检查是否可以通过交换二进制字符串中具有不相等字符的索引中的字符对来使字符串成为回文

给定一个长度为N的字符串S和一个二进制字符串B ，任务是检查给定的字符串S 是否可以通过在字符串B 中由不等字符组成的任何一对索引处重复交换字符来检查是否可以使给定的字符串S成为回文。

例子：

Input: S = “BAA”, B = “100”
Output: Yes
Explanation:
Swapping S[0] and S[1] modifies S to “ABA” and B to “010”.

Input: S = “ACABB”, B = “00000”
Output: No

方法：按照以下步骤解决此问题：

• 检查字符串S 是否可以重新排列以形成回文字符串。如果发现为false ，则打印“No” 。
• 否则，如果字符串S是回文，则打印“Yes” 。
• 如果0和1的计数至少为 1 ，那么总有一种方法可以交换字符以使给定的字符串S回文。因此，打印“是” 。否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Utility function to check if string
// S can be made palindromic or not
bool canBecomePalindromeUtil(string S, string B)
{
     
    // Stores the number of distinct
    // characters present in string S
    unordered_set set;
 
    // Traverse the characters of string S
    for(int i = 0; i < S.size(); i++)
    {
         
        // Insert current character in S
        set.insert(S[i]);
    }
 
    // Count frequency of each
    // character of string S
    map map;
 
    // Traverse the characters of string S
    for(int i = 0; i < S.length(); i++)
    {
        map[S[i]] += 1;
    }
 
    bool flag = false;
 
    // Check for the odd length string
    if (S.size() % 2 == 1)
    {
         
        // Stores the count of
        // even and odd frequenct
        // characters in the string S
        int count1 = 0, count2 = 0;
 
        for(auto e : map)
        {
            if (e.second % 2 == 1)
            {
                 
                // Update the count of
                // odd frequent characters
                count2++;
            }
            else
            {
                 
                // Update the count of
                // even frequent characters
                count1++;
            }
        }
 
        // If the conditions satisfies
        if (count1 == set.size() - 1 && count2 == 1)
        {
            flag = true;
        }
    }
 
    // Check for even length string
    else
    {
         
        // Stores the frequency of
        // even and odd characters
        // in the string S
        int count1 = 0, count2 = 0;
 
        for(auto e : map)
        {
            if (e.second % 2 == 1)
            {
                 
                // Update the count of
                // odd frequent characters
                count2++;
            }
            else
            {
 
                // Update the count of
                // even frequent characters
                count1++;
            }
        }
 
        // If the condition satisfies
        if (count1 == set.size() && count2 == 0)
        {
            flag = true;
        }
    }
 
    // If a palindromic string
    // cannot be formed
    if (!flag)
    {
        return false;
    }
 
    else
    {
 
        // Check if there is
        // atleast one '1' and '0'
        int count1 = 0, count0 = 0;
        for(int i = 0; i < B.size(); i++)
        {
 
            // If current character is '1'
            if (B[i] == '1')
            {
                count1++;
            }
            else
            {
                count0++;
            }
        }
 
        // If atleast one '1' and '0' is present
        if (count1 >= 1 && count0 >= 1)
        {
            return true;
        }
 
        else
        {
            return false;
        }
    }
}
 
// Function to determine whether
// string S can be converted to
// a palindromic string or not
void canBecomePalindrome(string S, string B)
{
    if (canBecomePalindromeUtil(S, B))
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver code
int main()
{
    string S = "ACABB";
    string B = "00010";
    canBecomePalindrome(S, B);
 
    return 0;
}
 
// This code is contributed by Kingash

Java

// Java program for the above approach
 
import java.io.*;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
 
class GFG {
 
    // Utility function to check if string
    // S can be made palindromic or not
    public static boolean
    canBecomePalindromeUtil(String S,
                            String B)
    {
        // Stores the number of distinct
        // characters present in string S
        HashSet set = new HashSet<>();
 
