## Time-Complexity

 Question 1

Consider the recurrence function Then T(n) in terms of Θ notation is

 A Θ(log⁡log⁡n) B Θ(log⁡n) C Θ(√n) D Θ(n)
Algorithms       Time-Complexity       GATE 2017 [Set-2]       Video-Explanation
Question 1 Explanation:
T(n) = 2T(√n) + 1
(or)
T(n) = 2T(n(1⁄2)) + 1
Here, Assume n = 2k
T(2k) = 2T(2k)(1⁄2) + 1
T(2k) = 2T(2(k/2) ) + 1
Assume T(2k) = S(k)
S(k) = 2S(k/2) + 1
Apply Master’s theorem Case-1
a=2, b=2
S(k) = k(log22 )
S(k) = θ(k')
but S(k) = T(2k)
T(2k) = θ(k')
but n = 2k
k = log⁡n
T(n) = θ(logn)
 Question 2

Consider the following C function.

```int fun (int n)   {
int i, j;
for (i = 1; i <= n; i++)   {
for (j = 1; j < n; j += i)   {
printf("%d %d",i,j);
}
}
}
```

Time complexity of fun in terms of Θ notation is

 A Θ(n√n) B Θ(n2) C Θ(n log⁡n) D Θ(n2 logn)
Algorithms       Time-Complexity       GATE 2017 [Set-2]       Video-Explanation
Question 2 Explanation:
We can solve iterative programs time complexity with the help of rollback method.
int fun(int n)
{
int i, j;
for (i = 1; i <= n ; i++)    /* It is independent loop. So, it takes O(n) time */
{
for (j = 1; j < n; j += i)     /* It is dependent loop, It always incrementing with the help of i. It will take approximately O(log n) times*/
{
printf("%d %d", i, j);     /* It takes O(1) time */
}
}
}
So, the overall complexity of the program is θ(n log⁡n) times.
 Question 3

The given diagram shows the ﬂowchart for a recursive function A(n). Assume that all statements, except for the recursive calls, have O(1) time complexity. If the worst case time complexity of this function is O(nα), then the least possible value (accurate upto two decimal  positions) of α is _________. A 2.32193 B 2.32193 C 2.32193 D 2.32193
Algorithms       Time-Complexity       GATE 2016 [Set-2]       Video-Explanation
Question 3 Explanation:
This type of problem, we have to consider worst case time complexity, it mean that all possibilities.
According to flow chart total 5 worst case possibilities of function calls. The remaining function calls/return statements will execute only constant amount of time.
So, total function calls 5.
The Recurrence will be
A(n) = 5A(n/2) + O(1)
Apply Master’s theorem,
A=5, b=2, k=0, p=0
a > bk case
A(n) = n(logba ) = n(log25) = n2.3219280
∴α = 2.3219280
 Question 4

An algorithm performs (log N)1/2 find operations, N insert operations, (log N)1/2 delete operations, and (log N)1/2 decrease-key operations on a set of data items with keys drawn from a linearly ordered set. For a delete operation, a pointer is provided to the record that must be deleted. For the decrease-key operation, a pointer is provided to the record that has its key decreased. Which one of the following data structures is the most suited for the algorithm to use, if the goal is to achieve the best total asymptotic complexity considering all the operations?

 A Unsorted array B Min-heap C Sorted array D Sorted doubly linked list
Algorithms       Time-Complexity       GATE 2015 [Set-1]
Question 4 Explanation: Question 5

Consider a rooted n node binary tree represented using pointers. The best upper bound on the time required to determine the number of subtrees having exactly 4 nodes is O(nalogbn). Then the value of a+10b is ________.

