将给定的二进制字符串转换为另一个二进制字符串，最少操作数为翻转除一个以外的所有位

str1 = "001001111", str2 = "011111000";

3

 str1 = "0000", str2 = "1111";

 str1 = "0111", str2 = "1000";

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

// counting 01 and 10 pairs
int totalOperations(string str1, string str2) {
    int len = str1.size();
    int count10 = 0, count01 = 0;
    for (int p = 0; p < len; p++) {
        // If characters at p index are not same
        if (str1[p] != str2[p]) {
            // Increase count01 if 0(str1)-1(str2), else count10 if 1(str1)-0(str2)
            if (str1[p] == '0')
                count01++;
            else
                count10++;
        }
    }
    // If we have euqal number of 01 and 10 pairs, we need 2 operations to flip one pair.
    if (count01 == count10)
        return 2 * count01;
    return -1;
}
bool StringModification(string &temp1, string &temp2, char ch) {
    int len = temp1.size();
    int index = -1;
    // Find the pair of 1c character. (1 -> temp1, c -> temp2)
    for (int p = 0; p < len; p++) {
        if (temp1[p] == '1' && temp2[p] == ch) {
            index = p;
            break;
        }
    }
    // return 0 if pair is not found
    if (index == -1)
        return false;
    // Flip other characters in both strings
    for (int p = 0; p < len; p++) {
        if (p != index) {
            if (temp1[p] == '1')
                temp1[p] = '0';
            else
                temp1[p] = '1';
        }
    }
    return true;
}
// finding minimum operations
int minimumOperations(string str1, string str2) {
    int ans = INT_MAX;
    // first case with initial strings
    int operation1 = totalOperations(str1, str2);
    if (operation1 != -1)
        ans = min(ans, operation1);
    string temp1 = str1, temp2 = str2;
    // Case 2, modification for 11 pair
    if (StringModification(temp1, temp2, '1')) {
        // get operations after modification
        int operation2 = totalOperations(temp1, temp2);
        // adding 1 to operation2 as we have done one modification initially
        if (operation2 != -1)
            ans = min(ans, 1 + operation2);
    }
    // Case 3 modification for 10 pair
    temp1 = str1, temp2 = str2;
    if (StringModification(temp1, temp2, '0')) {
        int operation3 = totalOperations(temp1, temp2);
        if (operation3 != -1)
            ans = min(ans, 1 + operation3);
    }
    if (ans == INT_MAX)
        return -1;
    else
        return ans;
}
int main() {
    string str1 = "001001111";
    string str2 = "011111000";
    int ans = minimumOperations(str1, str2);
    if (ans == -1){
        cout << "S1 to S2 conversion is not possible";
    }
    else{
        cout << "Minimum number of operations required are: " << ans << "\n";
    }
    return 0;
}

Minimum number of operations required are: 3