        // Traverse the characters of string S
        for (int i = 0; i < S.length(); i++) {
 
            // Insert current character in S
            set.add(S.charAt(i));
        }
 
        // Count frequency of each
        // character of string S
        HashMap map
            = new HashMap<>();
 
        // Traverse the characters of string S
        for (int i = 0; i < S.length(); i++) {
            Integer k = map.get(S.charAt(i));
            map.put(S.charAt(i),
                    (k == null) ? 1 : k + 1);
        }
 
        boolean flag = false;
 
        // Check for the odd length string
        if (S.length() % 2 == 1) {
 
            // Stores the count of
            // even and odd frequenct
            // characters in the string S
            int count1 = 0, count2 = 0;
 
            for (Map.Entry e :
                 map.entrySet()) {
 
                if (e.getValue() % 2 == 1) {
 
                    // Update the count of
                    // odd frequent characters
                    count2++;
                }
                else {
 
                    // Update the count of
                    // even frequent characters
                    count1++;
                }
            }
 
            // If the conditions satisfies
            if (count1 == set.size() - 1
                && count2 == 1) {
                flag = true;
            }
        }
 
        // Check for even length string
        else {
 
            // Stores the frequency of
            // even and odd characters
            // in the string S
            int count1 = 0, count2 = 0;
 
            for (Map.Entry e :
                 map.entrySet()) {
 
                if (e.getValue() % 2 == 1) {
 
                    // Update the count of
                    // odd frequent characters
                    count2++;
                }
                else {
 
                    // Update the count of
                    // even frequent characters
                    count1++;
                }
            }
 
            // If the condition satisfies
            if (count1 == set.size()
                && count2 == 0) {
                flag = true;
            }
        }
 
        // If a palindromic string
        // cannot be formed
        if (!flag) {
            return false;
        }
 
        else {
 
            // Check if there is
            // atleast one '1' and '0'
            int count1 = 0, count0 = 0;
            for (int i = 0;
                 i < B.length(); i++) {
 
                // If current character is '1'
                if (B.charAt(i) == '1') {
                    count1++;
                }
                else {
                    count0++;
                }
            }
 
            // If atleast one '1' and '0' is present
            if (count1 >= 1 && count0 >= 1) {
                return true;
            }
 
            else {
                return false;
            }
        }
    }
 
    // Function to determine whether
    // string S can be converted to
    // a palindromic string or not
    public static void
    canBecomePalindrome(String S,
                        String B)
    {
        if (canBecomePalindromeUtil(S, B))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "ACABB";
        String B = "00010";
        canBecomePalindrome(S, B);
    }
}

Python3

# Python3 program for the above approach
 
# Utility function to check if string
# S can be made palindromic or not
def canBecomePalindromeUtil(S, B):
     
    # Stores the number of distinct
    # characters present in string S
    s = set(S)
 
    # Count frequency of each
    # character of string S
    map = {}
 
    # Traverse the characters of string S
    for i in range(len(S)):
        if S[i] in map:
            map[S[i]] += 1
        else:
            map[S[i]] = 1
 
    flag = False
 
    # Check for the odd length string
    if (len(S) % 2 == 1):
         
        # Stores the count of
        # even and odd frequenct
        # characters in the string S
        count1 = 0
        count2 = 0
 
        for e in map:
            if (map[e] % 2 == 1):
                 
                # Update the count of
                # odd frequent characters
                count2 += 1
                 
            else:
                 
                # Update the count of
                # even frequent characters
                count1 += 1
 
        # If the conditions satisfies
        if (count1 == len(s) - 1 and
            count2 == 1):
            flag = True
 
    # Check for even length string
    else:
         
        # Stores the frequency of
        # even and odd characters
        # in the string S
        count1 = 0
        count2 = 0
 
        for e in map:
            if (map[e] % 2 == 1):
                 
                # Update the count of
                # odd frequent characters
                count2 += 1
                 
            else:
 