 A 1 B 2 C 3 D 4
Data-Structures       Time-Complexity       GATE 2014 [Set-1]
Question 5 Explanation:
Binary tree traversal takes O(n) time for reaching 4th level from the leaf node. Every node has to check their subtree having exactly 4 nodes. Checking time can be done in maximum constant time for each node. Nodes in the level is always less than n and it's complexity is O(n). Finally a value becomes 1 and b value becomes 0.
Substitute into a and b values in equation.
⇒ a+10b
⇒ a*1 + 10*0
⇒ 1+0
⇒ 1
 Question 6

Consider the following pseudo code. What is the total number of multiplications to be performed?

```D = 2
for i = 1 to n do
for j = i to n do
for k = j + 1 to n do
D = D * 3```
 A Half of the product of the 3 consecutive integers. B One-third of the product of the 3 consecutive integers. C One-sixth of the product of the 3 consecutive integers. D None of the above.
Data-Structures       Time-Complexity       GATE 2014 [Set-1]
Question 6 Explanation:
D = 2
for i = 1 to n do
for j = i to n do
for k = j + 1 to n do
D = D * 3; Also you can try for smaller ‘n’.
 Question 7

Which one of the following correctly determines the solution of the recurrence relation with T(1) = 1?

`     T(n) = 2T(n/2) + Logn  `
 A Θ(n) B Θ(nlog n) C Θ(n2) D Θ(log⁡n)
Algorithms       Time-Complexity       GATE 2014 [Set-2]
Question 7 Explanation:
T(n) = 2T(n/2)+logn
Apply Master’s theorem,
a=2, b=2, k=0, p=1
According to Master’s theorem, we can apply Case-I
(I) If a>bk, then T(n) = θ(n(logba )) = θ(n(log22)) = θ (n)
 Question 8

Suppose you want to move from 0 to 100 on the number line. In each step, you either move right by a unit distance or you take a shortcut. A shortcut is simply a pre-specified pair of integers i, j with i < j.  Given a shortcut i, j if you are at position i on the number line, you may directly move to j. Suppose T(k) denotes the smallest number of steps needed to move from k to 100. Suppose further that there is at most 1 shortcut involving any number, and in particular from 9 there is a shortcut to 15. Let y and z be such that T(9) = 1 + min(T(y),T(z)).  Then the value of the product yz is _____.

 A 150 B 151 C 152 D 153
Theory-of-Computation       Time-Complexity       GATE 2014 [Set-3]
Question 8 Explanation:
T(k) is the smallest no. of steps needed to move from k to 100.
Now, it is given that
T(9) = 1 + min(T(y),T(z))
where,
T(y) = steps from y to 100
T(z) = steps from z to 100
where y and z are two possible values that can be reached from 9.
One number that can be reached from 9 is 10. Another no. is 15, the shortcut path from 9, as given in the question.
∴ The value of 'y' and 'z' are 10 and 15.
So, y × z = 10 × 15 = 150
 Question 9

Consider the following operation along with Enqueue and Dequeue operations on queues, where k is a global parameter.

```MultiDequeue(Q){
m = k
while (Q is not empty and m  > 0) {
Dequeue(Q)
m = m - 1
}
}```

What is the worst case time complexity of a sequence of n MultiDequeue() operations on an initially empty queue?

 A Θ(n) B Θ(n + k) C Θ(nk) D Θ(n2)
Algorithms       Time-Complexity       GATE 2013
Question 9 Explanation:
Initially the queue is empty and we have to perform n operations.
i. One option is to perform all Enqueue operations i.e. n Enqueue operations. Complexity will be Ѳ(n).
or
ii. We can perform a mix of Enqueue and Dequeue operations. It can be Enqueue for first n/2 times and then Dequeue for next n/2, or Enqueue and Dequeue alternately, or any permutation of Enqueues and Dequeues totaling ‘n’ times. Completity will be Ѳ(n).
or iii. We can perform Enqueues and MultiDequeues. A general pattern could be as follows: Enqueue Enqueue …(ktimes) MultiDequeue Enqueue Enqueue …(ktimes) MultiDequeue …Up to total n …. K items enqueued ….k items deleted….k items enqueued…..k items deleted --- and so on.
The number of times this k-Enqueues, MultiDequeue cycle is performed = n/k+1
So, Complexity will be k times Enqueue + 1 MultiDequeue) ×n/k+1
Which is (2k×n/k+1) = (n)
or
iv. We can just perform n MultiDequeues (or n Dequeues for the matter):
Each time the while condition is false (empty queue), condition is checked just once for each of the ‘n’ operations. So Ѳ(n).
 Question 10

Let W(n) and A(n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n. Which of the following is ALWAYS TRUE?

