How do I solve this airplane seating problem in c/c++

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An airlines company has several planes of the same type.Each plane has a seating capacity of 24 X 3 rows and 8 seats in each row split has shown.

1 [a] [b] [c] [d] [e] [f] [g] [h]
2 [a] [b] [c] [d] [e] [f] [g] [h]
3 [a] [b] [c] [d] [e] [f] [g] [h]

If 4 people book - allocate 4 seats in middle row.Else 2 on the right and 2 on the left.

if 3 people book - id middle section is empty,allocate there continuously.Else go to next row middle section.

if 2 people book - allocate the edge seats.

if 1 person has booked - then allocate whatever free seat available.

eg-
input 4
output-1c 1d 1e 1f

input-3
output- 2c 2d 2e

input-2
output-1a 1b

This was asked to me in an interview. My logic doesn't work for all cases. Any help would be appreciated.

Posted 23-Aug-15 4:18am

Member 11929161

Updated 23-Aug-15 5:26am

PIEBALDconsult

v3

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Comments

PIEBALDconsult 23-Aug-15 11:40am

Just to be clear -- this plane has three decks with identical seating on all three?

Sergey Alexandrovich Kryukov 23-Aug-15 12:27pm

What have you tried so far?
—SA

PIEBALDconsult 23-Aug-15 13:22pm

See also, bin packing and knapsack problems.

2 solutions

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CPallini · Answer 1 · 2015-08-23T04:42:00

Solution 1

It doesn't look that difficult: The requirements already define the algorithm (that uses independent allocation strategies).
At each input, based on the number of passengers, you just have to find the first fit, considering only available seats. What is your doubt about?

Posted 23-Aug-15 4:42am

CPallini

Comments

Sergey Alexandrovich Kryukov 23-Aug-15 12:26pm

5ed.
—SA

CPallini 23-Aug-15 13:10pm

Thank you.

Member 16078047 · Answer 2 · 2023-08-24T03:11:00

JavaScript

const seatingChart = [
    ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'],
    ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'],
    ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'],
];

function allocateSeats(numPeople) {
    //X denotes booked seat
    if (numPeople === 4) {
        // Allocate 4 seats in the middle row or edge seats.
        for (let index = 0;index < seatingChart.length;index++) {
            if (seatingChart[index][2] !== 'X' && seatingChart[index][3] !== 'X' && seatingChart[index][4] !== 'X' && seatingChart[index][5] !== 'X') {
                seatingChart[index][2] = 'X';
                seatingChart[index][3] = 'X';
                seatingChart[index][4] = 'X';
                seatingChart[index][5] = 'X';
                break;
            } else if (seatingChart[index][0] !== 'X' && seatingChart[index][1] !== 'X' && seatingChart[index][6] !== 'X' && seatingChart[index][7] !== 'X') {
                seatingChart[index][0] = 'X';
                seatingChart[index][1] = 'X';
                seatingChart[index][6] = 'X';
                seatingChart[index][7] = 'X';
                break;
            }
        }
    }
    if (numPeople === 3) {
        for (let index = 0;index < seatingChart.length;index++) {
            // If the middle section is empty, allocate there continuously.
            if (seatingChart[index][2] !== 'X' && seatingChart[index][3] !== 'X' && seatingChart[index][4] !== 'X') {
                seatingChart[index][2] = 'X';
                seatingChart[index][3] = 'X';
                seatingChart[index][4] = 'X';
                break;
            }
        }
    }

    if (numPeople === 2) {
        // Allocate the edge seats.
        for (let index = 0;index < seatingChart.length;index++) {
            if (seatingChart[index][0] !== 'X' && seatingChart[index][1] !== 'X') {
                seatingChart[index][0] = 'X';
                seatingChart[index][1] = 'X';
                break;
            } else if (seatingChart[index][6] !== 'X' && seatingChart[index][7] !== 'X') {
                seatingChart[index][6] = 'X';
                seatingChart[index][7] = 'X';
                break;
            }
        }
    }

    if (numPeople === 1) {
        seatingChart.filter(i => { debugger; });
        let isBreak = false;
        // Allocate whatever free seat is available.
        for (let i = 0;i < seatingChart.length;i++) {
            for (let j = 0;j < seatingChart[i].length;j++) {
                if (seatingChart[i][j] !== 'X') {
                    seatingChart[i][j] = 'X';
                    isBreak = true;
                    break;
                }
            }
            if (isBreak) {
                break;
            }
        }
    }

    console.log(seatingChart);
}

Here's a solution in JS.