For the "longest common prefix" problem on Leetcode, Leetcode said that this solution is O(S) complexity where S is the sum of all characters. https://leetcode.com/problems/longest-common-prefix/solution/
Solution:
public String longestCommonPrefix(String[] strs) {
if (strs.length == 0) return "";
String prefix = strs[0];
for (int i = 1; i < strs.length; i++)
while (strs[i].indexOf(prefix) != 0) {
prefix = prefix.substring(0, prefix.length() - 1);
if (prefix.isEmpty()) return "";
}
return prefix;
}
Isn't a while loop inside of a for loop n^2 though? I thought all nested loops were n^2.
For O(n) time complexity we must define n. It is typically the length of the input array. However in this case, we instead define S to be the sum of all characters as the length of the input array cannot accurately describe our time complexity.
The complexity of O(n^2) will be if we are iterating over the same array of length n with both the for loop and while loop.
Isn't a while loop inside of a for loop n^2 though?
Not generally, but in this case you are right: the algorithm has a worst time complexity of O(𝑛²). The article is wrong.
Let's take as input the following two strings (respectively 10 and 20 characters long):
abbbbbbbbb
aaaaaaaaaacccccccccc
The outer loop will only make one iteration.
The inner loop will make 10 iterations, each time performing indexOf and then making the prefix string one character shorter, from its initial size of 10 until it is just "a".
Java's indexOf method has in general a O(𝑛𝑚) time complexity. The first time it executes, it will try to match the prefix string in the second string at index 0, 1, ..., 8, 9, each time failing at the second character comparison (of "b" with another letter). So we have 20 character-to-character comparisons here. The second time it executes, prefix has one less "b", and so there are now 18 comparisons, then 16, then 14, ... 4, and then the last iteration (when prefix is just one character): 1 (success!), making a total of 109 character-to-character comparisons!
Clearly this leads to a quadratic complexity.
The problem is with the indexOf method. The code should use the startsWith method instead, as it makes no sense to look at other indices than index 0. Then the algorithm will have a O(𝑛) complexity. This is easy to see: every character in the strings S2 ... Sn is compared at most once with a character from S1.
So the code should have:
while (!strs[i].startsWith(prefix)) {
I thought all nested loops were n^2.
No, that is not generally true. As 𝑛 here is the total number of characters, we can see that nor the outer loop, nor the inner loop has a number of iterations that corresponds to that 𝑛.
Here are some other examples of nested loops that do not represent a complexity of O(𝑛²)
for (int i = 0; i < n; i++)
for (int j = 1; j < n; j*=2)
//...
This is O(𝑛log𝑛)
Or:
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) {
if (j % 3 == i % 3) break;
//...
}
This is O(𝑛)
Related
I'm having some trouble calculating the bigO of this algorithm:
public void foo(int[] arr){
int count = 0;
for(int i = 0; i < arr.length; i++){
for(int j = i; j > 0; j--){
count++;
}
}
}
I know the first for loop is O(n) time but I can't figure out what nested loop is. I was thinking O(logn) but I do not have solid reasoning. I'm sure I'm missing out on something pretty easy but some help would be nice.
Let's note n the length of the array.
If you consider the second loop alone, it is just a function f(i), and since it will iterate on all elements from i to 1, its complexity will be O(i). Since you know that j<n, you can say that it is O(n). However, there is no logarithm involved, since in the worst case, i.e. j=n, you will perfrom n iterations.
As for evaluating the complexity of both loops, observe that for each value of i, the second loop goes throught i iterations, so the total number of iterations is
1+2+...+(n-1)= n*(n-1)/2=(1/2)*(n^2-n)
which is O(n^2).
If we consider c as a number of times count is incremented in the inner loop then the total number of times count is incremented can be represented by the formula below:
As you can see, the total time complexity of the algorithm is O(n^2).
These programs do the calculation ∑𝑖=0 𝑎𝑖 𝑥
I am trying to figure out big O calculations. I have done alot of study but I am having a problem getting this down. I understand that big O is worst case scenario or upper bounds. From what I can figure program one has two for loops one that runs for the length of the array and the other runs to the value of the first loop up to the length of the array. I think that if both ran the full length of the array then it would be quadratic O(N^2). Since the second loop only runs the length of the length of the array once I am thinking O(NlogN).
Second program has only one for loop so it would be O(N).
Am I close? If not please explain to me how I would calculate this. Since this is in the homework I am going to have to be able to figure something like this on the test.
Program 1
// assume input array a is not null
public static double q6_1(double[] a, double x)
{
double result = 0;
for (int i=0; i<a.length; i++)
{
double b = 1;
for (int j=0; j<i; j++)
{
b *= x;
}
result += a[i] * b;
}
return result;
}
Program 2
// assume input array a is not null
public static double q6_2(double[] a, double x)
{
double result = 0;
for (int i=a.length-1; i>=0; i--)
{
result = result * x + a[i];
}
return result;
}
I'm using N to refer to the length of the array a.