                # Update the count of
                # even frequent characters
                count1 += 1
 
        # If the condition satisfies
        if (count1 == len(s) and count2 == 0):
            flag = True
 
    # If a palindromic string
    # cannot be formed
    if (not flag):
        return False
 
    else:
 
        # Check if there is
        # atleast one '1' and '0'
        count1 = 0
        count0 = 0
        for i in range(len(B)):
 
            # If current character is '1'
            if (B[i] == '1'):
                count1 += 1
            else:
                count0 += 1
 
        # If atleast one '1' and '0' is present
        if (count1 >= 1 and count0 >= 1):
            return True
        else:
            return False
       
# Function to determine whether
# string S can be converted to
# a palindromic string or not
def canBecomePalindrome(S, B):
 
    if (canBecomePalindromeUtil(S, B)):
        print("Yes")
    else:
        print("No")
 
# Driver code
if __name__ == "__main__":
 
    S = "ACABB"
    B = "00010"
     
    canBecomePalindrome(S, B)
 
# This code is contributed by AnkThon

Javascript

 
// Javascript program for the above approach
 
// Utility function to check if string
// S can be made palindromic or not
function canBecomePalindromeUtil(S, B)
{
     
    // Stores the number of distinct
    // characters present in string S
    var st = new Set()
    var i;
     
    // Traverse the characters of string S
    for(i = 0; i < S.length; i++)
    {
         
        // Insert current character in S
        st.add(S[i]);
    }
 
    // Count frequency of each
    // character of string S
    var mp = new Map();
 
    // Traverse the characters of string S
    for(i = 0; i < S.length; i++)
    {
        if (mp.has(S[i]))
            mp.set(S[i], mp.get(S[i]))
        else
            mp.set(S[i], 1);
    }
 
    var flag = false;
 
    // Check for the odd length string
    if (S.length % 2 == 1)
    {
         
        // Stores the count of
        // even and odd frequenct
        // characters in the string S
        var count1 = 0, count2 = 0;
        for(const [key, value] of Object.entries(mp))
        {
            if (value % 2 == 1)
            {
                 
                // Update the count of
                // odd frequent characters
                count2++;
            }
            else
            {
                 
                // Update the count of
                // even frequent characters
                count1++;
            }
         }
 
        // If the conditions satisfies
        if (count1 == st.size - 1 && count2 == 1)
        {
            flag = true;
        }
    }
 
    // Check for even length string
    else
    {
         
        // Stores the frequency of
        // even and odd characters
        // in the string S
        var count1 = 0, count2 = 0;
 
        for (const [key, value] of Object.entries(mp))
        {
            if (value % 2 == 1)
            {
                 
                // Update the count of
                // odd frequent characters
                count2++;
            }
            else
            {
 
                // Update the count of
                // even frequent characters
                count1++;
            }
        }
 
        // If the condition satisfies
        if (count1 == st.size && count2 == 0)
        {
            flag = true;
        }
    }
 
    // If a palindromic string
    // cannot be formed
    if (!flag)
    {
        return false;
    }
 
    else
    {
         
        // Check if there is
        // atleast one '1' and '0'
        var count1 = 0, count0 = 0;
        for(i = 0; i < B.length; i++)
        {
 
            // If current character is '1'
            if (B[i] == '1')
            {
                count1++;
            }
            else
            {
                count0++;
            }
        }
 
        // If atleast one '1' and '0' is present
        if (count1 >= 1 && count0 >= 1)
        {
            return true;
        }
 
        else
        {
            return false;
        }
    }
}
 
// Function to determine whether
// string S can be converted to
// a palindromic string or not
function canBecomePalindrome(S, B)
{
    if (canBecomePalindromeUtil(S, B))
        document.write("No");
    else
        document.write("Yes");
}
 
// Driver code
var S = "ACABB";
var B = "00010";
 
canBecomePalindrome(S, B);
     
// This code is contributed by SURENDRA_GANGWAR
 

输出：

Yes

时间复杂度： O(N)
辅助空间： O(N)

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