So far, we analyzed linear search and binary search by counting the maximum number of guesses we need to make. But what we really want to know is how long these algorithms take. We're interested in time, not just guesses. The running times of linear search and binary search include the time needed to make and check guesses, but there's more to these algorithms.
The running time of an algorithm depends on how long it takes a computer to run the lines of code of the algorithm—and that depends on the speed of the computer, the programming language, and the compiler that translates the program from the programming language into code that runs directly on the computer, among other factors.
Let's think about the running time of an algorithm more carefully. We can use a combination of two ideas. First, we need to determine how long the algorithm takes, in terms of the size of its input. This idea makes intuitive sense, doesn't it? We've already seen that the maximum number of guesses in linear search and binary search increases as the length of the array increases. Or think about a GPS. If it knew about only the interstate highway system, and not about every little road, it should be able to find routes more quickly, right? So we think about the running time of the algorithm as a function of the size of its input.
The second idea is that we must focus on how fast a function grows with the input size. We call this the rate of growth of the running time. To keep things manageable, we need to simplify the function to distill the most important part and cast aside the less important parts. For example, suppose that an algorithm, running on an input of size n n , takes 6n2+100n+300 6n^2 + 100n + 300 machine instructions. The 6n2 6n^2 term becomes larger than the remaining terms, 100n+300 100 n + 300 , once n n becomes large enough, 20 in this case. Here's a chart showing values of 6n2 6n^2 and 100n+300 100n + 300 for values of n n from 0 to 100:
We would say that the running time of this algorithm grows as n2 n^2 , dropping the coefficient 6 and the remaining terms 100n+300 100n + 300 . It doesn't really matter what coefficients we use; as long as the running time is an2+bn+c an^2 + bn + c , for some numbers a>0 a > 0 , b b , and c c , there will always be a value of n n for which an2 an^2 is greater than bn+c bn + c , and this difference increases as n n increases. For example, here's a chart showing values of 0.6n2 0.6n^2 and 1000n+3000 1000n + 3000  so that we've reduced the coefficient of n2 n^2 by a factor of 10 and increased the other two constants by a factor of 10:
The value of n n at which 0.6n2 0.6n^2 becomes greater than 1000n+3000 1000n + 3000 has increased, but there will always be such a crossover point, no matter what the constants.
By dropping the less significant terms and the constant coefficients, we can focus on the important part of an algorithm's running time—its rate of growth—without getting mired in details that complicate our understanding. When we drop the constant coefficients and the less significant terms, we use asymptotic notation. We'll see three forms of it: big-Θ \Theta notation, big-O notation, and big-Ω \Omega notation.

This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom plus the Khan Academy computing curriculum team. The content is licensed CC-BY-NC-SA.