We have discussed efficient implementation of k stack in an array. In this post, same for queue is discussed. Following is the detailed problem statement.

Create a data structure kQueues that represents k queues. Implementation of kQueues should use only one array, i.e., k queues should use the same array for storing elements. Following functions must be supported by kQueues.

*enqueue(int x, int qn) –> adds x to queue number ‘qn’ where qn is from 0 to k-1
dequeue(int qn) –> deletes an element from queue number ‘qn’ where qn is from 0 to k-1
*

**Method 1 (Divide the array in slots of size n/k)**

A simple way to implement k queues is to divide the array in k slots of size n/k each, and fix the slots for different queues, i.e., use arr[0] to arr[n/k-1] for first queue, and arr[n/k] to arr[2n/k-1] for queue2 where arr[] is the array to be used to implement two queues and size of array be n.

The problem with this method is inefficient use of array space. An enqueue operation may result in overflow even if there is space available in arr[]. For example, consider k as 2 and array size n as 6. Let we enqueue 3 elements to first and do not enqueue anything to second second queue. When we enqueue 4th element to first queue, there will be overflow even if we have space for 3 more elements in array.

**Method 2 (A space efficient implementation)**

The idea is similar to the stack post, here we need to use three extra arrays. In stack post, we needed to extra arrays, one more array is required because in queues, enqueue() and dequeue() operations are done at different ends.

Following are the three extra arrays are used:

1) **front[]**: This is of size k and stores indexes of front elements in all queues.

2) **rear[]**: This is of size k and stores indexes of rear elements in all queues.

2) **next[]**: This is of size n and stores indexes of next item for all items in array arr[].

Here arr[] is actual array that stores k stacks.

Together with k queues, a stack of free slots in arr[] is also maintained. The top of this stack is stored in a variable ‘free’.

All entries in front[] are initialized as -1 to indicate that all queues are empty. All entries next[i] are initialized as i+1 because all slots are free initially and pointing to next slot. Top of free stack, ‘free’ is initialized as 0.

Following is C++ implementation of the above idea.

// A C++ program to demonstrate implementation of k queues in a single // array in time and space efficient way #include<iostream> #include<climits> using namespace std; // A C++ class to represent k queues in a single array of size n class kQueues { int *arr; // Array of size n to store actual content to be stored in queue int *front; // Array of size k to store indexes of front elements of queue int *rear; // Array of size k to store indexes of rear elements of queue int *next; // Array of size n to store next entry in all queues // and free list int n, k; int free; // To store beginning index of free list public: //constructor to create k queue in an array of size n kQueues(int k, int n); // A utility function to check if there is space available bool isFull() { return (free == -1); } // To enqueue an item in queue number 'qn' where qn is from 0 to k-1 void enqueue(int item, int qn); // To dequeue an from queue number 'qn' where qn is from 0 to k-1 int dequeue(int qn); // To check whether queue number 'qn' is empty or not bool isEmpty(int qn) { return (front[qn] == -1); } }; // Constructor to create k queues in an array of size n kQueues::kQueues(int k1, int n1) { // Initialize n and k, and allocate memory for all arrays k = k1, n = n1; arr = new int[n]; front = new int[k]; rear = new int[k]; next = new int[n]; // Initialize all queues as empty for (int i = 0; i < k; i++) front[i] = -1; // Initialize all spaces as free free = 0; for (int i=0; i<n-1; i++) next[i] = i+1; next[n-1] = -1; // -1 is used to indicate end of free list } // To enqueue an item in queue number 'qn' where qn is from 0 to k-1 void kQueues::enqueue(int item, int qn) { // Overflow check if (isFull()) { cout << "\nQueue Overflow\n"; return; } int i = free; // Store index of first free slot // Update index of free slot to index of next slot in free list free = next[i]; if (isEmpty(qn)) front[qn] = i; else next[rear[qn]] = i; next[i] = -1; // Update next of rear and then rear for queue number 'qn' rear[qn] = i; // Put the item in array arr[i] = item; } // To dequeue an from queue number 'qn' where qn is from 0 to k-1 int kQueues::dequeue(int qn) { // Underflow checkSAS if (isEmpty(qn)) { cout << "\nQueue Underflow\n"; return INT_MAX; } // Find index of front item in queue number 'qn' int i = front[qn]; front[qn] = next[i]; // Change top to store next of previous top // Attach the previous front to the beginning of free list next[i] = free; free = i; // Return the previous front item return arr[i]; } /* Driver program to test kStacks class */ int main() { // Let us create 3 queue in an array of size 10 int k = 3, n = 10; kQueues ks(k, n); // Let us put some items in queue number 2 ks.enqueue(15, 2); ks.enqueue(45, 2); // Let us put some items in queue number 1 ks.enqueue(17, 1); ks.enqueue(49, 1); ks.enqueue(39, 1); // Let us put some items in queue number 0 ks.enqueue(11, 0); ks.enqueue(9, 0); ks.enqueue(7, 0); cout << "Dequeued element from queue 2 is " << ks.dequeue(2) << endl; cout << "Dequeued element from queue 1 is " << ks.dequeue(1) << endl; cout << "Dequeued element from queue 0 is " << ks.dequeue(0) << endl; return 0; }

Output:

Dequeued element from queue 2 is 15 Dequeued element from queue 1 is 17 Dequeued element from queue 0 is 11

Time complexities of enqueue() and dequeue() is O(1).

The best part of above implementation is, if there is a slot available in queue, then an item can be enqueued in any of the queues, i.e., no wastage of space. This method requires some extra space. Space may not be an issue because queue items are typically large, for example queues of employees, students, etc where every item is of hundreds of bytes. For such large queues, the extra space used is comparatively very less as we use three integer arrays as extra space.

This article is contributed by **Sachin**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above