Skip to main content

Single Array to Implement three Stacks ( Program in java )

File: Stack.java
----------------------------------------------------------------------------------------
package org.developersbrain.solutions;

public class Stack {
int[] stackList;
int stackSize;
int s1Pos,s1End,s2Pos,s2End,s3Pos,s3End;
int s1Pop,s2Pop,s3Pop;
Stack(){
stackSize = 15;//Default Array Size
stackList=new int[stackSize];
setStacks();
}

Stack(int stackSize){
this.stackSize=stackSize;
stackList=new int[stackSize];
setStacks();
}

private void setStacks(){
int eachStackSize=stackSize/3;
s1Pop=s1Pos=0;
s1End=eachStackSize-1;
s2Pop=s2Pos=s1End+1;
s2End=s2Pos+eachStackSize-1;
s3Pop=s3Pos=s2End+1;
s3End=s3Pos+eachStackSize;
}

void pushS1Element(int item){
if(s1Pos>s1End){
System.out.println("Stack 1 Out of Memory");
}else{
stackList[s1Pos]=item;
s1Pos++;
}
}

void pushS2Element(int item){
if(s2Pos>s2End){
System.out.println("Stack 2 Out of Memory");
}else{
stackList[s2Pos]=item;
s2Pos++;
}
}

void pushS3Element(int item){
if(s3Pos>s3End){
System.out.println("Stack 3 Out of Memory");
}else{
stackList[s3Pos]=item;
s3Pos++;
}
}

void popS1Element(){
if(s1Pos<s1Pop){
System.out.println("Stack empty!");
s1Pos++;
}else{
System.out.println(stackList[s1Pos-1]);
s1Pos--;
}
}

void popS2Element(){
if(s2Pos<s2Pop){
System.out.println("Stack empty!");
s2Pos++;
}else{
System.out.println(stackList[s2Pos-1]);
s2Pos--;
}
}

void popS3Element(){
if(s3Pos<s3Pop){
System.out.println("Stack empty!");
s3Pos++;
}else{
System.out.println(stackList[s3Pos-1]);
s3Pos--;
}
}



}

File:MainClass.java
---------------------------------------------------------------------
package org.developersbrain.solutions;

public class MainClass {

public static void main(String[] args) {
Stack st=new Stack();
System.out.println("Adding Elements to Stack...");
System.out.println();
st.pushS1Element(10);
st.pushS1Element(11);
st.pushS1Element(12);
st.pushS1Element(13);
st.pushS1Element(14);
st.pushS2Element(1);
st.pushS2Element(2);
st.pushS2Element(3);
st.pushS2Element(4);
st.pushS2Element(5);
st.pushS3Element(6);
st.pushS3Element(7);
st.pushS3Element(8);
st.pushS3Element(9);
st.pushS3Element(10);

System.out.println("Pop Elements ( Last in First Out )");
System.out.println("--------------------------------");
st.popS1Element();
st.popS2Element();
st.popS3Element();

System.out.println();
System.out.println("Now Adding Elements to Stack");
System.out.println();
st.pushS1Element(15);
st.pushS2Element(55);
st.pushS3Element(100);

System.out.println("Pop Elements ( Last in First Out )");
System.out.println("--------------------------------");
st.popS1Element();
st.popS2Element();
st.popS3Element();
}

}

Output:
------------------------------------------------------------------------------------------------

Adding Elements to Stack...

Pop Elements ( Last in First Out )
--------------------------------
14
5
10

Now Adding Elements to Stack

Pop Elements ( Last in First Out )
--------------------------------
15
55
100



Comments

Popular posts from this blog

CODILITY: Determine whether given string of parentheses is properly nested.

Task description A string S consisting of N characters is called  properly nested  if: S is empty; S has the form " (U) " where U is a properly nested string; S has the form " VW " where V and W are properly nested strings. For example, string " (()(())()) " is properly nested but string " ()) " isn't. Write a function: class Solution { public int solution(String S); } that, given a string S consisting of N characters, returns 1 if string S is properly nested and 0 otherwise. For example, given S = " (()(())()) ", the function should return 1 and given S = " ()) ", the function should return 0, as explained above. Assume that: N is an integer within the range [ 0 .. 1,000,000 ]; string S consists only of the characters " ( " and/or " ) ". Complexity: expected worst-case time complexity is O(N); expected worst-case space complexity is O(1) (not counting the storage requi...

Java Interface

Problem Statement A Java interface can only contain method signatures and fields. Interface can be used to achieve polymorphism. In this problem you will practice your knowledge on interfaces. You are given an interface   AdvancedArithmetic   which contains a method signature   public abstract int divisorSum(int n) . You need to write a class called MyCalculator which implements the interface. divisorSum   function just takes an integer as input and return the sum of all its divisors. For example divisors of 6 are 1,2,3 and 6, so   divisorSum   should return 12. Value of n will be at most 1000. Read the partially completed code in the editor and complete it. You just need to write the MyCalculator class only.   Your class shouldn't be public. Sample Input 6 Sample Output I implemented: AdvancedArithmetic 12 Explanation Divisors of 6 are 1,2,3 and 6. 1+2+3+6=12. import java.util.*; interface AdvancedArithmetic{   p...

Distinct: Compute number of distinct values in an array.

Task description Write a function class Solution { public int solution(int[] A); } that, given a zero-indexed array A consisting of N integers, returns the number of distinct values in array A. Assume that: N is an integer within the range [ 0 .. 100,000 ]; each element of array A is an integer within the range [ −1,000,000 .. 1,000,000 ]. For example, given array A consisting of six elements such that: A[0] = 2 A[1] = 1 A[2] = 1 A[3] = 2 A[4] = 3 A[5] = 1 the function should return 3, because there are 3 distinct values appearing in array A, namely 1, 2 and 3. Complexity: expected worst-case time complexity is O(N*log(N)); expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments). Elements of input arrays can be modified. class Solution { public int solution ( int [] A) { // write your code in Java SE 8 int len=A.length; int count= 1 ; ...