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feat: add solutions to lc problem: No.3556 #4430

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166 changes: 162 additions & 4 deletions solution/3500-3599/3556.Sum of Largest Prime Substrings/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -71,32 +71,190 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3556.Su

<!-- solution:start -->

### 方法一
### 方法一：枚举 + 哈希表

我们可以枚举所有的子字符串，然后判断它们是否是质数。由于题目要求我们返回最大的 3 个不同质数的和，因此我们可以使用一个哈希表来存储所有的质数。

在遍历完所有的子字符串后，我们将哈希表中的质数按从小到大的顺序排序，然后取出最大的 3 个质数进行求和。

如果哈希表中质数的数量小于 3，则返回所有质数的和。

时间复杂度 $O(n^2 \times \sqrt{M})$，空间复杂度 $O(n^2)$，其中 $n$ 为字符串的长度，而 $M$ 为字符串中最大的子字符串的值。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def sumOfLargestPrimes(self, s: str) -> int:
def is_prime(x: int) -> bool:
if x < 2:
return False
return all(x % i for i in range(2, int(sqrt(x)) + 1))

st = set()
n = len(s)
for i in range(n):
x = 0
for j in range(i, n):
x = x * 10 + int(s[j])
if is_prime(x):
st.add(x)
return sum(sorted(st)[-3:])
```

#### Java

```java

class Solution {
public long sumOfLargestPrimes(String s) {
Set<Long> st = new HashSet<>();
int n = s.length();

for (int i = 0; i < n; i++) {
long x = 0;
for (int j = i; j < n; j++) {
x = x * 10 + (s.charAt(j) - '0');
if (is_prime(x)) {
st.add(x);
}
}
}

List<Long> sorted = new ArrayList<>(st);
Collections.sort(sorted);

long ans = 0;
int start = Math.max(0, sorted.size() - 3);
for (int idx = start; idx < sorted.size(); idx++) {
ans += sorted.get(idx);
}
return ans;
}

private boolean is_prime(long x) {
if (x < 2) return false;
for (long i = 2; i * i <= x; i++) {
if (x % i == 0) return false;
}
return true;
}
}
```

#### C++

```cpp

class Solution {
public:
long long sumOfLargestPrimes(string s) {
unordered_set<long long> st;
int n = s.size();

for (int i = 0; i < n; ++i) {
long long x = 0;
for (int j = i; j < n; ++j) {
x = x * 10 + (s[j] - '0');
if (is_prime(x)) {
st.insert(x);
}
}
}

vector<long long> sorted(st.begin(), st.end());
sort(sorted.begin(), sorted.end());

long long ans = 0;
int cnt = 0;
for (int i = (int) sorted.size() - 1; i >= 0 && cnt < 3; --i, ++cnt) {
ans += sorted[i];
}
return ans;
}

private:
bool is_prime(long long x) {
if (x < 2) return false;
for (long long i = 2; i * i <= x; ++i) {
if (x % i == 0) return false;
}
return true;
}
};
```

#### Go

```go
func sumOfLargestPrimes(s string) (ans int64) {
st := make(map[int64]struct{})
n := len(s)

for i := 0; i < n; i++ {
var x int64 = 0
for j := i; j < n; j++ {
x = x*10 + int64(s[j]-'0')
if isPrime(x) {
st[x] = struct{}{}
}
}
}

nums := make([]int64, 0, len(st))
for num := range st {
nums = append(nums, num)
}
sort.Slice(nums, func(i, j int) bool { return nums[i] < nums[j] })
for i := len(nums) - 1; i >= 0 && len(nums)-i <= 3; i-- {
ans += nums[i]
}
return
}

func isPrime(x int64) bool {
if x < 2 {
return false
}
sqrtX := int64(math.Sqrt(float64(x)))
for i := int64(2); i <= sqrtX; i++ {
if x%i == 0 {
return false
}
}
return true
}
```

#### TypeScript