 A A(n) = Ω (W(n)) B A(n) = Θ (W(n)) C A(n) = O (W(n)) D A(n) = o (W(n))
Algorithms       Time-Complexity       GATE 2012
Question 10 Explanation:
The average case time can be lesser than or even equal to the worst case.
So, A(n) would be upper bounded by W(n) and it will not be strict upper bound as it can even be same (e.g. Bubble Sort and merge sort).
A(n) = O(W(n))
Note: Option A is wrong because A(n) is not equal to Ω(w(n)) .
 Question 11

Two alternative packages A and B are available for processing a database having 10k records. Package A requires 0.0001n2 time units and package B requires 10nlog10n time units to process n records.What is the smallest value of k for which package B will be preferred over A?

 A 12 B 10 C 6 D 5
Algorithms        Time-Complexity       GATE 2010
Question 11 Explanation:
As per given information Package B 10nlog10n is lesser than or equals to Package A 0.0001n2 0 because n2 is asymptotically larger than nlogn. Finally, 10nlog10n ≤ 0.0001n2
Let n = 10k records. Substitute into 10nlog10n ≤ 0.0001n2
10(10k)log1010k ≤ 0.0001(10k)2
10k+1k ≤ 0.0001 × 102k
k ≤ 102k−k−1−4
k ≤ 10k−5
According to the problem value 6 is suitable for K.
 Question 12

The running time of an algorithm is represented by the following recurrence relation: Which one of the following represents the time complexity of the algorithm?

 A θ(n) B θ(n log n) C θ(n2) D θ(n2log n)
Algorithms        Time-Complexity       GATE 2009
Question 12 Explanation:
The above recurrence relation is form of master theorem. So, we can apply directly master's theorem T(n) = aT(n/b) + nk logp n)
The recurrence relation is T(n) = T(n/3) + cn
Here a=1 , b=3, k=1, p=0.
So, ak and p>=0. So, it comes under case 3.
Therefore T(n) = θ(nk(logpn)) => θ(n1(log0 n)) => θ(n)
 Question 13

Consider the following functions:

f (n) = 2n
g(n) = n!
h (n) = nlogn

Which of the following statements about the asymptotic behaviour of f(n), g(n), and h(n) is true?

 A f(n) = O(g(n)); g(n) = O(h(n)) B f(n) = Ω(g(n)); g(n) = O(h(n)) C g(n) = O(f(n)); h(n) = O(f(n)) D h(n) = O(f(n)); g(n) = Ω(f(n))
Algorithms        Time-Complexity       GATE 2008
Question 13 Explanation:
When we want to find asymptotically bigger/smaller functions, we have to follow basically 2 steps:
1. Check whether the function if polynomial time or exponential time.
2. Group them into polynomial functions and exponential function. The exponential functions are always bigger than polynomial functions.
As per the above 3 functions, the g(n) is greater than others. f(n) greater than h(n). So, the order of growth h(n) < f(n) < g(n).
Note: For computing, take always big number.
 Question 14

The minimum number of comparisons required to determine if an integer appears more than n/2 times in a sorted array of n integers is

 A θ(n) B θ(logn) C θ(log*n) D θ(1)
Algorithms        Time-Complexity       GATE 2008
Question 14 Explanation:
The Best way to find out whether an integer appears more than n/2 times in a sorted array(Ascending Order) of n integers, would be binary search approach.
1. The First occurrence of an element can be found out in O(log(n)) time using divide and conquer technique, lets say it is i.
2. The Last occurrence of an element can be found out in O(log(n)) time using divide and conquer technique,lets say it is j.
3. Now number of occurrence of that element(count) is (j-i+1).
Overall time complexity = log n +log n +1 = O(logn)
Program:
/* If x is present in arr[low...high] then returns the index of first occurrence of x, otherwise returns -1 */
int binarySearch(int arr[], int low, int high, int x){
if (high >= low) {
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* Check if arr[mid] is the first occurrence of x.
arr[mid] is first occurrence if x is one of the following is true:
(i) mid == 0 and arr[mid] == x
(ii) arr[mid-1] < x and arr[mid] == x */
if ( (mid == 0 || x > arr[mid-1]) && (arr[mid] == x) )
return mid;
else if (x > arr[mid]
) return_binarySearch(arr, (mid + 1), high, x);
else
return_binarySearch(arr, low, (mid - 1), x);
}
return -1;
}
Note: The code finds out the first occurrence of the element in the array and checks if the element at (n/2+1)th position is same(as the array is sorted). Takes O(logn) time.
 Question 15

