从任一端到给定字符串中最大和最小字符的最小跳转

给定一个字符串str ，任务是找到达到字典上最大和最小字符所需的最小移动次数。在一次移动中，可以从给定字符串的最左侧或最右侧进行跳跃。

例子：

Input: str = AEDCB, N = 5  
Output: 2
Explanation: take two steps from leftmost side to reach A and E

Input: str = BACDEFHG, N = 8
Output: 4
Explanation: take two steps from leftmost side to reach A and 2 steps from rightmost side to reach H(2+2=4)

Input: str = CDBA, N = 4
Output: 3
Explanation: take  three steps from rightmost side to reach A and  then D

方法：这个问题是基于实现的。请按照以下步骤解决给定的问题。

• 查找字符串中最大和最小元素的索引
• 计算这些索引的最小值和最大值，找到可以采取的min_steps和max_steps
• 只有三种可能的方式来达到这两个元素
  • 从头开始遍历并覆盖两个元素，即 min_steps+1
  • 从最后一个遍历并覆盖两个元素，即n-max_steps
  • 从开始和结束都遍历，即min_steps+1+n-max_steps
• 最终的答案是所有可能的三种遍历方式中的最小值。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Minimum number of moves required  to
// reach and largest and smallest ASCII values
int min_moves(string s, int n)
{
    int maxpos = 0, minpos = 0;
 
    // Finding index of maximum
    // and minimum element in string
    for (int i = 0; i < n; i++) {
 
        if (s[i] > s[maxpos])
            maxpos = i;
 
        if (s[i] < s[minpos])
            minpos = i;
    }
 
    // Calculating minimum ans maximum steps
    // that can be taken
 
    int min_steps = min(maxpos, minpos);
    int max_steps = max(maxpos, minpos);
 
    // Only three possible ways
    // to reach both elements
    int ans1, ans2, ans3;
 
    ans1 = n - min_steps;
 
    ans2 = max_steps + 1;
 
    ans3 = min_steps + 1 + n - max_steps;
 
    int result;
 
    // Minimum steps in all three ways
    result = min(ans1, min(ans2, ans3));
 
    // Return the final result
    return result;
}
 
// Driver code
int main()
{
 
    string str = "BACDEFHG";
    int N = str.length();
 
    cout << min_moves(str, N);
 
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG
{
   
// Minimum number of moves required  to
// reach and largest and smallest ASCII values
static int min_moves(String s, int n)
{
    int maxpos = 0, minpos = 0;
 
    // Finding index of maximum
    // and minimum element in string
    for (int i = 0; i < n; i++) {
 
        if (s.charAt(i) > s.charAt(maxpos))
            maxpos = i;
 
        if (s.charAt(i)  < s.charAt(minpos))
            minpos = i;
    }
 
    // Calculating minimum ans maximum steps
    // that can be taken
 
    int min_steps = Math.min(maxpos, minpos);
    int max_steps = Math.max(maxpos, minpos);
 
    // Only three possible ways
    // to reach both elements
    int ans1, ans2, ans3;
 
    ans1 = n - min_steps;
 
    ans2 = max_steps + 1;
 
    ans3 = min_steps + 1 + n - max_steps;
 
    int result;
 
    // Minimum steps in all three ways
    result = Math.min(ans1, Math.min(ans2, ans3));
 
    // Return the final result
    return result;
}
 
// Driver code
    public static void main (String[] args) {
       String str = "BACDEFHG";
    int N = str.length();
 
    
        System.out.println(min_moves(str, N));
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python code for the above approach
 
# Minimum number of moves required to
# reach and largest and smallest ASCII values
def min_moves(s, n):
    maxpos = 0;
    minpos = 0;
 
    # Finding index of maximum
    # and minimum element in string
    for i in range(n):
 
        if (s[i] > s[maxpos]):
            maxpos = i;
 
        if (s[i] < s[minpos]):
            minpos = i;
 
    # Calculating minimum ans maximum steps
    # that can be taken
    min_steps = min(maxpos, minpos);
    max_steps = max(maxpos, minpos);
 
    # Only three possible ways
    # to reach both elements
    ans1, ans2, ans3 = 0,0,0;
    ans1 = n - min_steps;
    ans2 = max_steps + 1;
    ans3 = min_steps + 1 + n - max_steps;
 
    result=0;
 
    # Minimum steps in all three ways
    result = min(ans1, min(ans2, ans3));
 
    # Return the final result
    return result;
 
# Driver code
if __name__ == '__main__':
    str = "BACDEFHG";
    N = len(str);
 
    print(min_moves(str, N));
 
# This code is contributed by shikhasingrajput

C#

// C# program for above approach
using System;
 
class GFG{
 
// Minimum number of moves required  to
// reach and largest and smallest ASCII values
static int min_moves(string s, int n)
{
    int maxpos = 0, minpos = 0;
 
    // Finding index of maximum
    // and minimum element in string
    for(int i = 0; i < n; i++)
    {
        if (s[i] > s[maxpos])
            maxpos = i;
 
        if (s[i] < s[minpos])
            minpos = i;
    }
 
    // Calculating minimum ans maximum steps
    // that can be taken
    int min_steps = Math.Min(maxpos, minpos);
    int max_steps = Math.Max(maxpos, minpos);
 
    // Only three possible ways
    // to reach both elements
    int ans1, ans2, ans3;
 
    ans1 = n - min_steps;
    ans2 = max_steps + 1;
    ans3 = min_steps + 1 + n - max_steps;
 
    int result;
 
    // Minimum steps in all three ways
    result = Math.Min(ans1, Math.Min(ans2, ans3));
 
    // Return the final result
    return result;
}
 
// Driver code
public static void Main(string[] args)
{
    String str = "BACDEFHG";
    int N = str.Length;
 
    Console.WriteLine(min_moves(str, N));
}
}
 
// This code is contributed by ukasp

Javascript

 
// JavaScript program for above approach
 
// Minimum number of moves required to
// reach and largest and smallest ASCII values
const min_moves = (s, n) => {
     
    let maxpos = 0, minpos = 0;
 
    // Finding index of maximum
    // and minimum element in string
    for(let i = 0; i < n; i++)
    {
        if (s[i] > s[maxpos])
            maxpos = i;
 
        if (s[i] < s[minpos])
            minpos = i;
    }
 
    // Calculating minimum ans maximum steps
    // that can be taken
    let min_steps = Math.min(maxpos, minpos);
    let max_steps = Math.max(maxpos, minpos);
 
    // Only three possible ways
    // to reach both elements
    let ans1, ans2, ans3;
 
    ans1 = n - min_steps;
    ans2 = max_steps + 1;
    ans3 = min_steps + 1 + n - max_steps;
 
    let result;
 
    // Minimum steps in all three ways
    result = Math.min(ans1, Math.min(ans2, ans3));
 
    // Return the final result
    return result;
}
 
// Driver code
let str = "BACDEFHG";
let N = str.length;
 
document.write(min_moves(str, N));
 
// This code is contributed by rakeshsahni
 

输出

4

时间复杂度： O(N)
辅助空间： O(1)