通过基于给定命令在 X 轴上移动来最大化从原点的绝对位移

给定一个长度为N的字符串S ，其中字符串的每个字符都等于“L”、“R”或“？” ，任务是通过从原点(0, 0)开始在 X 轴上按照给定命令移动来找到从原点的最大绝对位移：

• 'L'：在负 X 方向移动一个单位。
• 'R'：沿 X 正方向移动一个单位。
• '?'：可以在负 X 方向或正 X 方向移动一个单位。

例子：

Input: S = “LL??R” 
Output: 3
Explanation:
One of the possible way to move is:

1. S[0] = ‘L’, move one unit in -ve X direction, so displacement becomes equal to -1.
2. S[1] = ‘L’, move one unit in -ve X direction, so displacement becomes equal to -2.
3. S[2] = ‘?’, move one unit in -ve X direction, so displacement becomes equal to -3.
4. S[3] = ‘?’, move one unit in -ve X direction, so displacement becomes equal to -4.
5. S[4] = ‘R’, move one unit in +ve X direction, so displacement becomes equal to -3.

 Therefore, the absolute displacement is abs(-3)=3, and also it is the maximum absolute displacement possible.

Input: S = “?RRR?”
Output: 5

天真的方法：解决问题的最简单方法是尝试替换“？”使用递归使用“L”或“R” ，然后打印获得的最大绝对位移。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
// Recursive function to find the maximum
// absolute displacement from origin by
// performing the given set of moves
int DistRecursion(string S, int i, int dist)
{
    // If i is equal to N
    if (i == S.length())
        return abs(dist);
 
    // If S[i] is equal to 'L'
    if (S[i] == 'L')
        return DistRecursion(S, i + 1, dist - 1);
 
    // If S[i] is equal to 'R'
    if (S[i] == 'R')
        return DistRecursion(S, i + 1, dist + 1);
 
    // If S[i] is equal to '?'
    return max(DistRecursion(S, i + 1, dist - 1),
               DistRecursion(S, i + 1, dist + 1));
}
// Function to find the maximum absolute
// displacement from the origin
 
int maxDistance(string S)
{
 
    // Return the maximum absolute
    // displacement
    return DistRecursion(S, 0, 0);
}
 
// Driver Code
int main()
{
    // Input
    string S = "?RRR?";
 
    // Function call
 
    cout << maxDistance(S);
    return 0;
}
 
// This code is contributed by lokesh potta.

Java

// Java program for the above approach
import java.util.*;
class GFG {
 
    // Recursive function to find the maximum
    // absolute displacement from origin by
    // performing the given set of moves
    static int DistRecursion(String S, int i, int dist)
    {
        char[] ch = S.toCharArray();
        // If i is equal to N
        if (i == ch.length)
            return Math.abs(dist);
 
        // If S[i] is equal to 'L'
        if (ch[i] == 'L')
            return DistRecursion(S, i + 1, dist - 1);
 
        // If S[i] is equal to 'R'
        if (ch[i] == 'R')
            return DistRecursion(S, i + 1, dist + 1);
 
        // If S[i] is equal to '?'
        return Math.max(DistRecursion(S, i + 1,
 
                                      dist - 1),
                        DistRecursion(S, i + 1, dist + 1));
    }
 
    // Function to find the maximum absolute
    // displacement from the origin
    static int maxDistance(String S)
    {
 
        // Return the maximum absolute
        // displacement
        return DistRecursion(S, 0, 0);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Input
        String S = "?RRR?";
 
        // Function call
        System.out.print(maxDistance(S));
    }
}
 
// This code is contributed by ukasp.

Python3

# Python3 program for the above approach
 
# Recursive function to find the maximum
# absolute displacement from origin by
# performing the given set of moves
 
 
def DistRecursion(S, i, dist):
 
    # If i is equal to N
    if i == len(S):
        return abs(dist)
 
    # If S[i] is equal to 'L'
    if S[i] == 'L':
        return DistRecursion(S, i + 1, dist-1)
 
    # If S[i] is equal to 'R'
    if S[i] == 'R':
        return DistRecursion(S, i + 1, dist + 1)
 
    # If S[i] is equal to '?'
    return max(DistRecursion(S, i + 1, dist-1),
               DistRecursion(S, i + 1, dist + 1))
 
 
# Function to find the maximum absolute
# displacement from the origin
def maxDistance(S):
 
    # Return the maximum absolute
    # displacement
    return DistRecursion(S, 0, 0)
 
# Driver Code
 
 
# Input
S = "?RRR?"
 
# Function call
print(maxDistance(S))

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Recursive function to find the maximum
// absolute displacement from origin by
// performing the given set of moves
static int DistRecursion(string S, int i, int dist)
{
     
    // If i is equal to N
    if (i == S.Length)
        return Math.Abs(dist);
 
    // If S[i] is equal to 'L'
    if (S[i] == 'L')
        return DistRecursion(S, i + 1, dist - 1);
 
    // If S[i] is equal to 'R'
    if (S[i] == 'R')
        return DistRecursion(S, i + 1, dist + 1);
 
    // If S[i] is equal to '?'
    return Math.Max(DistRecursion(S, i + 1,
                                   
                                  dist - 1),
                    DistRecursion(S, i + 1, dist + 1));
}
 
// Function to find the maximum absolute
// displacement from the origin
static int maxDistance(string S)
{
     
