Easy
Intuition: The algorithm iterates through each student\'s preference and counts the number of students preferring circular sandwiches (count0) and square sandwi
- **Intuition:** The algorithm iterates through each student\'s preference and counts the number of students preferring circular sandwiches (`count0`) and square sandwiches (`count1`). Then, it iterates through the sandwiches and simulates the process of serving sandwiches to students. If there are still students preferring the current type of sandwich on top of the stack, the algorithm lets them take it. If no students prefer the current type of sandwich on top of the stack, it returns the count of students unable to eat.\n\n- **Approach:** The approach is to maintain counts of students preferring each type of sandwich and simulate the serving process by iterating through the sandwiches. At each step, it checks if there are still students preferring the current type of sandwich on top of the stack. If not, it returns the count of students unable to eat.\n\n- **Time complexity:** O(n), where n is the number of students or sandwiches. The algorithm iterates through the students and sandwiches once.\n\n- **Space complexity:** O(1), the algorithm uses only a constant amount of extra space for counting.\n\n# Code\n```c++ []\nclass Solution {\npublic:\n    int countStudents(vector<int>& students, vector<int>& sandwiches) {\n        int count1 = 0, count0 = 0;\n        \n        // Count the number of students preferring each type of sandwich\n        for (int x : students) {\n            if (x)\n                count1++;\n            else\n                count0++;\n        }\n\n        // Iterate through the sandwiches\n        for (int x : sandwiches) {\n            // If the sandwich on top is preferred by a student\n            if (x) {\n                // If there are still students who prefer this type of sandwich\n                if (count1 > 0) {\n                    count1--; // Reduce the count of students preferring this type of sandwich\n                } else {\n                    // If no students prefer this type of sandwich, return the count of students unable to eat\n                    return count0;\n                }\n            } else {\n                // If the sandwich on top is not preferred by a student\n                // If there are still students who prefer this type of sandwich\n                if (count0 > 0) {\n                    count0--; // Reduce the count of students preferring this type of sandwich\n                } else {\n                    // If no students prefer this type of sandwich, return the count of students unable to eat\n                    return count1;\n                }\n            }\n        }\n        return 0; // If all students can eat, return 0\n    }\n};\n\n```\n
['C++']