Last Updated : 21 Jul, 2024
Comments
Improve
We have discussed space-efficient implementation of 2 stacks in a single array. In this post, a general solution for k stacks is discussed. Following is the detailed problem statement. Create a data structure kStacks that represents k stacks. Implementation of kStacks should use only one array, i.e., k stacks should use the same array for storing elements.
The following functions must be supported by k Stacks. push(int x, int sn) –> pushes x to stack number ‘sn’ where sn is from 0 to k-1 pop(int sn) –> pops an element from stack number ‘sn’ where sn is from 0 to k-1
Method 1 (Divide the array in slots of size n/k) :
A simple way to implement k stacks is to divide the array in k slots of size n/k each, and fix the slots for different stacks, i.e., use arr[0] to arr[n/k-1] for first stack, and arr[n/k] to arr[2n/k-1] for stack2 where arr[] is the array to be used to implement two stacks and size of array be n. The problem with this method is inefficient use of array space. A stack push operation may result in stack overflow even if there is space available in arr[]. For example, say the k is 2 and array size (n) is 6 and we push 3 elements to first and do not push anything to second stack. When we push 4th element to first, there will be overflow even if we have space for 3 more elements in array.
Following is the implementation of the above idea.
#include <iostream>#include <vector>#include <stdexcept>class KStacks {private: int n; // Total size of array int k; // Number of stacks std::vector<int> arr; // Array to hold the k stacks std::vector<int> top; // Array to hold the indices of the top elements of the k stacks std::vector<int> next; // Array to hold the next entry in all stacks and free list int free; // To store the beginning index of the free listpublic: KStacks(int k, int n) : k(k), n(n), arr(n), top(k, -1), next(n) { // Initialize all spaces as free for (int i = 0; i < n - 1; ++i) { next[i] = i + 1; } next[n - 1] = -1; // -1 indicates end of free list free = 0; // First free index } // Check if there is space available bool isFull() { return free == -1; } // Check if a stack is empty bool isEmpty(int sn) { return top[sn] == -1; } // Push an item to stack number 'sn' void push(int item, int sn) { if (isFull()) { throw std::overflow_error("Stack Overflow"); } int i = free; // Get the first free index // Update free to index of the next slot in free list free = next[i]; // Update next of i to current top of stack number 'sn' next[i] = top[sn]; // Update top to new value top[sn] = i; // Put the item in array arr[i] = item; } // Pop an item from stack number 'sn' int pop(int sn) { if (isEmpty(sn)) { throw std::underflow_error("Stack Underflow"); } // Find index of top item in stack number 'sn' int i = top[sn]; // Update top to previous node in stack number 'sn' top[sn] = next[i]; // Add this slot to free list next[i] = free; free = i; // Return the previous top item return arr[i]; } // Peek at the top item of stack number 'sn' int peek(int sn) { if (isEmpty(sn)) { throw std::underflow_error("Stack Underflow"); } return arr[top[sn]]; }};int main() { int k = 3, n = 10; KStacks ks(k, n); // Let us put some items in stack number 2 ks.push(15, 2); ks.push(45, 2); // Let us put some items in stack number 1 ks.push(17, 1); ks.push(49, 1); ks.push(39, 1); // Let us put some items in stack number 0 ks.push(11, 0); ks.push(9, 0); ks.push(7, 0); std::cout << "Popped element from stack 2 is " << ks.pop(2) << std::endl; std::cout << "Popped element from stack 1 is " << ks.pop(1) << std::endl; std::cout << "Popped element from stack 0 is " << ks.pop(0) << std::endl; return 0;}
Output
Popped element from stack 2 is 45Popped element from stack 1 is 39Popped element from stack 0 is 7
Time Complexity: O(1), as each operation (push, pop, peek) is performed in constant time.
Auxiliary Space: O(n + k), as the space required is linear with respect to the total number of elements (n) and the number of stacks (k).
Method 2 (A space-efficient implementation) :
The idea is to use two extra arrays for efficient implementation of k stacks in an array. This may not make much sense for integer stacks, but stack items can be large for example stacks of employees, students, etc where every item is of hundreds of bytes. For such large stacks, the extra space used is comparatively very less as we use two integer arrays as extra space.
Following are the two extra arrays are used:
1) top[]: This is of size k and stores indexes of top elements in all stacks.
2) next[]: This is of size n and stores indexes of next item for the items in array arr[].
Here arr[] is actual array that stores k stacks. Together with k stacks, a stack of free slots in arr[] is also maintained. The top of this stack is stored in a variable ‘free’. All entries in top[] are initialized as -1 to indicate that all stacks 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.