The first one is O(N^2). The inner loop runs 1, 2, 3, 4, ..., N - 1 times. This sum is approx N(N-1)/2 which is O(N^2).
The second one is O(N). It is simply iterating through the length of the array.
Complexity of a program is basically number of instructions executed.
When we talk about the upper bound, it means we are considering the things in worst case which should be taken in consideration by every programmer.
Let n = a.length;
Now coming back to your question, you are saying that the time complexity of the first program should be O(nlogn), which is wrong. As when i = a.length-1 the inner loop will also iterate from j = 0 to j = i. Hence the complexity would be O(n^2).
You are correct in judging the time complexity of the second program which is O(n).
From a popular definition ,a loop or recursion that runs a constant number of times is also considered as O(1).
For example the following loop is O(1)
// Here c is a constant
for (int i = 1; i <= c; i++) {
// some O(1) expressions
}
Time Complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount.
For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
I got a little confused with the following example here lets take c = 5 and according to the O(1) definition the below code becomes - O(1)
for(int i = 0; i < 5 ; i++){
cout<<"Hello<<endl";
}
Function 1:
for(int i = 0; i < len(array); i+=2){
if(key == array[i])
cout<<"Element found";
}
Function 2:
for(int i =0;i < len(array) ; i++){
if(key == array[i])
cout<<"Element found";
}
But when we compare the above 2 examples will they both become O(n) or first function is O(1) from definition.What exaclty does a loop running constant number of times means?
Assuming that len(array) is the b we're talking about [*], both your functions are O(n).
Function 2 will execute the if n times (once for each element of the array), making it obviously O(n).
Function 1, on the other hand, will execute the if n/2 times (once for every other element in the array), leading to a run time of O(n*1/2), and since constant factors (1/2 in this case) are usually omitted in O notation, you'll again end up with O(n).
[*] For the sake of completeness, if your array were of a fixed size, ie. len(array) were a constant, than both functions would be O(1).
"Loop running a costant number of times" means the loop runs a number of times that is limited from above by a constant, i.e. a given number that is indipendent from the input of your program.
Both in function 1 and 2 (unless the lenghts of the arrays are fixed or you can prove they'll never be grater than a specific constant, indipendently of the input) the if will be execute a number of time that depends on the size of the input so the time complexity can't be O(1).
"Time Complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount" is a misleading definition
I have the below pseudocode that takes a given unsorted array of length size and finds the range by finding the max and min values in the array. I'm just learning about the various time efficiency methods, but I think the below code is Θ(n), as a longer array adds a fixed number of actions (3).
For example, ignoring the actual assignments to max and min (as the unsorted array is arbitrary and these assignments are unknown in advance), an array of length 2 would only require 5 actions total (including the final range calculation). An array of length 4 only uses 9 actions total, again adding the final range calculation. An array of length 12 uses 25 actions.
This all points me to Θ(n), as it is a linear relationship. Is this correct?
Pseudocode:
// Traverse each element of the array, storing the max and min values
// Assuming int size exists that is size of array a[]
// Assuming array is a[]
min = a[0];
max = a[0];
for(i = 0; i < size; i++) {
if(min > a[i]) { // If current min is greater than val,
min = a[i]; // replace min with val
}
if(max < a[i]) { // If current max is smaller than val,
max = a[i]; // replace max with val
}
}
range = max – min; // range is largest value minus smallest
You're right. It's O(n).
An easy way to tell in simple code (like the one above) is to see how many for() loops are nested, if any. For every "normal" loop (from i = 0 -> n), you add a factor of n.
[Edit2]: That is, if you have code like this:
array a[n]; //Array with n elements.
for(int i = 0; i < n; ++i){ //Happens n times.
for(int j = 0; j < n; ++j){ //Happens n*n times.
//something //Happens n*n times.
}
}
//Overall complexity is O(n^2)
Whereas
array a[n]; //Array with n elements.
for(int i = 0; i < n; ++i){ //Happens n times.
//something //Happens n times.
}
for(int j = 0; j < n; ++j){ //Happens n times.
//something //Happens n times.
}
//Overall complexity is O(2n) = O(n)
This is pretty rudimentary, but useful if someone has not taken an Algorithm course.
The procedures within your for() loop are irrelevant in a complexity question.
[Edit]: This assumes that size actually means the size of array a.
Yes, this would be Θ(n). Your reasoning is a little skewed though.
You have to look at every item in your loop so you're bounded above by a linear function. Conversely, you are also bounded below by a linear function (the same one in fact), because you can't avoid looking at every element.
O(n) only requires that you bound above, Omega(n) requires that you bound below.
Θ(n) says you're bounded on both sides.
