Array Max Integer Finder (With Big O Analysis)

This tutorial shows you an implementation of three different methods of searching for the maximum number in an Integer array. The array is assumed to hold positive integers.

  1. First method findMaxCompareMax() assumes the first number on the array to be maximum and the loops through other numbers in sequence and replaces the current maximum with the new maximum. When all the elements are checked, we have a maximum number.
  2. The second method findMaxCompareAll() takes an array element and compares it with every other element in the array to see if it is maximum. If it is not, then it proceeds with the next element.
  3. The third method findMaxCompareAfterIndex() This method takes an array element and compare it with other elements in the array to see if it is maximum. If it is not, then it proceeds with the next element. However, this method does not compare every element from the beginning. To avoid redundant comparing, it only compares with the remaining elements in the array (i.e. position + 1 onwards)

Check the code comments for the Big O Analysis of these three methods.

package com.icodejava.blog.published.search;

import java.util.Arrays;

/**
 * 
 * This class is used to demonstrate 3 different ways of finding a maximum
 * number in an Integer Array holding positive integer numbers.
 * 
 * Also provides a Big O analysis.
 */
public class MaxIntegerFinder {

    public static void main(String args[]) {

        int[] numbers = {5, 4, 9, 2, 2, 1, 10, 21, 6 };
        
        System.out.println("Source Array " + Arrays.toString(numbers));

        findMaxCompareMax(numbers);

        findMaxCompareAll(numbers);

        findMaxCompareAfterIndex(numbers);

    }


    /**
     * This method assumes the first number on the array to be the maximum
     * You can also assume the last one to be the max and compare it in reverse.
     * It then loops through other numbers in the array and replaces the 
     * current maximum with the new maximum it finds.
     * 
     * Complexity: O(n) where n is the number of elements in the array
     */
    private static int findMaxCompareMax(int[] numbers) {

        if (numbers.length < 1 ) {
            return 0;
        }
        
        int currentMax = numbers[0];
        for (int i = 1 ; i < numbers.length; i++) { if(numbers[i] > currentMax) {
                currentMax = numbers[i];
            }
        }
        
        System.out.println("Max Number using findMaxCompareMax -> " + currentMax);
        return currentMax;
        
    }
    
    /**
     * This method takes an array element and compare it with 
     * every other element in the array to see if it is maximum.
     * 
     * If it is not, then it proceeds with the next element
     * 
     * Complexity: O(n^2) in worst case and O(n) in best case 
     * where n is the number of elements in the array
     * 
     * Best case scenario is where the first element itself is maximum
     * Worst case scenario is where the last element is maximum.
     */
    private static int findMaxCompareAll(int[] numbers) {

        if (numbers.length < 1) {
            return 0;
        }

        int max = -1;

        for (int i = 0; i < numbers.length; i++) {
            for (int j = 0; j < numbers.length; j++) {
                if (max < numbers[j]) { max = numbers[j]; break; } } } System.out.println("Max Number using findMaxCompareAll -> " + max);
        return max;

    }
    
    /**
     * This method takes an array element and compare it with 
     * other elements in the array to see if it is maximum.
     * 
     * If it is not, then it proceeds with the next element. 
     * 
     * However, this method does not compare every element from the beginning. 
     * To avoid redundant comparing, it only compares with the remaining elements
     * in the array (i.e. position + 1 onwards)
     * 
     * Complexity: O(n^2) in worst case and O(n) in best case 
     * where n is the number of elements in the array
     * 
     * Best case scenario is where the first element itself is maximum
     * Worst case scenario is where the last element is maximum.
     */
    private static int findMaxCompareAfterIndex(int[] numbers) {

        if (numbers.length < 1 ) {
            return 0;
        }
        
        int max=-1;
        
        for (int i = 0 ; i < numbers.length; i++) {
            for(int j = i ; j< numbers.length; j++) {
                if(max < numbers [j]) { max = numbers[j]; break; } } } System.out.println("Max Number using findMaxCompareAfterIndex -> " + max);
        return max;
    }

}

OUTPUT:

Source Array [5, 4, 9, 2, 2, 1, 10, 21, 6]
Max Number using findMaxCompareMax -> 21
Max Number using findMaxCompareAll -> 21
Max Number using findMaxCompareAfterIndex -> 21

Splunk Cheat Sheet – Some Commonly Used Splunk Commands

Get N number of results
index=myIndex sourcetype=myFileType SearchTermOrExpression | head 2

Exclude a Phrase
index=myIndex sourcetype=myFileType PhraseToSearch NOT PhraseNotToSearch

Search Exceptions or Errors
index=myIndex sourcetype=myFileType *Exception* OR *Error*

Stats – Count by Host
index=myIndex sourcetype=myFileType SearchTerm | status count by host

Stats – Count per day
index=myIndex sourcetype=myFileType SearchTerm |convert timeformat=”%Y-%m-%d” ctime(_time) AS date | stats count by date

Table of Extracted Fields
index=myIndex sourcetype=myFileType SearchTerm |table myExtractedField1, myExtractedField2, myExtractedField3

Status – Sort Descending
index=myIndex sourcetype=myFileType SearchTerm | status count by someField | sort by someField

Status – Sort Ascending
index=myIndex sourcetype=myFileType SearchTerm | status count by someField | sort by -someField

Items per hour (date_hour)
index=myIndex sourcetype=myFileType SearchTerm | top date_hour

Remove Duplicates
index=myIndex sourcetype=myFileType SearchTerm |table myExtractedField1, myExtractedField2, myExtractedField3 | dedup myExtractedField1

A Basic Implementation of Binary Tree in Java

A Binary Tree simply can be defined in Java as a node that holds a value for itself and holds reference to a left child node and a right child node. The binary tree can have nodes that have a maximum of 2 children. As you can see in the example below, I have defined the value to be of type Comparable, so that it can be any object that can be compare for less than, greater than or equals. However, if you are going to only store either integers or Strings, for example, you can change the data type to int or String as needed.

The following is a very simple implementation of this concept.

Note: – if the binary tree has left nodes only or right nodes only, it becomes a singly linked list.

package com.icodejava.blog.published.datastructure;
/**
* @author Kushal Paudyal
* Created on 12/5/2016
* Last Modified on 12/5/2016
*
* Binary Tree Node representation.
* - Node has a value, a left node and a right node.
* - Single node, when created, has left and right node as null.
*/
@SuppressWarnings("rawtypes")
class BinaryTreeNode {
Comparable value;
BinaryTreeNode left;
BinaryTreeNode right;

public BinaryTreeNode(Comparable value) {
this.value = value;
this.left = null;
this.right = null;
}

public Comparable getValue() {
return value;
}

public void setValue(Comparable value) {
this.value = value;
}

public BinaryTreeNode getLeft() {
return left;
}

public void setLeft(BinaryTreeNode left) {
this.left = left;
}

public BinaryTreeNode getRight() {
return right;
}

public void setRight(BinaryTreeNode right) {
this.right = right;
}

}