Algorithm:
- Initialize an array of size k to keep track of the top element of each stack.
- Initialize an array next of size n, where n is the total size of the array that will hold the k stacks. Set the value of next[i] to i+1 for all 0 ? i < n-1, and next[n-1] to -1. This array will be used to keep track of the next element in the stack.
- Initialize an array top of size k to store the index of the top element of each stack. Set the value of top[i] to -1 for all 0 ? i < k.
- To push an element onto the i-th stack, do the following:
- Check if the array is full by checking if next[0] is -1. If it is, return an error message indicating that the stack is full.
- Set the value of next[0] to top[i].
- Set the value of top[i] to 0.
- Set the value of next[top[i]] to the new element’s index.
- Increment the value of top[i] by the block size.
- To pop an element from the i-th stack, do the following:
- Check if the stack is empty by checking if top[i] is -1. If it is, return an error message indicating that the stack is empty.
- Decrement the value of top[i] by the block size.
- Set the value of next[top[i]] to -1.
- Return the element at the index top[i] + block size – 1.
Following is the implementation of the above idea.
// A C++ program to demonstrate implementation of k stacks in a single // array in time and space efficient way#include<bits/stdc++.h>using namespace std;// A C++ class to represent k stacks in a single array of size nclass kStacks{ int *arr; // Array of size n to store actual content to be stored in stacks int *top; // Array of size k to store indexes of top elements of stacks int *next; // Array of size n to store next entry in all stacks // and free list int n, k; int free; // To store beginning index of free listpublic: //constructor to create k stacks in an array of size n kStacks(int k, int n); // A utility function to check if there is space available bool isFull() { return (free == -1); } // To push an item in stack number 'sn' where sn is from 0 to k-1 void push(int item, int sn); // To pop an from stack number 'sn' where sn is from 0 to k-1 int pop(int sn); // To check whether stack number 'sn' is empty or not bool isEmpty(int sn) { return (top[sn] == -1); }};//constructor to create k stacks in an array of size nkStacks::kStacks(int k1, int n1){ // Initialize n and k, and allocate memory for all arrays k = k1, n = n1; arr = new int[n]; top = new int[k]; next = new int[n]; // Initialize all stacks as empty for (int i = 0; i < k; i++) top[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 push an item in stack number 'sn' where sn is from 0 to k-1void kStacks::push(int item, int sn){ // Overflow check if (isFull()) { cout << "\nStack 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]; // Update next of top and then top for stack number 'sn' next[i] = top[sn]; top[sn] = i; // Put the item in array arr[i] = item;}// To pop an element from stack number 'sn' where sn is from 0 to k-1int kStacks::pop(int sn){ // Underflow check if (isEmpty(sn)) { cout << "\nStack Underflow\n"; return INT_MAX; } // Find index of top item in stack number 'sn' int i = top[sn]; top[sn] = next[i]; // Change top to store next of previous top // Attach the previous top to the beginning of free list next[i] = free; free = i; // Return the previous top item return arr[i];}/* Driver program to test twoStacks class */int main(){ // Let us create 3 stacks in an array of size 10 int k = 3, n = 10; kStacks ks(k, n); // Let us put some items in stack number 2 ks.push(15, 2); ks.push(45, 2); // Let us put some items in stack number 1 ks.push(17, 1); ks.push(49, 1); ks.push(39, 1); // Let us put some items in stack number 0 ks.push(11, 0); ks.push(9, 0); ks.push(7, 0); cout << "Popped element from stack 2 is " << ks.pop(2) << endl; cout << "Popped element from stack 1 is " << ks.pop(1) << endl; cout << "Popped element from stack 0 is " << ks.pop(0) << endl; return 0;}
// Java program to demonstrate implementation of k stacks in a single // array in time and space efficient waypublic class GFG { // A Java class to represent k stacks in a single array of size n static class KStack { int arr[]; // Array of size n to store actual content to be stored in stacks int top[]; // Array of size k to store indexes of top elements of stacks int next[]; // Array of size n to store next entry in all stacks // and free list int n, k; int free; // To store beginning index of free list //constructor to create k stacks in an array of size n KStack(int k1, int n1) { // Initialize n and k, and allocate memory for all arrays k = k1; n = n1; arr = new int[n]; top = new int[k]; next = new int[n]; // Initialize all stacks as empty for (int i = 0; i < k; i++) top[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 } // A utility