    // Return the maximum absolute
    // displacement
    return DistRecursion(S, 0, 0);
}
 
// Driver Code
public static void Main()
{
     
    // Input
    string S = "?RRR?";
 
    // Function call
    Console.Write(maxDistance(S));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

 
// JavaScript program for the above approach
 
 
// Recursive function to find the maximum
// absolute displacement from origin by
// performing the given set of moves
function DistRecursion(S, i, dist) {
    // If i is equal to N
    if (i == S.length)
        return Math.abs(dist);
 
    // If S[i] is equal to 'L'
    if (S[i] == 'L')
        return DistRecursion(S, i + 1, dist - 1);
 
    // If S[i] is equal to 'R'
    if (S[i] == 'R')
        return DistRecursion(S, i + 1, dist + 1);
 
    // If S[i] is equal to '?'
    return Math.max(DistRecursion(S, i + 1, dist - 1),
        DistRecursion(S, i + 1, dist + 1));
}
// Function to find the maximum absolute
// displacement from the origin
 
function maxDistance(S) {
 
    // Return the maximum absolute
    // displacement
    return DistRecursion(S, 0, 0);
}
 
// Driver Code
 
// Input
let S = "?RRR?";
 
// Function call
 
document.write(maxDistance(S));
 
 
// This code is contributed by _saurabh_jaiswal
 

输出

5

时间复杂度： O(2 ^N )
辅助空间： O(1)

有效方法：上述方法可以根据观察得到优化，即在“？”时将获得最大绝对位移。替换为最大出现字符。请按照以下步骤解决问题：

• 在S中查找字符'L'的计数，并将其存储在变量l中。
• 在S中查找字符'R'的计数，并将其存储在变量 say r中。
• 将答案打印为N – min(l, r)

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
int count(string s, char c) {
    int ans = 0;
    for(int i = 0; i < s.length(); i++)
    {
        if (c == s[i])
        {
            ans++;
        }
    }
    return ans;
}
 
// Function to find the maximum absolute
// displacement from the origin
int maxDistance(string S) {
  
    // Stores the count of 'L'
    int l = count(S, 'L');
 
    // Stores the count of 'R'
    int r = count(S, 'R');
 
    // Stores the length of S
    int N = S.length();
 
    // Return the answer
    return abs(N - min(l, r));
}
     
int main()
{
    // Input
    string S = "?RRR?";
 
    // Function call
    cout << maxDistance(S);
 
    return 0;
}
 
// This code is contributed by divyesh072019.

Java

// Java program for the above approach
 
// Function to find the maximum absolute
// displacement from the origin
class GFG {
    static int maxDistance(String S) {
 
        // Stores the count of 'L'
        int l = count(S, 'L');
 
        // Stores the count of 'R'
        int r = count(S, 'R');
 
        // Stores the length of S
        int N = S.length();
 
        // Return the answer
        return Math.abs(N - Math.min(l, r));
 
    }
 
    private static int count(String s, char c) {
        int ans = 0;
        for (char i : s.toCharArray())
            if (c == i)
                ans++;
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args) {
 
        // Input
        String S = "?RRR?";
 
        // Function call
        System.out.println(maxDistance(S));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach
 
# Function to find the maximum absolute
# displacement from the origin
 
 
def maxDistance(S):
 
    # Stores the count of 'L'
    l = S.count('L')
 
    # Stores the count of 'R'
    r = S.count('R')
 
    # Stores the length of S
    N = len(S)
 
    # Return the answer
    return abs(N - min(l, r))
 
 
# Driver Code
 
# Input
S = "?RRR?"
 
# Function call
print(maxDistance(S))

C＃

// C# program for the above approach
 
// Function to find the maximum absolute
// displacement from the origin
using System; 
 
public class GFG {
    static int maxDistance(String S) {
 
        // Stores the count of 'L'
        int l = count(S, 'L');
 
        // Stores the count of 'R'
        int r = count(S, 'R');
 
        // Stores the length of S
        int N = S.Length;
 
        // Return the answer
        return Math.Abs(N - Math.Min(l, r));
 
    }
 
    private static int count(String s, char c) {
        int ans = 0;
        foreach (char i in s.ToCharArray())
            if (c == i)
                ans=ans+1;
        return ans;
    }
 
    // Driver Code
    public static void Main(String[] args) {
 
        // Input
        String S = "?RRR?";
 
        // Function call
        Console.WriteLine(maxDistance(S));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program for the above approach
 
// Function to find the maximum absolute
// displacement from the origin
    function maxDistance(S)
    {
 
        // Stores the count of 'L'
        var l = count(S, 'L');
 
        // Stores the count of 'R'
        var r = count(S, 'R');
 
        // Stores the length of S
        var N = S.length;
 
        // Return the answer
        return Math.abs(N - Math.min(l, r));
 
    }
 
    function count(s, c) {
        var ans = 0;
        for (var i in s.split(''))
            if (c == i)
                ans++;
        return ans;
    }
 
// Driver Code
 
// Input
var S = "?RRR?";
 
// Function call
document.write(maxDistance(S));
 
// This code is contributed by 29AjayKumar

输出

5

时间复杂度： O(N)
辅助空间： O(1)