给定数字字符串中最小非素数子序列的长度
给定一个由数字[1, 9]组成的大小为N的字符串S ，任务是找到字符串中最小子序列的长度，使其不是素数。
例子：
Input: S = “237”
Output: 2
Explanation:
There are 7 non empty subsequence {“2”, “3”, “7”, “23”, “27”, “37”, “237”}. Among these subsequence there are two subsequence that are not prime, i.e., 27 and 237. Thus as 27 has a length of 2. So print 2.
Input: S = “44444”
Output: 1
方法：解决这个问题的想法是基于以下观察：如果从字符串S中删除j或多于j个字符，则字符串是素数，那么答案应该大于j 。基于这一事实，形成所有字符串，使得从该字符串中删除元素给出一个素数。上述直觉中所有可能的字符串都是{2, 3, 5, 7, 23, 37, 53, 73}因此子序列的最大可能大小是3 。因此，想法是遍历大小为1和大小为2的所有子序列，如果发现子序列包含至少 1 个不存在于列表中的元素，则大小可能是1或2 。否则，大小将为3 。请按照以下步骤解决问题：
- 将布尔变量标志初始化为假。
- 初始化一个空字符串dummy。
- 使用变量j迭代范围[0, N)并执行以下任务：
- 如果第j 个位置的字符不等于2或3或5或7，则打印答案为1，将flag的值设置为true并中断。
- 如果flag为true ，则返回，否则执行以下任务。
- 使用变量j迭代范围[0, N)并执行以下任务：
- 使用变量j1迭代范围[j+1, N)并执行以下任务：
- 用j和j1位置的字符构建一个虚拟字符串。
- 如果虚拟字符串不等于2、3、5、7、23、37、53和73 ，则将答案打印为2 ，并将flag的值设置为true并中断。
- 使用变量j1迭代范围[j+1, N)并执行以下任务：
- 如果标志为false并且字符串S的长度大于等于3 ，则打印3作为答案，否则打印-1 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the smallest
// length of resultant subsequence
void findMinimumSubsequence(
string S)
{
bool flag = false;
string dummy;
// Check for a subsequence of size 1
for (int j = 0; j < S.length(); j++)
{
if (S[j] != '2' && S[j] != '3' && S[j] != '5' && S[j] != '7')
{
cout << 1;
flag = true;
break;
}
}
// Check for a subsequence of size 2
if (!flag)
{
for (int j = 0;
j < S.length() - 1; j++)
{
for (int j1 = j + 1;
j1 < S.length(); j1++)
{
dummy = S[j] + S[j1];
if (dummy != "23" && dummy != "37" && dummy != "53" && dummy != "73")
{
cout << 2;
}
if (flag = true)
break;
}
if (flag = true)
break;
}
}
// If none of the above check is
// successful then subsequence
// must be of size 3
if (!flag)
{
if (S.length() >= 3)
{
// Never executed
cout << 3;
}
else
{
cout << -1;
}
}
}
// Driver Code
int main()
{
string S = "237";
findMinimumSubsequence(S);
return 0;
}
// This code is contributed by Potta Lokesh Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to find the smallest
// length of resultant subsequence
public static void findMinimumSubsequence(
String S)
{
boolean flag = false;
StringBuilder dummy = new StringBuilder();
// Check for a subsequence of size 1
for (int j = 0; j < S.length(); j++) {
if (S.charAt(j) != '2' && S.charAt(j) != '3'
&& S.charAt(j) != '5'
&& S.charAt(j) != '7') {
System.out.println(1);
flag = true;
break;
}
}
// Check for a subsequence of size 2
if (!flag) {
loop:
for (int j = 0;
j < S.length() - 1; j++) {
for (int j1 = j + 1;
j1 < S.length(); j1++) {
dummy = new StringBuilder(
Character.toString(S.charAt(j)));
dummy.append(S.charAt(j1));
if (!dummy.toString().equals("23")
&& !dummy.toString().equals("37")
&& !dummy.toString().equals("53")
&& !dummy.toString().equals("73")) {
System.out.println(2);
flag = true;
break loop;
}
}
}
}
// If none of the above check is
// successful then subsequence
// must be of size 3
if (!flag) {
if (S.length() >= 3) {
// Never executed
System.out.println(3);
}
else {
System.out.println(-1);
}
}
}
// Driver Code
public static void main(String[] args)
{
String S = "237";
findMinimumSubsequence(S);
}
}C#
// C# program for the above approach
using System;
class GFG
{
// Function to find the smallest
// length of resultant subsequence
static void findMinimumSubsequence(string S)
{
bool flag = false;
string dummy = "";
// Check for a subsequence of size 1
for (int j = 0; j < S.Length; j++) {
if (S[j] != '2' && S[j] != '3' && S[j] != '5'
&& S[j] != '7') {
Console.WriteLine(1);
flag = true;
break;
}
}
// Check for a subsequence of size 2
if (!flag) {
for (int j = 0; j < S.Length - 1; j++) {
for (int j1 = j + 1; j1 < S.Length; j1++) {
dummy = S[j].ToString()
+ S[j1].ToString();
if (dummy != "23" && dummy != "37"
&& dummy != "53" && dummy != "73") {
Console.WriteLine(2);
}
if (flag == true)
break;
else
flag = true;
}
if (flag == true)
break;
}
}
// If none of the above check is
// successful then subsequence
// must be of size 3
if (flag == false) {
if (S.Length >= 3) {
// Never executed
Console.WriteLine(3);
}
else {
Console.WriteLine(-1);
}
}
}
// Driver Code
public static void Main()
{
string S = "237";
findMinimumSubsequence(S);
}
}
// This code is contributed by ukasp.Python3
# Python 3 program for the above approach
# Function to find the smallest
# length of resultant subsequence
def findMinimumSubsequence(S):
flag = False
dummy = ''
# Check for a subsequence of size 1
for j in range(len(S)):
if (S[j] != '2' and S[j] != '3' and S[j] != '5' and S[j] != '7'):
print(1)
flag = True
break
# Check for a subsequence of size 2
if (flag == False):
for j in range(len(S)):
for j1 in range(j + 1,len(S),1):
dummy = S[j] + S[j1]
if (dummy != "23" and dummy != "37" and dummy != "53" and dummy != "73"):
print(2)
if (flag == True):
break
else:
flag = True
if (flag == True):
break
# If none of the above check is
# successful then subsequence
# must be of size 3
if (flag == False):
if (len(S) >= 3):
# Never executed
print(3)
else:
print(-1)
# Driver Code
if __name__ == '__main__':
S = "237"
findMinimumSubsequence(S)
# This code is contributed by ipg2016107.Javascript
输出：
2时间复杂度： O(N 2 )
辅助空间： O(1)