将相同索引字符的交换次数最小化，使得两个字符串中字符的ASCII值之和为奇数

#include <bits/stdc++.h>
using namespace std;

int minSwaps(string str1, string str2) {
   int len = str1.length();
   int ascii_sum1 = 0, ascii_sum2 = 0;
   
   for (int i = 0; i < len; i++) {
      ascii_sum1 += str1[i];
      ascii_sum2 += str2[i];
   }
   
   // If total sum is odd, it's impossible to have both sums odd
   if ((ascii_sum1 + ascii_sum2) % 2 != 0) return -1;
   
   // If both sums are odd already, no swaps are needed
   if (ascii_sum1 % 2 != 0 && ascii_sum2 % 2 != 0) return 0;
   
   // If both sums are even, we just need to make one of them odd
   if (ascii_sum1 % 2 == 0 && ascii_sum2 % 2 == 0) {
      for (int i = 0; i < len; i++) {
         // If we find an odd character in str1 and an even character in str2, or vice versa, swap them
         if ((str1[i] - '0') % 2 != (str2[i] - '0') % 2) return 1;
      }
   }

   // If we reach here, it means no eligible swaps were found
    return -1;
}

int main() {
   string str1 = "abc";
   string str2 = "def";
   
   int result = minSwaps(str1, str2);
   if(result == -1) {
      cout << "No valid swaps found.\n";
   } else {
      cout << "Minimum swaps required: " << result << endl;
   }
   
   return 0;
}

No valid swaps found.

str1 = "abc", str2 = "def"