Code Logo

Count Empty Seats

Published at19 Apr 2026
Created by
Loops & Iteration Easy 10 views
Like0

Given an array representing seats in a venue where 0 means empty and 1 means occupied, count how many seats are empty. Return the total number of empty seats.

For example, seats [0, 1, 0, 0, 1] has 3 empty seats. All occupied [1, 1, 1] has 0 empty seats. All empty [0, 0, 0] has 3 empty seats. An empty array returns 0.

Counting elements that match a value is a fundamental aggregation pattern. It applies to inventory counting, survey response tallying, error log analysis, and many data processing tasks.

The solution initializes a counter to 0 and iterates through each seat value. Whenever the value is 0, increment the counter. After the loop, return the total count.

Edge cases include an empty array (return 0), all seats full (return 0), and all seats empty (count equals array length).

Learn about our pseudocode specification
Guide

Example Input & Output

Example 1
Input
seats = [1,0,1,0,0]
Output
3
Explanation

There are three empty seats in the row.

Example 2
Input
seats = [1,1,1]
Output
0
Explanation

All seats are occupied, so none are empty.

Example 3
Input
seats = []
Output
0
Explanation

With no seats in the list, the empty count stays 0.

Algorithm Flow

Recommendation Algorithm Flow for Count Empty Seats

Solution Approach

Iterate through the array and count elements equal to 0.

function countEmpty(seats)
  count = 0
  for each s in seats
    if s == 0 then count = count + 1
  return count

Initialize count to 0. Loop through each seat value. If the value is 0 (meaning the seat is empty), increment the counter. After the loop finishes, return the total count of empty seats found. This works correctly for empty arrays because the loop body never executes and count remains 0.

Time complexity is O(n), space complexity is O(1).

Best Answers

Pseudocode - Approach 1
program count_empty_seats
dictionary
   seats: array[1..100] of integer
   empty_count, i, n: integer
algorithm
   input(seats)
   empty_count <- 0
   n <- seats.length
   for i <- 0 to n - 1 do
      if seats[i] = 0 then
         empty_count <- empty_count + 1
      endif
   endfor
   output(empty_count)
endprogram