通过重复删除具有不同开头和结尾的子字符串，最小移动以清空字符串

给定一个字符串str ，任务是通过删除任何开始和结束字符的子字符串来找到使str为空所需的最小移动次数 是不同的。如果无法清空字符串，则返回“ -1 ”。

例子：

Input: str = “abba”
Output: 2
Explanation: Follow the following steps to get the empty string: “abba” -> (delete ab) -> “ba” -> (delete ba) -> “”

Input: aba
Output: -1
Explanation: It is not possible to empty the given string.

Input: abc
Output: 1

方法：可以在观察的基础上解决任务。

• 如果第一个和最后一个字符不相等，则只需要一个操作
• 否则，检查是否存在与第一个和最后一个字符不匹配的任何2 个索引。如果它们存在，则所需的最小操作数为2 ，否则不可能将字符串缩减为空字符串。

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 number of moves
// required to make the string empty.
void numberOfMoves(string& str, int n)
{
    // If the first and the last
    // character are different
    if (str[0] != str[n - 1]) {
        cout << "1";
        return;
    }
 
    // Check if there is a index i such that
    // s[i] != s[0] and s[i+1] != s[n-1]
    for (int i = 0; i < n - 1; i++) {
        if (str[i] != str[0]
            && str[i + 1] != str[n - 1]) {
            cout << "2";
            return;
        }
    }
 
    // If no such index exists then
    // the answer is -1
    cout << -1;
    return;
}
 
// Driver Code
int main()
{
    string str = "abba";
    int n = 4;
    numberOfMoves(str, n);
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG
{
 
  // Function to find the number of moves
  // required to make the string empty.
  static void numberOfMoves(String str, int n)
  {
 
    // If the first and the last
    // character are different
    if (str.charAt(0) != str.charAt(n-1)) {
      System.out.println("1");
      return;
    }
 
    // Check if there is a index i such that
    // s[i] != s[0] and s[i+1] != s[n-1]
    for (int i = 0; i < n - 1; i++) {
      if (str.charAt(i) != str.charAt(0)
          && str.charAt(i+1) != str.charAt(n-1)) {
        System.out.println("2");
        return;
      }
    }
 
    // If no such index exists then
    // the answer is -1
    System.out.println(-1);
    return;
  }
 
  // Driver Code
  public static void main (String[] args) {
    String str = "abba";
    int n = 4;
    numberOfMoves(str, n);
 
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Pyhton program for the above approach
 
# Function to find the number of moves
# required to make the string empty.
def numberOfMoves(str, n):
   
    # If the first and the last
    # character are different
    if (str[0] != str[n - 1]):
        print("1")
        return
 
    # Check if there is a index i such that
    # s[i] != s[0] and s[i+1] != s[n-1]
    for i in range(0, len(str)):
        if (str[i] != str[0] and
            str[i + 1] != str[n - 1]):
                 
            print("2")
            return
 
    # If no such index exists then
    # the answer is -1
    print(-1)
    return
 
# Driver Code
if __name__ == '__main__':
     
    str = "abba"
    n = 4;
    numberOfMoves(str, n);
 
    # This code is contributed by Samim Hossain Mondal.

C#

// C# program for the above approach
using System;
 
public class GFG
{
 
// Function to find the number of moves
// required to make the string empty.
static void numberOfMoves(string str, int n)
{
   
    // If the first and the last
    // character are different
    if (str[0] != str[n - 1]) {
        Console.WriteLine("1");
        return;
    }
 
    // Check if there is a index i such that
    // s[i] != s[0] and s[i+1] != s[n-1]
    for (int i = 0; i < n - 1; i++) {
        if (str[i] != str[0]
            && str[i + 1] != str[n - 1]) {
            Console.WriteLine("2");
            return;
        }
    }
 
    // If no such index exists then
    // the answer is -1
    Console.WriteLine(-1);
    return;
}
 
// Driver Code
public static void Main(String []args) {
     
    string str = "abba";
    int n = 4;
    numberOfMoves(str, n);
}
}
 
// This code is contributed by target_2.

Javascript

// Javascript program for the above approach
 
 
// Function to find the number of moves
// required to make the string empty.
function numberOfMoves(str, n) {
  // If the first and the last
  // character are different
  if (str[0] != str[n - 1]) {
    document.write("1");
    return;
  }
 
  // Check if there is a index i such that
  // s[i] != s[0] and s[i+1] != s[n-1]
  for (let i = 0; i < n - 1; i++) {
    if (str[i] != str[0]
      && str[i + 1] != str[n - 1]) {
      document.write("2");
      return;
    }
  }
 
  // If no such index exists then
  // the answer is -1
  document.write(-1);
  return;
}
 
// Driver Code
 
let str = "abba";
let n = 4;
numberOfMoves(str, n);
 
// This code is contributed by gfgking.

输出

2

时间复杂度： O(N)
辅助空间： O(1)。