function to check if there is space available boolean isFull() { return (free == -1); } // To push an item in stack number 'sn' where sn is from 0 to k-1 void push(int item, int sn) { // Overflow check if (isFull()) { System.out.println("Stack Overflow"); 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]; // Update next of top and then top for stack number 'sn' next[i] = top[sn]; top[sn] = i; // Put the item in array arr[i] = item; } // To pop an element from stack number 'sn' where sn is from 0 to k-1 int pop(int sn) { // Underflow check if (isEmpty(sn)) { System.out.println("Stack Underflow"); return Integer.MAX_VALUE; } // Find index of top item in stack number 'sn' int i = top[sn]; top[sn] = next[i]; // Change top to store next of previous top // Attach the previous top to the beginning of free list next[i] = free; free = i; // Return the previous top item return arr[i]; } // To check whether stack number 'sn' is empty or not boolean isEmpty(int sn) { return (top[sn] == -1); } } // Driver program public static void main(String[] args) { // Let us create 3 stacks in an array of size 10 int k = 3, n = 10; KStack ks = new KStack(k, n); ks.push(15, 2); ks.push(45, 2); // Let us put some items in stack number 1 ks.push(17, 1); ks.push(49, 1); ks.push(39, 1); // Let us put some items in stack number 0 ks.push(11, 0); ks.push(9, 0); ks.push(7, 0); System.out.println("Popped element from stack 2 is " + ks.pop(2)); System.out.println("Popped element from stack 1 is " + ks.pop(1)); System.out.println("Popped element from stack 0 is " + ks.pop(0)); }}// This code is Contributed by Sumit Ghosh
# Python 3 program to demonstrate implementation # of k stacks in a single array in time and space # efficient way class KStacks: def __init__(self, k, n): self.k = k # Number of stacks. self.n = n # Total size of array holding # all the 'k' stacks. # Array which holds 'k' stacks. self.arr = [0] * self.n # All stacks are empty to begin with # (-1 denotes stack is empty). self.top = [-1] * self.k # Top of the free stack. self.free = 0 # Points to the next element in either # 1. One of the 'k' stacks or, # 2. The 'free' stack. self.next = [i + 1 for i in range(self.n)] self.next[self.n - 1] = -1 # Check whether given stack is empty. def isEmpty(self, sn): return self.top[sn] == -1 # Check whether there is space left for # pushing new elements or not. def isFull(self): return self.free == -1 # Push 'item' onto given stack number 'sn'. def push(self, item, sn): if self.isFull(): print("Stack Overflow") return # Get the first free position # to insert at. insert_at = self.free # Adjust the free position. self.free = self.next[self.free] # Insert the item at the free # position we obtained above. self.arr[insert_at] = item # Adjust next to point to the old # top of stack element. self.next[insert_at] = self.top[sn] # Set the new top of the stack. self.top[sn] = insert_at # Pop item from given stack number 'sn'. def pop(self, sn): if self.isEmpty(sn): return None # Get the item at the top of the stack. top_of_stack = self.top[sn] # Set new top of stack. self.top[sn] = self.next[self.top[sn]] # Push the old top_of_stack to # the 'free' stack. self.next[top_of_stack] = self.free self.free = top_of_stack return self.arr[top_of_stack] def printstack(self, sn): top_index = self.top[sn] while (top_index != -1): print(self.arr[top_index]) top_index = self.next[top_index]# Driver Codeif __name__ == "__main__": # Create 3 stacks using an # array of size 10. kstacks = KStacks(3, 10) # Push some items onto stack number 2. kstacks.push(15, 2) kstacks.push(45, 2) # Push some items onto stack number 1. kstacks.push(17, 1) kstacks.push(49, 1) kstacks.push(39, 1) # Push some items onto stack number 0. kstacks.push(11, 0) kstacks.push(9, 0) kstacks.push(7, 0) print("Popped element from stack 2 is " + str(kstacks.pop(2))) print("Popped element from stack 1 is " + str(kstacks.pop(1))) print("Popped element from stack 0 is " + str(kstacks.pop(0))) kstacks.printstack(0)# This code is contributed by Varun Patil
using System;// C# program to demonstrate implementation of k stacks in a single // array in time and space efficient way public class GFG{ // A c# class to represent k stacks in a single array of size n public class KStack { public int[] arr; // Array of size n to store actual content to be stored in stacks public int[] top; // Array of size k to store indexes of top elements of stacks public int[] next; // Array of size n to store next entry in all stacks // and free list public int n, k; public int free; // To store beginning index of free list //constructor to create k stacks in an array of size n public KStack(int k1, int n1) { // Initialize n and k, and allocate memory for all arrays k = k1; n = n1; arr = new int[n]; top = new