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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.