查找在相应索引处具有不同字符的给定字符串的字谜

给定一个由N个字符组成的字符串S ，任务是找到给定字符串S的字谜，使得相同索引处的字符与原始字符串不同。

例子：

Input: S = “geek”
Output: egke
Explanation:
The anagram of the given string such that all the characters at all the corresponding indices are not same is “egke”.  

Input: S = “aaaa”
Output: -1

方法：给定的问题可以通过使用两个指针的方法来解决。这个想法是交换字符串末尾的不同字符，如果这些索引处的字符相同，则检查下一个可能具有不同字符的索引对。执行上述操作后，检查 anagram 是否在每个索引处生成不同的字符并相应地打印结果。请按照以下步骤解决给定的问题：

• 初始化一个字符串，比如存储给定字符串S的T。
• 初始化两个指针，比如i和j分别为0和(N – 1) 。
• 迭代直到i的值小于N并且j为非负数并执行以下步骤：
  1. 如果字符S[i]和S[j]不相同并且字符对(S[i], T[j])和(T[i], S[j])也不相同（确保交换操作后的字符与原来的相同），然后交换字符(S[i], S[j])并将j的值更新为(N – 1) 。
  2. 否则，将i的值增加1 。
• 如果字符串有奇数个字符并且中间索引处的字符相等，则执行上述步骤所示的交换操作。
• 完成上述步骤后，如果字符S和T对应索引处的字符串不相同，则打印字符串S作为结果字符串。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 anagram of string
// such that characters at the same
// indices are different
void findAnagram(string s)
{
    // Copying our original string
    // for comparison
    string check = s;
 
    // Declaring the two pointers
    int i = 0, j = s.length() - 1;
 
    while (i < s.length() && j >= 0) {
 
        // Checking the given condition
        if (s[i] != s[j] && check[i] != s[j]
            && check[j] != s[i]) {
            swap(s[i], s[j]);
            i++;
 
            j = s.length() - 1;
        }
        else {
            j--;
        }
    }
 
    // When string length is odd
    if (s.length() % 2 != 0) {
 
        // The mid element
        int mid = s.length() / 2;
 
        // If the characters are the
        // same, then perform the swap
        // operation as illustrated
        if (check[mid] == s[mid]) {
            for (int i = 0; i < s.length(); i++) {
                if (check[i] != s[mid]
                    && s[i] != s[mid]) {
                    swap(s[i], s[mid]);
                    break;
                }
            }
        }
    }
 
    // Check if the corresponding indices
    // has the same character or not
    bool ok = true;
    for (int i = 0; i < s.length(); i++) {
        if (check[i] == s[i]) {
            ok = false;
            break;
        }
    }
 
    // If string follows required
    // condition
    if (ok)
        cout << s;
    else
        cout << -1;
}
 
// Driver Code
int main()
{
    string S = "geek";
    findAnagram(S);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
// Function to find anagram of string
// such that characters at the same
// indices are different
static void findAnagram(String s)
{
    
    // Copying our original string
    // for comparison
    String check = s;
  
    // Declaring the two pointers
    int i = 0, j = s.length() - 1;
  
    while (i < s.length() && j >= 0) {
  
        // Checking the given condition
        if (s.charAt(i) != s.charAt(j) && check.charAt(i) != s.charAt(j)
            && check.charAt(j) != s.charAt(i)) {
            char temp = s.charAt(i);
            s = s.substring(0, i) + s.charAt(j) + s.substring(i + 1);
            s = s.substring(0, j) + temp + s.substring(j + 1);
            i++;
  
            j = s.length() - 1;
        }
        else {
            j--;
        }
    }
  
    // When string length is odd
    if (s.length() % 2 != 0) {
  
        // The mid element
        int mid = s.length() / 2;
  
        // If the characters are the
        // same, then perform the swap
        // operation as illustrated
        if (check.charAt(mid) == s.charAt(mid)) {
            for (i = 0; i < s.length(); i++) {
                if (check.charAt(i) != s.charAt(mid)
                    && s.charAt(i) != s.charAt(mid)) {
                    char temp = s.charAt(i);
                    s = s.substring(0, i) + s.charAt(mid) + s.substring(i + 1);
                    s = s.substring(0, mid) + temp + s.substring(mid + 1);
                    break;
                }
            }
        }
    }
  
    // Check if the corresponding indices
    // has the same character or not
    boolean ok = true;
     
    for (i = 0; i < s.length(); i++) {
        if (check.charAt(i) == s.charAt(i) ) {
            ok = false;
            break;
        }
    }
  
    // If string follows required
    // condition
    if (ok)
        System.out.println(s);
    else
        System.out.println(-1);
}
 
// Driver Code
public static void main (String[] args)
{
    String S = "geek";
    findAnagram(S);
}
}
 
// This code is contributed by sanjoy_62.