```ts
function sumOfLargestPrimes(s: string): number {
const st = new Set<number>();
const n = s.length;

for (let i = 0; i < n; i++) {
let x = 0;
for (let j = i; j < n; j++) {
x = x * 10 + Number(s[j]);
if (isPrime(x)) {
st.add(x);
}
}
}

const sorted = Array.from(st).sort((a, b) => a - b);
const topThree = sorted.slice(-3);
return topThree.reduce((sum, val) => sum + val, 0);
}

function isPrime(x: number): boolean {
if (x < 2) return false;
for (let i = 2; i * i <= x; i++) {
if (x % i === 0) return false;
}
return true;
}
```

<!-- tabs:end -->
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166 changes: 162 additions & 4 deletions solution/3500-3599/3556.Sum of Largest Prime Substrings/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -69,32 +69,190 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3556.Su

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration + Hash Table

We can enumerate all substrings and check whether they are prime numbers. Since the problem requires us to return the sum of the largest 3 distinct primes, we can use a hash table to store all the primes.

After traversing all substrings, we sort the primes in the hash table in ascending order, and then take the largest 3 primes to calculate the sum.

If there are fewer than 3 primes in the hash table, return the sum of all primes.

The time complexity is $O(n^2 \times \sqrt{M})$, and the space complexity is $O(n^2)$, where $n$ is the length of the string and $M$ is the value of the largest substring.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def sumOfLargestPrimes(self, s: str) -> int:
def is_prime(x: int) -> bool:
if x < 2:
return False
return all(x % i for i in range(2, int(sqrt(x)) + 1))

st = set()
n = len(s)
for i in range(n):
x = 0
for j in range(i, n):
x = x * 10 + int(s[j])
if is_prime(x):
st.add(x)
return sum(sorted(st)[-3:])
```

#### Java

```java

class Solution {
public long sumOfLargestPrimes(String s) {
Set<Long> st = new HashSet<>();
int n = s.length();

for (int i = 0; i < n; i++) {
long x = 0;
for (int j = i; j < n; j++) {
x = x * 10 + (s.charAt(j) - '0');
if (is_prime(x)) {
st.add(x);
}
}
}

List<Long> sorted = new ArrayList<>(st);
Collections.sort(sorted);

long ans = 0;
int start = Math.max(0, sorted.size() - 3);
for (int idx = start; idx < sorted.size(); idx++) {
ans += sorted.get(idx);
}
return ans;
}

private boolean is_prime(long x) {
if (x < 2) return false;
for (long i = 2; i * i <= x; i++) {
if (x % i == 0) return false;
}
return true;
}
}
```

#### C++

```cpp

class Solution {
public:
long long sumOfLargestPrimes(string s) {
unordered_set<long long> st;
int n = s.size();

for (int i = 0; i < n; ++i) {
long long x = 0;
for (int j = i; j < n; ++j) {
x = x * 10 + (s[j] - '0');
if (is_prime(x)) {
st.insert(x);
}
}
}

vector<long long> sorted(st.begin(), st.end());
sort(sorted.begin(), sorted.end());

long long ans = 0;
int cnt = 0;
for (int i = (int) sorted.size() - 1; i >= 0 && cnt < 3; --i, ++cnt) {
ans += sorted[i];
}
return ans;
}

private:
bool is_prime(long long x) {
if (x < 2) return false;
for (long long i = 2; i * i <= x; ++i) {
if (x % i == 0) return false;
}
return true;
}
};
```

#### Go

```go
func sumOfLargestPrimes(s string) (ans int64) {
st := make(map[int64]struct{})
n := len(s)

for i := 0; i < n; i++ {
var x int64 = 0
for j := i; j < n; j++ {
x = x*10 + int64(s[j]-'0')
if isPrime(x) {
st[x] = struct{}{}
}
}
}

nums := make([]int64, 0, len(st))
for num := range st {
nums = append(nums, num)
}
sort.Slice(nums, func(i, j int) bool { return nums[i] < nums[j] })
for i := len(nums) - 1; i >= 0 && len(nums)-i <= 3; i-- {
ans += nums[i]
}
return
}

func isPrime(x int64) bool {
if x < 2 {
return false
}
sqrtX := int64(math.Sqrt(float64(x)))
for i := int64(2); i <= sqrtX; i++ {
if x%i == 0 {
return false
}
}
return true
}
```

#### TypeScript

```ts
function sumOfLargestPrimes(s: string): number {
const st = new Set<number>();
const n = s.length;

for (let i = 0; i < n; i++) {
let x = 0;
for (let j = i; j < n; j++) {
x = x * 10 + Number(s[j]);
if (isPrime(x)) {
st.add(x);
}
}
}

const sorted = Array.from(st).sort((a, b) => a - b);
const topThree = sorted.slice(-3);
return topThree.reduce((sum, val) => sum + val, 0);
}

function isPrime(x: number): boolean {
if (x < 2) return false;
for (let i = 2; i * i <= x; i++) {
if (x % i === 0) return false;
}
return true;
}
```

<!-- tabs:end -->
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