int[k]; next = new int[n]; // Initialize all stacks as empty for (int i = 0; i < k; i++) { top[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 } // A utility function to check if there is space available public virtual bool Full { get { return (free == -1); } } // To push an item in stack number 'sn' where sn is from 0 to k-1 public virtual void push(int item, int sn) { // Overflow check if (Full) { Console.WriteLine("Stack Overflow"); 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]; // Update next of top and then top for stack number 'sn' next[i] = top[sn]; top[sn] = i; // Put the item in array arr[i] = item; } // To pop an element from stack number 'sn' where sn is from 0 to k-1 public virtual int pop(int sn) { // Underflow check if (isEmpty(sn)) { Console.WriteLine("Stack Underflow"); return int.MaxValue; } // Find index of top item in stack number 'sn' int i = top[sn]; top[sn] = next[i]; // Change top to store next of previous top // Attach the previous top to the beginning of free list next[i] = free; free = i; // Return the previous top item return arr[i]; } // To check whether stack number 'sn' is empty or not public virtual bool isEmpty(int sn) { return (top[sn] == -1); } } // Driver program public static void Main(string[] args) { // Let us create 3 stacks in an array of size 10 int k = 3, n = 10; KStack ks = new KStack(k, n); ks.push(15, 2); ks.push(45, 2); // Let us put some items in stack number 1 ks.push(17, 1); ks.push(49, 1); ks.push(39, 1); // Let us put some items in stack number 0 ks.push(11, 0); ks.push(9, 0); ks.push(7, 0); Console.WriteLine("Popped element from stack 2 is " + ks.pop(2)); Console.WriteLine("Popped element from stack 1 is " + ks.pop(1)); Console.WriteLine("Popped element from stack 0 is " + ks.pop(0)); }}// This code is contributed by Shrikant13
<script>// javascript program to demonstrate implementation of k stacks in a single // array in time and space efficient way // A javascript class to represent k stacks in a single array of size n class KStack { // constructor to create k stacks in an array of size n constructor(k1 , n1) { // Initialize n and k, and allocate memory for all arrays this.k = k1; this.n = n1; this.arr = Array(n).fill(0); this.top = Array(k).fill(-1); this.next = Array(n).fill(0); // Initialize all spaces as free this.free = 0; for (var i = 0; i < n - 1; i++) this.next[i] = i + 1; this.next[n - 1] = -1; // -1 is used to indicate end of free list } // A utility function to check if there is space available isFull() { return (this.free == -1); } // To push an item in stack number 'sn' where sn is from 0 to k-1 push(item , sn) { // Overflow check if (this.isFull()) { document.write("Stack Overflow"); return; } var i = this.free; // Store index of first free slot // Update index of free slot to index of next slot in free list this.free = this.next[i]; // Update next of top and then top for stack number 'sn' this.next[i] = this.top[sn]; this.top[sn] = i; // Put the item in array this.arr[i] = item; } // To pop an element from stack number 'sn' where sn is from 0 to k-1 pop(sn) { // Underflow check if (this.isEmpty(sn)) { document.write("Stack Underflow"); return Number.MAX_VALUE; } // Find index of top item in stack number 'sn' var i = this.top[sn]; this.top[sn] = this.next[i]; // Change top to store next of previous top // Attach the previous top to the beginning of free list this.next[i] = this.free; this.free = i; // Return the previous top item return this.arr[i]; } // To check whether stack number 'sn' is empty or not isEmpty(sn) { return (this.top[sn] == -1); } } // Driver program // Let us create 3 stacks in an array of size 10 var k = 3; n = 10; var ks = new KStack(k, n); ks.push(15, 2); ks.push(45, 2); // Let us put some items in stack number 1 ks.push(17, 1); ks.push(49, 1); ks.push(39, 1); // Let us put some items in stack number 0 ks.push(11, 0); ks.push(9, 0); ks.push(7, 0); document.write("Popped element from stack 2 is " + ks.pop(2)); document.write("<br/>Popped element from stack 1 is " + ks.pop(1)); document.write("<br/>Popped element from stack 0 is " + ks.pop(0));// This code is contributed by gauravrajput1</script>
Output
Popped element from stack 2 is 45Popped element from stack 1 is 39Popped element from stack 0 is 7
Time complexities of operations push() and pop() is O(1). The best part of above implementation is, if there is a slot available in stack, then an item can be pushed in any of the stacks, i.e., no wastage of space.
Time Complexity of top() operation is also O(1)
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(N), as we are using extra space for the stack.
S
Sachin
Improve
Previous Article
Implement Stack using Queues
Next Article
Design a stack that supports getMin() in O(1) time and O(1) extra space