Let size be n, then it's clear to see that you always have 2n comparisons and of course the single assignment at the end. So you always have 2n + 1 operations in this algorithm.
In the worst case scenario, you have 2n assignments, thus 2n + 1 + 2n = 4n + 1 = O(n).
In the best case scenrio, you have 0 assignments, thus 2n + 1 + 0 = 2n + 1 = Ω(n).
Therefore, we have that both the best and worst case perform in linear time. Hence, Ɵ(n).
Yeah this surely is O(n) algorithm. I don't think you really need to drill down to see number of comparisons to arrive on the conclusion about the complexity of the algorithm. Just try to see how the number of comparisons will change with the increasing size of the input. For O(n) the comparisons should have a linear increase with the increase in input. For O(n^2) it increases by some multiple of n and so on.
I am reading some information on time complexity and I'm quite confused as to how the following time complexities are achieved and if there is a particular set of rules or methods for working this out?
1)
Input: int n
for(int i = 0; i < n; i++){
print("Hello World, ");
}
for(int j = n; j > 0; j--){
print("Hello World");
}
Tight: 6n + 5
Big O: O(n)
2)
Input: l = array of comparable items
Output: l = array of sorted items
Sort:
for(int i = 0; i < l.length; i++){
for(int j = 0; j < l.length; j++){
if(l{i} > l{j}){
} }
Swap(l{i},l{j});
}
return ls;
Worst Case Time Complexity: 4n2 +3n+2 = O(n2)
For a given algorithm, time complexity or Big O is a way to provide some fair enough estimation of "total elementary operations performed by the algorithm" in relationship with the given input size n.
Type-1
Lets say you have an algo like this:
a=n+1;
b=a*n;
there are 2 elementary operations in the above code, no matter how big your n is, for the above code a computer will always perform 2 operations, as the algo does not depend on the size of the input, so the Big-O of the above code is O(1).
Type-2
For this code:
for(int i = 0; i < n; i++){
a=a+i;
}
I hope you understand the Big-O in O(n), as elementary operation count directly depend on the size of n
Type-3
Now what about this code:
//Loop-1
for(int i = 0; i < n; i++){
print("Hello World, ");
}
//Loop-2
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++) {
x=x+j;
}
}
As you can see loop-1 is O(n) and loop-2 is O(n^2). So it feel like total complexity should be O(n)+O(n^2). But no, the time complexity of the above code is O(n^2). Why? Because we are trying to know the fair enough count of elementary operations performed by the algorithm for a given input size n, which will be comparatively easy to understand by another person. With this logic, O(n)+O(n^2) become O(n^2), or O(n^2)+O(n^3)+O(n^4) become O(n^4)!
Again, you may ask: But how? How all the lower power of Big-O become so insignificant as we add it with a higher power of Big-O, that we can completely omit them (lower powers) when we are describing the complexity of our algorithm to another human?
I will try show the reason for this case: O(n)+O(n^2)=O(n^2).
Lets say n=1000 then the exact count for O(n) is 1000 operations and the exact count for O(n^2) is 1000*1000=1000000, so O(n^2) is 1000 time bigger than O(n), which means your program will spend most of the execution time in O(n^2) and thus it is not worth to mention that your algorithm also has some O(n).
PS. Pardon my English :)
In the first example, the array has n elements, and you go through these elements Twice. The first time you start from index 0 until i, and the second time you start from index n to 0. So, to simplify this, we can say that it took you 2n. When dealing with Big O notation, you should keep in mind that we care about the bounds:
As a result, O(2n)=O(n)
and O(an+b)=O(n)
Input: int n // operation 1
for(int i = 0; i < n; i++){ // operation 2
print("Hello World, "); // Operation 3
}
for(int j = n; j > 0; j--) // Operation 4
{
print("Hello World"); //Operation 5
}
As you can see, we have a total of 5 operations outside the loops.
Inside the first loop, we do three internal operations: checking if i is less than n, printing "Hello World", and incrementing i .
Inside the second loop, we also have three internal operations.
So, the total number of of opetations that we need is: 3n ( for first loop) + 3n ( second loop) + 5 ( operations outside the loop). As a result, the total number of steps required is 6n+5 ( that is your tight bound).
As I mentioned before, O( an +b )= n because once an algorithm is linear, a and b do not have a great impact when n is very large.
So, your time complexity will become : O(6n+5) =O(n).
You can use the same logic for the second example keeping in mind that two nested loops take n² instead of n.
I will slightly modify Johns answer. Defining n is one constant operation, defining integer i and assigning it to 0 is 2 constant operations. defining integer j and assigning with n is another 2 constant operations. checking the conditions for i,j inside for loop,increment,print statement depends on n so the total will be 3n+3n+5 which is equal to 6n+5. Here we cannot skip any of the statements during execution so its average case running time will also be its worst case running time which is O(n)