GATE CS Mock 2018 | Question 45

This mock test problem is useful for programmers who are preparing for the GATE (Graduate Aptitude Test in Engineering) Computer Science exam.

The question deals with determining the time complexity of a given algorithm. The algorithm is described as follows:

function myst(a: array of integers):
   n:= length(a)
   for i:= 1 to n−2 do
      for j:= i+1 to n−1 do
         for k:= j+1 to n do
            if a[i]+a[j]=a[k] then
               return (i, j, k)
   return (−1, −1, −1)

The above algorithm takes an array of integers as input and returns a tuple (i, j, k) if there exists three integers in the array such that a[i] + a[j] = a [k]. Otherwise, it returns (-1,-1,-1).

To analyze the time complexity of the algorithm, we must count the number of basic operations executed. Here, we have three nested loops and an if-else statement. The time complexity of the algorithm is therefore O(n^3), where n is the length of the input array.

Programmers preparing for the GATE Computer Science exam should be familiar with analyzing the time complexity of algorithms. This knowledge will be helpful in designing and optimizing algorithms to run efficiently.