## Does the time complexity of code depend on loops or number of operations?

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I was making a code in which I needed to compare elements of my array with another array.

But I was thinking that this would increase my time complexity from O(n) to O(n^2) as it was already inside one for loop, for the first array.

So I came up with a code like this, inside the parent for loop (with parameter i):

int m = 0; int ele = array2[m]; if(array1[i] == ele) count++; m++;

But since the thing that is being done is same, only I have emitted a for loop, I was wondering that the time complexity indeed was O(n) or became O(n^2).

I also understand that this would only compare the same indexed elements and not all. I would appreciate if someone can part more information about this.

The time complexity of an algorithm depends on the way you write your loops and also the time complexity of the operation you do in the algorithm.

If your algorithm consists of any number of O(1) operations, your algorithm is O(1).

If your algorithm consists of one or more loops (not nested) from a constant to n, the time complexity is O(n * the_time_complexity_of_the_operation_in_the_loop)

Every statement below takes O(1) time:

int m = 0; int ele = array2[m]; if(array1[i] == ele) count++; m++;

so the whole thing is O(1). This block is inside a for loop that loops through your array, so the time complexity is O(n * 1) = O(n), where n is the array's length.

Here's an intuitive way to think about why the time complexity became less complex when you compare only the elements at the same index.

If you want to compare every possible pair of the two arrays, how many pairs are there? The first element of first array can form a pair with n other elements from the second array, The second element of the first array can form a pair with n elements from the second array, and so on. In general, for each element in the first array, n pairs can be formed. Assuming the two arrays are of the same length, the total number of pairs is n^2. Therefore, that many comparisons needs to be made, which means the time complexity is O(n^2).

If you want to compare the pairs elements at the same index, how many pairs can you have? n pairs, because each index correspond to exactly 1 pair. Therefore you need to make n comparisons, which means the time complexity is O(n).

**Complexity and Big-O Notation,** An important question is: How efficient is an algorithm or piece of code? Other methods may perform different numbers of operations, depending on the Since we assume the statements are O(1), the total time for the for loop is N * O(1), The time complexity of the loop depends upon the number of times the loop is being executed. Let’s take an example of a simple loop which is executing 0 till n as shown below. For better understanding, I am writing the same condition with both while and for loop. The time complexity of simple loops

Yes Time complexity depends on Loop and number of operations done within it.

In your case there is one assignment int `ele = array2[m]`

and one conditional checking `if(array1[i] == ele)`

along with 2 increments of `m`

and `count`

.

All the operations will take constant time and will depend on the number of time the loop is executing.

**What is time complexity of while loops?,** The number of operations is proportional to the size of the task squared. nested loops where the number of iterations of the inner loop depends on the value. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Thus, the amount of time taken and the number of elementary operations performed by the algorithm are taken to differ by at most a constant factor .

In this case, you are always comparing the `array[m] (in your case array[0])`

element with `i^th`

index of `array1`

array. So this is actually same as **traversing one array and count the number of elements matching given specific number**

Comparing one array to another usually takes O(n^2) time, though there are ways to do in lesser time complexity. Once such solution could be of `O(n+m) time complexity and O(min(n,m)) space complexity`

. Where you Traverse the lesser sized array and store the elements in one hash table, then you traverse the second array to match the elements from hash table.

**8 time complexities that every programmer should know,** Time complexity analysis estimates the time to run an algorithm. This can be achieved by choosing an elementary operation, which the algorithm performs repeatedly, and define the time In fact, the outer for loop is executed n - 1 times. Big O notation is a commonly used metric used in computer science to classify algorithms based on their time and space complexity. The time and space here is not based on the actual number of…

**[PDF] Time complexity :,** Time complexity represents the number of times a statement is The time complexity of an algorithm is NOT the actual time required to execute a particular code, since that depends on other factors like programming language, operating To remain constant, these algorithms shouldn't contain loops, You can get the time complexity by “counting” the number of operations performed by your code. This time complexity is defined as a function of the input size n using Big-O notation. n indicates the size of the input, while O is the worst-case scenario growth rate function.

**How to analyze time complexity: Count your steps · YourBasic,** 3) O(nc): Time complexity of nested loops is equal to the number of times the code is O(m) + O(n) which is O(m+n) If m == n, the time complexity becomes The time complexity for the above algorithm will be Linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time. for(i=0; i N; i++) { for(j=0; j . N;j++) { statement; } } This time, the time complexity for the above code will be Quadratic. The running time of the two loops is proportional to the square of N.

**Essential Programming,** Time Complexity of algorithm/code is not equal to the actual time required to execute a So, we can say that actual time requires to execute code is machine dependent amount of time require to execute code, no matter which operating system or Time Complexity of a Loop when Loop variable “Expands or Shrinks” In actual cases, the performance (Runtime) of an algorithm depends on n, that is the size of the input or the number of operations is required for each input item. The algorithms can be classified as follows from the best-to-worst performance (Running Time Complexity): A logarithmic algorithm – O (logn)

##### Comments

- "this would only compare the same indexed elements and not all", therefore "the thing that is being done is"
*not*same. Comparing same-indexed elements is O(n), comparing each element with each element is O(n^2) (even though the latter is possible with a single loop - appropriately, from`0`

to`n^2-1`

). - you should post both of your two pieces of code
- What will
`array2`

length is greater that`array1`

length? - Thank you, this explanation was great :)