Python3

# Python 3 program for the above approach
 
# Function to find anagram of string
# such that characters at the same
# indices are different
def findAnagram(s):
 
    # Copying our original string
    # for comparison
    check = s
    st = list(s)
 
    # Declaring the two pointers
    i = 0
    j = len(st) - 1
 
    while (i < len(st) and j >= 0):
 
        # Checking the given condition
        if (st[i] != st[j] and check[i] != st[j]
                and check[j] != st[i]):
            st[i], st[j] = st[j], st[i]
            i += 1
 
            j = len(st) - 1
 
        else:
            j -= 1
 
    # When string length is odd
    if (len(st) % 2 != 0):
 
        # The mid element
        mid = len(st) / 2
 
        # If the characters are the
        # same, then perform the swap
        # operation as illustrated
        if (check[mid] == st[mid]):
            for i in range(len(st)):
                if (check[i] != st[mid]
                        and st[i] != st[mid]):
                    st[i], st[mid] = st[mid], st[i]
                    break
 
    # Check if the corresponding indices
    # has the same character or not
    ok = True
    for i in range(len(st)):
        if (check[i] == st[i]):
            ok = False
            break
 
    # If string follows required
    # condition
    if (ok):
        print("".join(st))
    else:
        print(-1)
 
# Driver Code
if __name__ == "__main__":
 
    S = "geek"
    findAnagram(S)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find anagram of string
// such that characters at the same
// indices are different
static void findAnagram(string s)
{
   
    // Copying our original string
    // for comparison
    string check = s;
 
    // Declaring the two pointers
    int i = 0, j = s.Length - 1;
 
    while (i < s.Length && j >= 0) {
 
        // Checking the given condition
        if (s[i] != s[j] && check[i] != s[j]
            && check[j] != s[i]) {
            char temp = s[i];
            s = s.Substring(0, i) + s[j] + s.Substring(i + 1);
            s = s.Substring(0, j) + temp + s.Substring(j + 1);
            i++;
 
            j = s.Length - 1;
        }
        else {
            j--;
        }
    }
 
    // When string length is odd
    if (s.Length % 2 != 0) {
 
        // The mid element
        int mid = s.Length / 2;
 
        // If the characters are the
        // same, then perform the swap
        // operation as illustrated
        if (check[mid] == s[mid]) {
            for (i = 0; i < s.Length; i++) {
                if (check[i] != s[mid]
                    && s[i] != s[mid]) {
                    char temp = s[i];
                    s = s.Substring(0, i) + s[mid] + s.Substring(i + 1);
                    s = s.Substring(0, mid) + temp + s.Substring(mid + 1);
                    break;
                }
            }
        }
    }
 
    // Check if the corresponding indices
    // has the same character or not
    bool ok = true;
    for (i = 0; i < s.Length; i++) {
        if (check[i] == s[i]) {
            ok = false;
            break;
        }
    }
 
    // If string follows required
    // condition
    if (ok)
        Console.Write(s);
    else
        Console.Write(-1);
}
 
// Driver Code
public static void Main()
{
    string S = "geek";
    findAnagram(S);
}
}
 
// This code is contributed by ipg2016107.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find anagram of string
        // such that characters at the same
        // indices are different
        function swap(str, i, j) {
            if (i == j) {
                return str;
            }
 
            if (j < i) {
                var temp = j;
                j = i;
                i = temp;
            }
 
            if (i >= str.length) {
                return str;
 
            }
            return str.substring(0, i) +
                str[j] +
 
                str.substring(i + 1, j) +
 
                str[i] +
                str.substring(j + 1);
        }
 
        function findAnagram(s)
        {
         
            // Copying our original string
            // for comparison
            let check = s.slice();
 
            // Declaring the two pointers
            let i = 0, j = s.length - 1;
 
            while (i < s.length && j >= 0) {
 
                // Checking the given condition
                if (s[i] != s[j] && check[i] != s[j]
                    && check[j] != s[i]) {
                    s = swap(s, i, j)
                    i++;
 
                    j = s.length - 1;
                }
                else {
                    j--;
                }
            }
 
            // When string length is odd
            if (s.length % 2 != 0) {
 
                // The mid element
                let mid = s.length / 2;
 
                // If the characters are the
                // same, then perform the swap
                // operation as illustrated
                if (check[mid] == s[mid]) {
                    for (let i = 0; i < s.length; i++) {
                        if (check[i] != s[mid]
                            && s[i] != s[mid]) {
                            s = swap(s, i, mid)
                            break;
                        }
                    }
                }
            }
             
            // Check if the corresponding indices
            // has the same character or not
            let ok = true;
            for (let i = 0; i < s.length; i++) {
                if (check[i] == s[i]) {
                    ok = false;
                    break;
                }
            }
 
            // If string follows required
            // condition
            if (ok)
                document.write(s);
            else
                document.write(-1);
        }
 
        // Driver Code
 
        let S = "geek";
        findAnagram(S);
 
// This code is contributed by Potta Lokesh
    

输出：

egke

时间复杂度： O(N*log N)
辅助空间： O(N)