Consider the following C program

```int f1 (int n)
{
if(n == 0||n == 1)
return n;
else
return (2*f1(n-1)+3*f1(n-2));
}

int f2 (int n)
{
int i;
int X[N], Y[N], Z[N];

X=Y=Z=0;
X=1; Y=2; Z=3;
for(i=2; i<=n; i++) {
X[i] = Y[i-1] + Z[i-2];
Y[i] = 2*X[i];
Z[i] = 3*X[i];
}
return X[n];
}```

The running time of f1(n) and f2(n) are

 A θ(n) and θ(n) B θ(2n) and θ(n) C θ(n) and θ(2n) D θ(2n) and θ(2n)
Algorithms        Time-Complexity       GATE 2008
Question 15 Explanation:
Time complexity of f1 is given by
T(n) = T(n-1) + T(n-2) (multiplication by 2 and 3 won't affect complexity as it is a constant time operation)
T(0) = T(1) = 1
The recurrence is similar to Fibonacci series. The time complexity is O(2n)
T(n) = T(n-1) + T(n-2) + c
T(0) = 1
T(n-1) ≈ T(n-2)
But in reality, T(n-1) > T(n-2)
So to find upper bound the expression will be
T(n) = 2T(n-1) + c
= 4T(n-2) + 3c
= 8T(n-3) + 7c
= 2kT(n-k) + (2k-1)c
n-k=0 ⇒ k = n
T(n) = 2nT(0) + (2n-1)c
= 2n + (2n-1)c
⇒ T(n) = O(2n)
The time required for f2 is O(n) (because it consists of only one loop).
 Question 16

Which of the following sorting algorithms has the lowest worst-case complexity?

 A Merge sort B Bubble Sort C Quick Sort D Selection Sort
Algorithms        Time-Complexity       GATE 2007
Question 16 Explanation:
Worst case time complexities are
Merge sort→ O(nlogn)
Bubble sort→ O(n2)
Quick sort→ O(n2)
Selection sort→ O(n2)
 Question 17

Consider the following segment of C-code:

```  int j, n;
j = 1;
while (j <= n)
j = j*2;```

The number of comparisons made in the execution of the loop for any n > 0 is: Base of Log is 2 in all options.

 A ⌈log2n⌉ + 1 B n C ⌈log2n⌉ D ⌊log2n⌋ + 1
Algorithms        Time-Complexity       GATE 2007
Question 17 Explanation:
Let us consider n=6, then
1<=6 (✔️)
2<=6 (✔️)
4<=6 (✔️)
8<=6 (❌)
4 comparisons required
Option A:
[log n]+1
[log 6]+1
3+1 = 4 (✔)
Option B:
n=6 (❌)
Option C:
[log n]
[log 6]=3 (❌)
Option D:
[log2n]+1
[log26]+1 = 2+1 = 3 (❌)
 Question 18

In an unweighted, undirected connected graph, the shortest path from a node S to every other node is computed most efficiently, in terms of time complexity, by

 A Dijkstra’s algorithm starting from S. B Warshall’s algorithm. C Performing a DFS starting from S. D Performing a BFS starting from S.
Algorithms        Time-Complexity       GATE 2007
Question 18 Explanation:
→ Time Complexity of the Dijkstra’s algorithm : It depends on your implementation of Dijkstra's Algorithm. Simple algorithm is given below with Time complexity of O(V2). There are also some time-efficient Algorithms: Graph represented using adjacency list can be reduced to O(E log V) with the help of binary heap.
→ Time Complexity of the Warshall’s algorithm: O(n3). Warshall’s algorithm basically we are using to find all pair shortest path.
→ DFS cannot be used for finding shortest paths.
→ Time Complexity for BFS : O(E+V)
 Question 19

