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