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Commit e9cfffd

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feat: add solutions to lc problem: No.2870 (#1732)
No.2870.Minimum Number of Operations to Make Array Empty
1 parent 675a1f8 commit e9cfffd

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7 files changed

+174
-11
lines changed

7 files changed

+174
-11
lines changed

‎solution/2800-2899/2870.Minimum Number of Operations to Make Array Empty/README.md‎

Lines changed: 60 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -53,14 +53,28 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:哈希表 + 贪心**
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我们用一个哈希表 $count$ 统计数组中每个元素出现的次数,然后遍历哈希表,对于每个元素 $x,ドル如果 $x$ 出现的次数为 $c,ドル那么我们可以进行 $\lfloor \frac{c+2}{3} \rfloor$ 次操作,将 $x$ 删除,最后我们返回所有元素的操作次数之和即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
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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]) -> int:
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count = Counter(nums)
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ans = 0
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for c in count.values():
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if c == 1:
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return -1
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ans += (c + 2) // 3
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return ans
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```
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### **Java**
@@ -76,7 +90,7 @@ class Solution {
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count.merge(num, 1, Integer::sum);
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}
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int ans = 0;
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for (Integer c : count.values()) {
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for (int c : count.values()) {
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if (c < 2) {
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return -1;
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}
@@ -99,19 +113,60 @@ class Solution {
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### **C++**
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```cpp
102-
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class Solution {
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public:
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int minOperations(vector<int>& nums) {
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unordered_map<int, int> count;
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for (int num : nums) {
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++count[num];
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}
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int ans = 0;
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for (auto& [_, c] : count) {
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if (c < 2) {
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return -1;
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}
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ans += (c + 2) / 3;
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}
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return ans;
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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) (ans int) {
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count := map[int]int{}
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for _, num := range nums {
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count[num]++
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}
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for _, c := range count {
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if c < 2 {
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return -1
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}
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ans += (c + 2) / 3
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}
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return
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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[]): number {
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const count: Map<number, number> = new Map();
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for (const num of nums) {
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count.set(num, (count.get(num) ?? 0) + 1);
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}
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let ans = 0;
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for (const [_, c] of count) {
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if (c < 2) {
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return -1;
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}
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ans += ((c + 2) / 3) | 0;
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}
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return ans;
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}
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```
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### **...**

‎solution/2800-2899/2870.Minimum Number of Operations to Make Array Empty/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -47,12 +47,26 @@ It can be shown that we cannot make the array empty in less than 4 operations.
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## Solutions
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**Solution 1: Hash Table + Greedy**
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We use a hash table $count$ to count the number of occurrences of each element in the array. Then we traverse the hash table. For each element $x,ドル if it appears $c$ times, we can perform $\lfloor \frac{c+2}{3} \rfloor$ operations to delete $x$. Finally, we return the sum of the number of operations for all elements.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(n)$.
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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]) -> int:
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count = Counter(nums)
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ans = 0
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for c in count.values():
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if c == 1:
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return -1
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ans += (c + 2) // 3
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return ans
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```
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### **Java**
@@ -66,7 +80,7 @@ class Solution {
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count.merge(num, 1, Integer::sum);
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}
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int ans = 0;
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for (Integer c : count.values()) {
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for (int c : count.values()) {
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if (c < 2) {
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return -1;
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}
@@ -89,19 +103,60 @@ class Solution {
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### **C++**
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```cpp
92-
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class Solution {
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public:
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int minOperations(vector<int>& nums) {
109+
unordered_map<int, int> count;
110+
for (int num : nums) {
111+
++count[num];
112+
}
113+
int ans = 0;
114+
for (auto& [_, c] : count) {
115+
if (c < 2) {
116+
return -1;
117+
}
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ans += (c + 2) / 3;
119+
}
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return ans;
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}
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};
93123
```
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### **Go**
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```go
98-
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func minOperations(nums []int) (ans int) {
129+
count := map[int]int{}
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for _, num := range nums {
131+
count[num]++
132+
}
133+
for _, c := range count {
134+
if c < 2 {
135+
return -1
136+
}
137+
ans += (c + 2) / 3
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}
139+
return
140+
}
99141
```
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### **TypeScript**
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```ts
104-
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function minOperations(nums: number[]): number {
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const count: Map<number, number> = new Map();
148+
for (const num of nums) {
149+
count.set(num, (count.get(num) ?? 0) + 1);
150+
}
151+
let ans = 0;
152+
for (const [_, c] of count) {
153+
if (c < 2) {
154+
return -1;
155+
}
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ans += ((c + 2) / 3) | 0;
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}
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return ans;
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}
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```
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### **...**
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
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class Solution {
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public:
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int minOperations(vector<int>& nums) {
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unordered_map<int, int> count;
5+
for (int num : nums) {
6+
++count[num];
7+
}
8+
int ans = 0;
9+
for (auto& [_, c] : count) {
10+
if (c < 2) {
11+
return -1;
12+
}
13+
ans += (c + 2) / 3;
14+
}
15+
return ans;
16+
}
17+
};
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
func minOperations(nums []int) (ans int) {
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count := map[int]int{}
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for _, num := range nums {
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count[num]++
5+
}
6+
for _, c := range count {
7+
if c < 2 {
8+
return -1
9+
}
10+
ans += (c + 2) / 3
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}
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return
13+
}

‎solution/2800-2899/2870.Minimum Number of Operations to Make Array Empty/Solution.java‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -6,7 +6,7 @@ public int minOperations(int[] nums) {
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count.merge(num, 1, Integer::sum);
77
}
88
int ans = 0;
9-
for (Integer c : count.values()) {
9+
for (int c : count.values()) {
1010
if (c < 2) {
1111
return -1;
1212
}
Lines changed: 9 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,9 @@
1+
class Solution:
2+
def minOperations(self, nums: List[int]) -> int:
3+
count = Counter(nums)
4+
ans = 0
5+
for c in count.values():
6+
if c == 1:
7+
return -1
8+
ans += (c + 2) // 3
9+
return ans
Lines changed: 14 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,14 @@
1+
function minOperations(nums: number[]): number {
2+
const count: Map<number, number> = new Map();
3+
for (const num of nums) {
4+
count.set(num, (count.get(num) ?? 0) + 1);
5+
}
6+
let ans = 0;
7+
for (const [_, c] of count) {
8+
if (c < 2) {
9+
return -1;
10+
}
11+
ans += ((c + 2) / 3) | 0;
12+
}
13+
return ans;
14+
}

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