最小化相同索引字符的交换，使两个字符串的字符的 ASCII 值之和为奇数

给定由小写字母组成的两个长度为N 的字符串S和T ，任务是最小化使两个字符串的字符的 ASCII 值之和为奇数所需的相同索引元素的交换次数。如果不可能使 ASCII 值的总和为奇数，则打印“-1” 。

例子：

Input:S = ”acd”, T = ”dbf”
Output: 1
Explanation:
Swapping S[1] and T[1] modifies S to “abd” and T to “dcf”.
Sum of ASCII value of characters of the string S = 97 + 98 + 100 = 297 (Odd).
Sum of ASCII value of characters of the string T = 100 + 99 + 102 = 301 (Odd).

Input: S = “aey”, T = “cgj”
Output: -1

处理方法：按照以下步骤解决问题：

• 计算字符串S和T的字符的 ASCII 值之和，分别存储在变量sum1和sum2 中。
• 如果sum1和sum2已经是奇数，则打印0 ，因为不需要交换。
• 如果sum1和sum2具有不同的奇偶校验，则打印-1 ，因为两个字符串的和不能具有相同的奇偶校验。
• 如果sum1和sum2都是偶数，则遍历给定的字符串S和T。如果存在具有奇数ASCII值的任何字符，这两个字符串的字符的ASCII值的总和可以通过奇数仅1交换制备。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 count the number of swaps
// required to make the sum of ASCII values
// of the characters of both strings odd
void countSwaps(string S, string T)
{
    // Initialize alphabets with value
    int value[26];
 
    // Initialize values for each
    // alphabet
    for (int i = 0; i < 26; i++)
        value[i] = i + 1;
 
    // Size of the string
    int N = S.size();
 
    // Sum of string S
    int sum1 = 0;
 
    // Sum of string T
    int sum2 = 0;
 
    // Stores whether there is any
    // index i such that S[i] and
    // T[i] have different parities
    bool flag = false;
 
    // Traverse the strings
    for (int i = 0; i < N; i++) {
 
        // Update sum1 and sum2
        sum1 += value[S[i] - 'a'];
        sum2 += value[T[i] - 'a'];
 
        // If S[i] and T[i] have
        // different parities
        if ((value[S[i] - 'a'] % 2 == 0
             && value[T[i] - 'a'] % 2 == 1)
 
            || (value[S[i] - 'a'] % 2 == 1
                && value[T[i] - 'a'] % 2 == 0))
 
            flag = false;
    }
 
    // If sum1 and sum2 are both odd
    if (sum1 % 2 == 1
        && sum2 % 2 == 1)
        cout << "0\n";
 
    // If sum1 and sum2 are both even
    else if (sum1 % 2 == 0
             && sum2 % 2 == 0) {
 
        // If exists print 1
        if (flag)
            cout << "1";
 
        // Otherwise
        else
            cout << "-1";
    }
 
    // If sum1 and sum2 are
    // of different parities
    else {
        cout << "-1";
    }
}
 
// Driver Code
int main()
{
    string S = "acd";
    string T = "dbf";
 
    // Function Call
    countSwaps(S, T);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
 
  // Function to count the number of swaps
  // required to make the sum of ASCII values
  // of the characters of both strings odd
  static void countSwaps(String S, String T)
  {
 
    // Initialize alphabets with value
    int[] value = new int[26];
 
    // Initialize values for each
    // alphabet
    for (int i = 0; i < 26; i++)
      value[i] = i + 1;
 
    // Size of the string
    int N = S.length();
 
    // Sum of string S
    int sum1 = 0;
 
    // Sum of string T
    int sum2 = 0;
 
    // Stores whether there is any
    // index i such that S[i] and
    // T[i] have different parities
    boolean flag = false;
 
    // Traverse the strings
    for (int i = 0; i < N; i++) {
 
      // Update sum1 and sum2
      sum1 += value[S.charAt(i) - 'a'];
      sum2 += value[T.charAt(i) - 'a'];
 
      // If S[i] and T[i] have
      // different parities
      if ((value[S.charAt(i) - 'a'] % 2 == 0
           && value[T.charAt(i) - 'a'] % 2 == 1)
 
          || (value[S.charAt(i) - 'a'] % 2 == 1
              && value[T.charAt(i) - 'a'] % 2 == 0))
 
        flag = false;
    }
 
    // If sum1 and sum2 are both odd
    if (sum1 % 2 == 1
        && sum2 % 2 == 1)
      System.out.println("0\n");
 
    // If sum1 and sum2 are both even
    else if (sum1 % 2 == 0
             && sum2 % 2 == 0) {
 
      // If exists print 1
      if (flag)
        System.out.println("1");
 
      // Otherwise
      else
        System.out.println("-1");
    }
 
    // If sum1 and sum2 are
    // of different parities
    else {
      System.out.println("-1");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    String S = "acd";
    String T = "dbf";
 
    // Function Call
    countSwaps(S, T);
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Python3

# Python3 program for the above approach
 
# Function to count the number of swaps
# required to make the sum of ASCII values
# of the characters of both strings odd
def countSwaps(S, T):
   
