Runway reservation system

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Runway reservation system

Given an airport with a single runway, we have to design a runway reservation system of that airport. Add to details: each reservation request comes with requested landing time let’s say t. Landing can go through if there is no landing scheduled within k minutes of requested time, that means t can be added to the set of scheduling landings. K can vary and depends on external conditions. This system helps with reservations for the future landings.
Once the plane lands safely, we have to remove the plane for landing sets.

This is perfectly possible with airports with multiple runways where only one runway can be used because of weather conditions, maintenance etc. Also, one landing cannot follow another immediately due safety reasons, that’s why there has to be some minimum time before another landing takes place. We have to build this system with given constraints.

In nutshell, we have create a set of landings which are compatible with each other, i.e they do not violate the constraint put.  There are two operations to be performed on this set : insertion and removal. Insertion involves checking of constraints. 

Example: let’s say below is the timeline of all the landing currently scheduled and k = 3 min

reservation system

Now, if the new request comes for landing at 48.5, it should be added to the set as it does not violate the constraint of k mins. However, if a request comes for landing at 53, it cannot be added as it violates the constraint.
If a new request comes for 35, it is invalid as the request is in past.

Reservation system: thoughts

What is the most brute force solution which comes to mind? We can store all the incoming requests in an unsorted array.  Insertion should be O(1) operation as it is at the end. However, checking the constrain that it is satisfied will take O(n) because we have to scan through the entire array.
Same is true even if we use unsorted linked list. 

How about a sorted array? We can use a binary search algorithm to find the constraint, which will take O(log n) complexity. However, the insertion will still be O(n) as we have to move all the elements to the right from the position of insertion.

Sorted list solves the problem of insertion in O(1) but then search for the constraint will be O(n) complexity. It just moves the problem from one place to another.

Reservation system using binary search tree

We have to optimize two things : first check if the new request meets the constraints, second insert the new request into the set. 

Let’s think of binary search tree. To check a constraint, we have to check each node of the binary tree, but based on the relationship of the current node and the new request time, we can discard half of the tree. (Binary search tree property, where all the nodes on the left side are smaller than the root node and all the nodes on the right side are greater than the current node)

When a new request comes, we check with the root node and it does not violate the constraints, then we check if the requested time is less than the root. If yes, we go to left subtree and check there. If requested landing time is greater than the root node, we go to right subtree.
When we reach the leaf node, we add a new node with new landing time as the value of that node.

If at any given node, the constraint is violated, i.e not new landing time is within k minutes of time in the node, then we just return stating it is not possible to add new landing.

What will be complexity of checking the constraint? It will be O(h) where h is height of binary search tree. Insert is then O(1) operation.

Reservation system implementation

struct node{
    int value;
    struct node *left;
    struct node *right;
typedef struct node Node;

void inoderTraversal(Node * root){
    if(!root) return;
    printf("%d ", root->value);

Node *createNode(int value){
    Node * newNode =  (Node *)malloc(sizeof(Node));
    newNode->value = value;
    newNode->right= NULL;
    newNode->left = NULL;
    return newNode;

Node *addNode(Node *node, int value, int K){
        return createNode(value);
    if ( node->value + K > value && node->value - K < value ){
        return node;
    if (node->value > value)
        node->left = addNode(node->left, value, K);
        node->right = addNode(node->right, value, K);
    return node;

/* Driver program for the function written above */
int main(){
    Node *root = NULL;
    //Creating a binary tree
    root = addNode(root, 30, 3);
    root = addNode(root, 20, 3);
    root = addNode(root, 15, 3);
    root = addNode(root, 25, 3);
    root = addNode(root, 40, 3);
    root = addNode(root, 38, 3);
    root = addNode(root, 45, 3);
    return 0;

Let’s say a new requirement comes which is to find how many flights are scheduled till time t?

This problem can easily be solved using binary search tree, by keeping track of size of subtree at each node as shown in figure below.

runway reservation system

While inserting a new node, update counter of all the nodes on the nodes. When query is run, just return the count of node of that time or closest and smaller than that value of t. 

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