solution:
2+
def minOperations(self, nums: List[int], k: int) -> int:
3+
is_added = [False] * k
4+
count = 0
5+
n = len(nums)
6+
for i in range(n - 1, -1, -1):
7+
if nums[i] > k or is_added[nums[i] - 1]:
8+
continue
9+
is_added[nums[i] - 1] = True
10+
count += 1
11+
if count == k:
12+
return n - i

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/Solution.ts

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
1+
function minOperations(nums: number[], k: number): number {
2+
const n = nums.length;
3+
const isAdded = Array(k).fill(false);
4+
let count = 0;
5+
for (let i = n - 1; ; --i) {
6+
if (nums[i] > k || isAdded[nums[i] - 1]) {
7+
continue;
8+
}
9+
isAdded[nums[i] - 1] = true;
10+
++count;
11+
if (count === k) {
12+
return n - i;
13+
}
14+
}
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}

‎solution/2800-2899/2872.Maximum Number of K-Divisible Components/images/example1.jpg ‎solution/2800-2899/2872.Maximum Number of K-Divisible Components/images/example12-cropped2svg.jpg

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‎solution/2800-2899/2872.Maximum Number of K-Divisible Components/images/example2.jpg ‎solution/2800-2899/2872.Maximum Number of K-Divisible Components/images/example21svg-1.jpg

File renamed without changes.

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