大O结构

Question

Could anyone point me in the direction to where the Big O notation is in this coding I found? http://jsfiddle.net/neekit_rama/6qmoLbex/

<h1 id="title">Array operations with random functions & Big o notation Analysis</h1>

Enter an element to search(1-100):

<input type="text" id="myTextField1"/>

<input type="submit" id="byBtn" value="SearchArray" onclick="StoreArray()"/> <br>

Number of comparisions:<input type="text" id="myTextField2"/>

JavaScript:

function randomIntInc (low, high) {

    return Math.floor(Math.random() * (high - low + 1) + low);

}

function StoreArray(){

var numbers = new Array(100);

for (var i = 0; i < numbers.length; i++) {

    numbers[i] = randomIntInc(1, 100)

}

var element = document.getElementById('myTextField1').value;

    var c=SearchArray(numbers,element);

   var t= document.getElementById('myTextField2');

   myTextField2.value=c;

}

function SearchArray(numbers, Element)

{



    var count=0;

   for (var i = 0; i < numbers.length; i++) {

               count++;

       if(numbers[i]==Element)

       {

           alert('Element found!!.');

           return count;

       }

   }

    alert('Element not found!!.');

    return count;

}

Answer 1

The big-O complexity is O(n) where n is the number of elements in the numbers array.

O(n) complexity means that the algorithm scales proportionate to n . As n gets larger, the function will take a proportionately longer time to execute.

An easy way to figure out the complexity (for a simple loop-based algorithm like this) is by counting the number of comparisons:

for (var i = 0; i < numbers.length; i++) {
    count++;
    if(numbers[i]==Element)
    ...

The comparison numbers[i]==Element is likely to be the most expensive operation in the loop, and it will be executed numbers.length times in the worst case.

In the best case, the Element will be the first item in the array, and so the comparison will only be executed 1 time, or O(1) . This happens because of the return statement that terminates the loop when a match is found.

In the worst case, the Element will be the last item in the array, or will not exist at all, and so the comparison will be executed n times.