SP Contest 1

  • Last Updated : 11 Jul, 2018

1
Question 1
How many times is the below loop executed?
 for(int i=0; i < n; i++)
 {
    for(int j=0; j < (2*i); j+=(i/2))
    {
	cout<<"Hello Geeks";
    }
 }
A
O(n)
B
Infinite times
C
O(n2)
D
O(nlogn)
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Question 1 Explanation: 
At the second iteration of the outer loop, i.e. when i = 1, the inner loop will become an infinite loop as the increment condition is j = j+(i/2) and for i = 1, i/2 = 0.
Question 2
The Postorder traversal of a Binary Search Tree is {35, 30, 45, 40, 70, 85, 90, 80, 50}. What is its Preorder traversal?
A
50, 40, 30, 35, 45, 80, 70, 90, 85
B
50, 40, 45, 30, 35 ,80 , 90, 85, 70
C
50, 80, 90, 85, 70, 40, 45, 30, 35
D
30, 35, 40, 45, 50, 70, 80, 85, 90
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Question 2 Explanation: 
You may create a BST from it's given post order traversal and then find it's preorder traversal. Please refer to the post on Construct a Binary Search Tree from given postorder.
Question 3
Which statement is true about initializing a variable to a default value (i.e. integer to 0, boolean to false) using the default constructor?
A
It initializes the data member variables to default values in C++ but not in JAVA.
B
It initializes the data member variables to default values neither in C++ nor in JAVA.
C
It initializes the data member variables to default values in C++ and JAVA both.
D
It initializes the data member variables to default values in JAVA but not in C++.
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Question 4
Below are some operators separated by comma in C++. Which of these operators cannot be overloaded?
!, ::, ->*, ~, .*, sizeof, new
A
->*, .*, sizeof
B
.*, ~, new
C
::, sizeof, !
D
sizeof, .*, ::
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Question 5
What is the output of below C program?
#include <stdio.h>

void print(int c){
    
    if (c < 0) {
        return;
    }
    
    printf("%d ", c);
    c--;
    print(c);
    
    c++;
    
    printf("%d ", c);

}

int main() {
   int c = 5;
   print(c);
   return 0;	
}
A
5 4 3 2 1 0 0 0 1 2 3 4
B
5 4 3 2 1 0 0 1 2 3 4 5
C
1 2 3 4 5 0 0 5 4 3 2 1
D
5 4 3 2 1 1 0 1 2 3 4 5
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Question 6
A can finish an IT project in 15 days, B can finish the same project in 12 days. In how many days, both of them can finish 60% of the project? It may be assumed that working together does not impact their productivity.
A
2 days
B
4 days
C
6 days
D
7 days
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Question 6 Explanation: 
A finish the work in one day = 1 / 15 B finish the work in one day = 1 / 12 If A and B work together to finish the project, then one day's work = 1 /15 + 1 / 12. Then 60% of work can be finished: (1 /15 + 1 / 12 ) * x = 6 / 10. X = 4 days.
Question 7
How many numbers of 6 digits are possible with the property that the sum of their digits is 5?
A
96
B
104
C
120
D
126
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Question 7 Explanation: 
There are 6 digits and their sum is 5. So,
a + b + c + d + e + f = 5
where,
1 ≤ a ≤ 9, and 0 ≤ b, c, d, e, f ≤ 9
Therefore,
(a'+1) + b + c + d + e + f = 5
a' + b + c + d + e + f = 4
Hence,
(4+5)C5 = 9C5 = 126
So, option (D) is correct. Refer method-2: Indistinguishable balls and Distinguishable boxes .
Question 8
Find the missing term in the below sequence:
1, 1, 4, 25, 196, __ , 17424
A
1561
B
1764
C
1600
D
None of these
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Question 8 Explanation: 
Each term is the square of numbers of Catalan Series
Question 9
Here are the two concurrent process A, B with respective codes: Code A:
while (true) // infinite condition
{
    M :____;
    printf("%c", b);
    printf("%c", b);
    N:____;
}
Code B:
while (true) // infinite condition
{
    O:____;
    printf("%c", a);
    printf("%c", a);
    P:____;
}
What should be the binary semaphore operation on M, N, O, P respectively and what must be the initial values of semaphore X, Y in order to get the output bbaabbaabbaa . . . ? Where P is down and V is up operation respectively.
A
M = P(Y), N = V(X), O = P(X), P = V(Y); X = 0, Y = 1;
B
M = P(Y), N = V(X), O = P(X), P = P(Y); X = Y = 1;
C
M = P(Y), N = V(Y), O = P(X), P = V(X); X = 1, Y = 0;
D
M = P(Y), N = V(Y), O = P(X), P = V(X); X = Y = 1;
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Question 9 Explanation: 
In semaphore up is always a successful operation but down is not always successful. In following concurrent process Operations are: A: code
while (true) // infinite condition
{
M :P(Y); // Y become 0 successful down operation.
printf("%c", b);
printf("%c", b);
N:V(X); // X become 1 successful up operation.
}
B code:
while (true) // infinite condition
{
O:P(X); // X  become 0 successful down operation.
printf("%c", a);
printf("%c", a);
P:V(Y); // Ybecome 1 successful up operation.
}
Here all operation are successful with intial values of X and Y are 0 and 1 respectively. So, option (A) is correct.
There are 9 questions to complete.
1
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