In the following C function, let n >= m.

```int gcd(n,m)
{
if (n%m ==0) return m;
n = n%m;
return gcd(m,n);
} ```

How many recursive calls are made by this function?

 A θ (log2 n) B Ω (n) C θ (log2log2 n) D θ(√n)
Algorithms        Time-Complexity       GATE 2007
Question 19 Explanation:
The given code is to calculate the greatest common divisor (GCD) using Euclidean algorithm.
Then, time complexity is = O(log(a∙b) = O(log(a+b)) = O(log n)
 Question 20

What is the time complexity of the following recursive function:

```int DoSomething (int n) {
if (n <= 2)
return 1;
else
return (DoSomething (floor(sqrt(n))) + n);
}```
 A Ω (n2) B θ (nlog2 n) C θ (log2 n) D θ (log2log2 n)
Algorithms        Time-Complexity       GATE 2007
Question 20 Explanation: Now apply Master's theorem,
a=1, b=2, k=0, p=0
a = bk, so Case-2 will be applied, and p>-1, so Question 21

An array of n numbers is given, where n is an even number. The maximum as well as the minimum of these n numbers needs to be determined. Which of the following is TRUE about the number of comparisons needed?

 A At least 2n - c comparisons, for some constant c, are needed. B At most 1.5n - 2 comparisons are needed. C At least nlog2n comparisons are needed. D None of the above.
Algorithms        Time-Complexity       GATE 2007
Question 21 Explanation:
→ Take first two elements of an array and compare them and mark one of them as max and min for another one. So for now one comparision required.
→ Now take other two element and compare between them and we will get one minimum element which will be compared with min and we will get one maximum element which will be compared with max and accordingly we will mark them. So in total 3 comparisions are required, and we will do this with all (n-1)/2 pairs.
So in total, no. of comparisions required,
= 3(n-2)/2 + 2
= 3n/2 - 3 + 2
= 3n/2 - 1
= 1.5n - 1
 Question 22

