1700. Number of Students Unable to Eat Lunch

Time: O(n)

Space: O(1)

Intuition

Count students who cannot get their sandwich preference. After sandwiches in queue are exhausted for students wanting front sandwich, simulation ends.

Algorithm

1Count preference[0] and preference[1] students.
2Process sandwiches in order. If top sandwich matches count > 0: count--. Else: if both counts > 0, cycle. Return remaining students.
3Simplified: remaining = students who want opposite of remaining sandwiches.

Common Pitfalls

•When count for current top sandwich is 0 and count for other is >0: all remaining students are unsatisfied.

1700.cs

// Approach: Count student preferences; simulate serving sandwiches top-of-stack; stop when no match possible.
// Time: O(n) Space: O(1)

public class Solution
{
    public int CountStudents(int[] students, int[] sandwiches)
    {
        int[] count = new int[2];

        foreach (int student in students)
            ++count[student];

        for (int i = 0; i < sandwiches.Length; ++i)
        {
            if (count[sandwiches[i]] == 0)
                return sandwiches.Length - i;
            --count[sandwiches[i]];
        }

        return 0;
    }
}

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