    # Initialize alphabets with value
    value = [0]*26
 
    # Initialize values for each
    # alphabet
    for i in range(26):
        value[i] = i + 1
 
    # Size of the string
    N = len(S)
 
    # Sum of S
    sum1 = 0
 
    # Sum of T
    sum2 = 0
 
    # Stores whether there is any
    # index i such that S[i] and
    # T[i] have different parities
    flag = False
 
    # Traverse the strings
    for i in range(N):
 
        # Update sum1 and sum2
        sum1 += value[ord(S[i]) - ord('a')]
        sum2 += value[ord(T[i]) - ord('a')]
 
        # If ord(S[i]) anord('a)rd(T[i]) haord('a)
        # different parities
        if (value[ord(S[i]) - ord('a')] % 2 == 0
            and value[ord(T[i]) - ord('a')] % 2 == 1
            or value[ord(S[i]) - ord('a')] % 2 == 1
            and value[ord(T[i]) - ord('a')] % 2 == 0):
             
            flag = False
 
    # If sum1 and sum2 are both odd
    if (sum1 % 2 == 1 and sum2 % 2 == 1):
        print("0")
 
    # If sum1 and sum2 are both even
    elif (sum1 % 2 == 0 and sum2 % 2 == 0):
 
        # If exists pr1
        if (flag):
            print("1")
 
        # Otherwise
        else:
            print("-1")
 
    # If sum1 and sum2 are
    # of different parities
    else:
        print("-1")
 
# Driver Code
if __name__ == '__main__':
    S = "acd"
    T = "dbf"
 
    # Function Call
    countSwaps(S, T)
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
 
public class GFG
{
   
// Function to count the number of swaps
// required to make the sum of ASCII values
// of the characters of both strings odd
static void countSwaps(string S, string T)
{
   
    // Initialize alphabets with value
    int[] value = new int[26];
 
    // Initialize values for each
    // alphabet
    for (int i = 0; i < 26; i++)
        value[i] = i + 1;
 
    // Size of the string
    int N = S.Length;
 
    // Sum of string S
    int sum1 = 0;
 
    // Sum of string T
    int sum2 = 0;
 
    // Stores whether there is any
    // index i such that S[i] and
    // T[i] have different parities
    bool flag = false;
 
    // Traverse the strings
    for (int i = 0; i < N; i++) {
 
        // Update sum1 and sum2
        sum1 += value[S[i] - 'a'];
        sum2 += value[T[i] - 'a'];
 
        // If S[i] and T[i] have
        // different parities
        if ((value[S[i] - 'a'] % 2 == 0
             && value[T[i] - 'a'] % 2 == 1)
 
            || (value[S[i] - 'a'] % 2 == 1
                && value[T[i] - 'a'] % 2 == 0))
 
            flag = false;
    }
 
    // If sum1 and sum2 are both odd
    if (sum1 % 2 == 1
        && sum2 % 2 == 1)
        Console.Write("0\n");
 
    // If sum1 and sum2 are both even
    else if (sum1 % 2 == 0
             && sum2 % 2 == 0) {
 
        // If exists print 1
        if (flag)
            Console.Write("1");
 
        // Otherwise
        else
            Console.Write("-1");
    }
 
    // If sum1 and sum2 are
    // of different parities
    else {
        Console.Write("-1");
    }
}
 
 
// Driver Code
public static void Main(String[] args)
{
    string S = "acd";
    string T = "dbf";
 
    // Function Call
    countSwaps(S, T);
}
}
 
// This code is contributed by code_hunt.

Javascript

      // JavaScript program for the above approach
 
      // Function to count the number of swaps
      // required to make the sum of ASCII values
      // of the characters of both strings odd
      function countSwaps(S, T)
      {
       
        // Initialize alphabets with value
        var value = [...Array(26)];
 
        // Initialize values for each
        // alphabet
        for (var i = 0; i < 26; i++) value[i] = i + 1;
 
        // Size of the string
        var N = S.length;
 
        // Sum of string S
        var sum1 = 0;
 
        // Sum of string T
        var sum2 = 0;
 
        // Stores whether there is any
        // index i such that S[i] and
        // T[i] have different parities
        var flag = false;
 
        // Traverse the strings
        for (var i = 0; i < N; i++) {
          // Update sum1 and sum2
          sum1 += value[S[i] - "a"];
          sum2 += value[T[i] - "a"];
 
          // If S[i] and T[i] have
          // different parities
          if (
            (value[S[i] - "a"] % 2 === 0 && value[T[i] - "a"] % 2 === 1) ||
            (value[S[i] - "a"] % 2 === 1 && value[T[i] - "a"] % 2 === 0)
          )
            flag = false;
        }
 
        // If sum1 and sum2 are both odd
        if (sum1 % 2 === 1 && sum2 % 2 === 1) document.write("0 <br>");
        // If sum1 and sum2 are both even
        else if (sum1 % 2 === 0 && sum2 % 2 === 0) {
          // If exists print 1
          if (flag) document.write("1");
          // Otherwise
          else document.write("-1");
        }
 
        // If sum1 and sum2 are
        // of different parities
        else {
          document.write("-1");
        }
      }
 
      // Driver Code
      var S = "aey";
      var T = "cgj";
       
      // Function Call
      countSwaps(S, T);
       
      // This code is contributed by rdtank.
    

输出：

-1

时间复杂度： O(N)
辅助空间： O(26)

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