Consider the following C code segment:

```   int IsPrime(n)
{
int i,n;
for(i=2;i<=sqrt(n);i++)
if(n%i == 0)
{printf(“Not Prime\n”); return 0;}
return 1;
}```

Let T(n) denotes the number of times the for loop is executed by the program on input n. Which of the following is TRUE?>/p>

 A T(n) = O(√n) and T(n) = Ω(√n) B T(n) = O(√n) and T(n) = Ω(1) C T(n) = O(n) and T(n) = Ω(√n) D None of the above
Algorithms        Time-Complexity       GATE 2007
Question 22 Explanation:
Here in best case time complexity is omega(1) and in worst case time complexity is O(sqrt(n)).
 Question 23

Consider the following C-program fragment in which i, j and n are integer variables.

` for (i = n, j = 0; i >0; i /= 2, j += i);  `

Let val(j) denote the value stored in the variable j after termination of the for loop. Which one of the following is true?

 A val(j) = θ(logn) B val(j) = θ(√n) C val(j) = θ(n) D val(j) = θ(n logn)
Algorithms        Time-Complexity       GATE 2006
Question 23 Explanation:
The loop following series is n/2 + n/4 + n/8 + ... + 1 = Θ(n).
 Question 24

An element in an array X is called a leader if it is greater than all elements to the right of it in X. The best algorithm to find all leaders in an array

 A Solves it in linear time using a left to right pass of the array B Solves it in linear time using a right to left pass of the array C Solves it using divide and conquer in time θ(nlogn) D Solves it in time θ(n2)
Algorithms        Time-Complexity       GATE 2006
Question 24 Explanation:
Solves it in linear time using a right to left pass of the array will takes time complexity is O(n).
 Question 25

Consider the following recurrence: Which one of the following is true?

 A T(n) = θ(log logn) B T(n) = θ(logn) C T(n) = θ(√n) D T(n) = θ(n)
Algorithms        Time-Complexity       GATE 2006
Question 25 Explanation:
T(n) = 2T(√n)+1
Let n =2m
T(2m) = 2T(2m/2) + 1 --------(1)
Let T(2m) = S(m)
Then equation (1) becomes,
S(m) = 2S(m/2) + 1
So now lets apply Master's theorem,
a=2, b=2, k=0
Since, a>bk, so Case 1 applied here But m = log n
O(log n)
 Question 26

The median of n elements can be found in O(n) time. Which one of the following is correct about the complexity of quick sort, in which median is selected as pivot?

 A θ(n) B θ(n logn) C θ(n2) D θ(n3)
Algorithms        Time-Complexity       GATE 2006
Question 26 Explanation:
If we choose pivot element element as median then the recurrence relation will be
T(n) = 2T(n/2) +O(n)
The above recurrence in the form of masters theorem: a=2, b=2, k=1, p=0. Since, a=bk, so Case-2 is applied
and also p > -1, so Question 27

Given two arrays of numbers a1, a2, a3,...an and b1, b2, .. bn where each number is 0 or 1, the fastest algorithm to find the largest span(i, j) such that ai + ai+1, ....aj = bi + bi+1, ... bj. or report that there is not such span,

 A Takes O(3n) and Ω(2n) time if hashing is permitted B Takes O(n3) and Ω(n2.5) time in the key comparison model C Takes θ(n) time and space D Takes O(√n) time only if the sum of the 2n elements is an even number
Algorithms        Time-Complexity       GATE 2006
Question 27 Explanation:
As per the above question, the algorithms follows in
1. Since there are total n elements, maximum sum is n for both arrays.
2. Difference between two sums varies from -n to n. So there are total 2n + 1 possible values of difference.
3. If differences between prefix sums of two arrays become same at two points, then subarrays between these two points have same sum.
 Question 28

The time complexity of computing the transitive closure of a binary relation on a set of n elements is known to be:

 A O(n) B O(n log n) C O(n3/2) D O(n3)
Algorithms        Time-Complexity       GATE 2005
Question 28 Explanation:
First of all, In mathematics the transitive closure of a binary relation R on a set X is defined as the smallest relation on X that contains R and is transitive. More formally, the transitive closure of a binary relation R on a set X is the transitive relation R+ on set X such that R+ contains R and R+ is minimal.
For example, if X is a set of airports and x R y means "there is a direct flight from airport x to airport y" (for x and y in X), then the transitive closure of R on X is the relation R+ such that x R+ y means "it is possible to fly from x to y in one or more flights". Informally, the transitive closure gives you the set of all places you can get to from any starting place.
To find the transitive closure of given relation, we can represent the given relation by the graph such that if x R y then there should be directed edge between x and y in a graph. For this graph, we have to apply modified Floyd_Warshall Algorithm to find the transitive closure of the graph.
The time complexity of this algorithm is O(V3) where V is the number of vertices in the given graph. You can take here V as the number of elements is set i.e., N. Therefore time complexity for the given question is O(n3).
 Question 29

Suppose T(n) = 2T (n/2) + n, T(0) = T(1) = 1

Which one of the following is FALSE?

 A T(n) = O(n2) B T(n) = θ(n log n) C T(n) = Ω(n2) D T(n) = O(n log n)
Algorithms        Time-Complexity       GATE 2005
Question 29 Explanation:
Apply masters theorem to the above example,
a=2, b=2, k=1, p=0.
The recurrence relation result will be O(n logn).
 Question 30

Suppose there are ⌈log n⌉ sorted lists of ⌊n/log n⌋ elements each. The time complexity of producing a sorted list of all these elements is: (Hint: Use a heap data structure)

 A O(nlog log n) B θ(nlog n) C Ω(nlog n) D Ω(n3/2)
Algorithms        Time-Complexity       GATE 2005
Question 30 Explanation:
We can merge P sorted arrays of each size Q in O(PQ LogP)
P = log n
Q = n/log n
∴ Putting values we get,
O(n log log n)
 Question 31

The tightest lower bound on the number of comparisons, in the worst case, for comparison-based sorting is of the order of

 A n B n2 C nlogn D nlog2n
Algorithms        Time-Complexity       GATE 2004
Question 31 Explanation:
In comparison based sorting, the tightest bound of number of comparisons is θ(n log n).
→ Tightest upper bound is (big O).
Tightest lower bound is (big Ω).
There are 31 questions to complete.

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