Know How to Implement Stack in Java From Our Free Sample

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Fixed Division of the Array for the Stack Implementation

What we need to do is to implement three stacks using the arrays in Java. Like in many other tasks, the solution depends on how we are going to solve the problem. If we need to have a separate space for each stack, we can do it using the fixed division.

We can divide the array into three equal parts and place stacks within a limited space. Please note that we will further describe ranges in the array by using brackets: square brackets [] indicate that the limit values are included in the range, and parentheses mean the values are not included.

  • Stack 1: [0, n / 3).
  • Stack 2: [n / 3, 2n / 3).
  • Stack 3: [2n / 3, n].

Here is the code of the implementation with detailed comments:

int size = 100;

int[] buf = new int [size * 3]; //buffer for three stacks

int[] ptr = {0,0,0};    //pointers for tracking the top elements


void push(int num, int val) throws Exception { //push operation

   /* check for the free space */

   if (ptr[num] >= size){

       throw new Exception(“The space is not enough”);


   /* find index of the top element +1

   and increase the pointer of the stack */

   int ind = num * size + ptr[num] + 1; //index of the element

   ptr[num]++; //increase the size of the stack

   buf[ind] = val; //write the value to the stack



int pop(int num) throws Exception { //pop operation

   if (ptr[num] == 0) {//check if the stack is empty

       throw new Exception(“The stack is empty”);


   int ind = num * size + ptr[num];//index of the element

   ptr[num]–; //decrease the size of the stack

   int val = buf[ind]; //temporary value

   buf[ind] = 0; //delete the value of the stack element

   return val; //return the tmp value



int peek(int num) { //peek operation

   int ind = num * size + ptr[num]; //index of the element

   return buf[ind]; //return the element



boolean isEmpty(int num) { //check if the stack is empty

   return ptr[num] ==0;


If we will have more information about the appointment of the stack, we can modify the algorithm. For example, if we know that stack 1 will be larger than stack 2, we can reallocate the space in favor of the second stack.


Thanks for your attention!

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