每次字符替换查询后检查回文

给定一个字符串str和Q查询。每个查询包含一对整数 (i1, i2) 和一个字符“ch”。我们需要在指标I1和新的字符“频道” I2，然后告诉我们，如果字符串str是回文或不替换字符。 (0 <= i1, i2 < string_length)
例子：

Input : str = "geeks"  Q = 2
        query 1: i1 = 3 ,i2 = 0, ch = 'e'
        query 2: i1 = 0 ,i2 = 2, ch = 's'
Output : query 1: "NO"
         query 2: "NO"
Explanation :
        In query 1 : i1 = 3 , i2 = 0 ch = 'e'
                    After replacing char at index i1, i2
                    str[3] = 'e', str[0] = 'e'
                    string become "eeees" which is not
                    palindrome so output "NO"
        In query 2 : i1 = 0 i2 = 2  ch = 's'
                    After replacing char at index i1 , i2
                     str[0] = 's', str[2] = 's'
                    string become "sesks" which is
                    palindrome so output "NO"

Input : str = "jasonamat"  Q = 3
        query 1: i1 = 3, i2 = 8 ch = 'j'
        query 2: i1 = 2, i2 = 6 ch = 'n'
        query 3: i1 = 3, i2 = 7 ch = 'a'
Output :
       query 1: "NO"
       query 2: "NO"
       query 3: "YES"

一个简单的办法是，对于每个查询，我们在指标替代字符（I1和I2）有一个新的字符“CH”，然后检查是否字符串是回文与否。
以下是上述想法的实现

C++

// C++ program to find if string becomes palindrome
// after every query.
#include
using namespace std;
 
// Function to check if string is Palindrome or Not
bool IsPalindrome(string &str)
{
    int n = strlen(str);
    for (int i = 0; i < n/2 ; i++)
        if (str[i] != str[n-1-i])
            return false;
    return true;
}
 
// Takes two inputs for Q queries. For every query, it
// prints Yes if string becomes palindrome and No if not.
void Query(string &str, int Q)
{
    int i1, i2;
    char ch;
 
    // Process all queries one by one
    for (int q = 1 ; q <= Q ; q++ )
    {
        cin >> i1 >> i2 >> ch;
 
        // query 1: i1 = 3 ,i2 = 0, ch = 'e'
        // query 2: i1 = 0 ,i2 = 2 , ch = 's'
        // replace character at index i1 & i2 with new 'ch'
        str[i1] = str[i2] = ch;
 
        // check string is palindrome or not
        (isPalindrome(str)== true) ? cout << "YES" << endl :
                                     cout << "NO" << endl;
    }
}
 
// Driver program
int main()
{
    char str[] = "geeks";
    int Q = 2;
    Query(str, Q);
    return 0;
}

Python3

# Python3 program to find if
# string becomes palindrome
# after every query.
 
# Function to check if string
# is Palindrome or Not
def isPalindrome(string: list) -> bool:
    n = len(string)
    for i in range(n // 2):
        if string[i] != string[n - 1 - i]:
            return False
    return True
 
# Takes two inputs for Q queries.
# For every query, it prints Yes
# if string becomes palindrome
# and No if not.
def Query(string: list, Q: int) -> None:
 
    # Process all queries one by one
    for i in range(Q):
 
        # To get space separated
        # input from user
        inp = list(input().split())
 
        # parsing user inputs as integers
        # and strings/char
        i1 = int(inp[0])
        i2 = int(inp[1])
        ch = inp[2]
 
        # query 1: i1 = 3 ,i2 = 0, ch = 'e'
        # query 2: i1 = 0 ,i2 = 2 , ch = 's'
        # replace character at index
        # i1 & i2 with new 'ch'
        string[i1] = string[i2] = ch
 
        # check string is palindrome or not
        if isPalindrome(string):
            print("Yes")
        else:
            print("No")
 
# Driver Code
if __name__ == "__main__":
    string = list("geeks")
    Q = 2
    Query(string, Q)
 
# This code is contributed by
# sanjeev2552

输入：

3 0 e
0 2 s

输出：

"NO"
"YES"

时间复杂度 O(Q*n)（n 是字符串的长度）
一个有效的解决方案是使用散列。我们创建了一个空的哈希集，用于存储回文中不相等的索引（注意：“我们必须仅存储字符串的前半部分不相等的索引”）。

Given string "str" and length 'n'.
Create an empty set S and store unequal indexes in first half.
Do following for each query :
   1. First replace character at indexes i1 & i2 with 
      new char "ch"

   2. If i1 and/or i2 are/is greater than n/2 then convert 
      into first half index(es)

   3. In this step we make sure that S contains maintains 
      unequal indexes of first half.
      a) If str[i1] == str [n - 1 - i1] means i1 becomes 
         equal after replacement, remove it from S (if present)
         Else add i1 to S 
      b) Repeat step a) for i2 (replace i1 with i2)  

   4. If S is empty then string is palindrome else NOT

下面是上述想法的 C++ 实现

CPP

// C++/c program check if given string is palindrome
// or not after every query
#include
using namespace std;
 
// This function makes sure that set S contains
// unequal characters from first half. This is called
// for every character.
void addRemoveUnequal(string &str, int index, int n,
                              unordered_set &S)
{
    // If character becomes equal after query
    if (str[index] == str[n-1-index])
    {
        // Remove the current index from set if it
        // is present
        auto it = S.find(index);
        if (it != S.end())
            S.erase(it) ;
    }
 
    // If not equal after query, insert it into set
    else
        S.insert(index);
}
 
// Takes two inputs for Q queries. For every query, it
// prints Yes if string becomes palindrome and No if not.
void Query(string &str, int Q)
{
    int n = str.length();
 
    // create an empty set that store indexes of
    // unequal location in palindrome
    unordered_set S;
 
    // we store indexes that are unequal in palindrome
    // traverse only first half of string
    for (int i=0; i> i1 >> i2 >> ch;
 
        // Replace characters at indexes i1 & i2 with
        // new char 'ch'
        str[i1] = str [i2] = ch;
 
        // If i1 and/or i2 greater than n/2
        // then convert into first half index
        if (i1 > n/2)
            i1 = n- 1 -i1;
        if (i2 > n/2)
            i2 = n -1 - i2;
 
        // call addRemoveUnequal function to insert and remove
        // unequal indexes
        addRemoveUnequal(str, i1 , n, S );
        addRemoveUnequal(str, i2 , n, S );
 
        // if set is not empty then string is not palindrome
        S.empty()? cout << "YES\n" : cout << "NO\n";
    }
}
 
// Driver program
int main()
{
    string str = "geeks";
    int Q = 2 ;
    Query(str, Q);
    return 0;
}

输入：

3 0 e
0 2 s

输出：

"NO"
"YES"

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