Solved Placement Papers of Societe Generale
81 / 224
In what way the Symmetry Sparse Matrix can be stored efficiently?
AAdjacency List
BBinary tree
CHash table
DHeap
Answer: Option B
Explanation:Since Symmetry Sparse Matrix arises as the adjacency matrix of the undirected graph. Hence it can be stored efficiently as an adjacency list.
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82 / 224
If column-major order is used, how is the following matrix stored in memory?
Aadgbehcfi
Babcdefghi
Ccfibehadg
Dihgfedcba
Answer: Option A
Explanation:It starts with the first element and continues in the same column until the end of column is reached and then proceeds with the next column. Fortran follows column-major order.
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83 / 224
If the array is already sorted, then the running time for merge sort is: ?
AO(1)
BO(n*log n)
CO(n)
DO(n^2)
Answer: Option B
Explanation:Here is no explanation for this answer
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84 / 224
quicksort algorithm is used to sort an array of N elements. If all the N values w
complexity of quicksort that uses first element as the pivot. ?
AO(N*N)
BO(1)
CO(N)
DO(N log N)
Answer: Option A
Explanation:Here is no explanation for this answer
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85 / 224
The complexity of merge sort algorithm is
AO(n)
BO(log n)
CO(n2)
DO(n log n)
Answer: Option D
Explanation:Here is no explanation for this answer
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86 / 224
What is the time taken to delete a minimum element in a leftist heap?
AO(N)
BO(N log N
CO(log N)
DO(M log N)
Answer: Option C
Explanation:The time taken to delete a minimum element in a leftist heap is mathematically found to be O(log N).
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87 / 224
The amortized time efficiency for performing deletion of a minimum element is?
AO(N)
BO(log N)
CO(N2)
DO(M log N)
Answer: Option B
Explanation:The amortized time efficiency for performing deletion of a minimum element is mathematically found to be O(log N).
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88 / 224
Consider the C function given below. Assume the array listA contains (n>0) elements, sorted in ascending order.
int Process array (int * list A, int x, int n)
{
int i, j, k;
i =0;j=n-1;
do {
k = (i+j)/2;
if (x<=list A[k])
j=k-1;
if (list A[k] <=x)
i =k+1;
} while (i <=j);
if (list A[k] == x)
return (k);
else
return -1;
}
AIt will run into an infinite loop when x is not in listA
BIt is an implementation of binary search.
CIt will always find the maximum element in listA.
DIt will return -1 even when x is present in listA.
Answer: Option B
Explanation:Here is no explanation for this answer
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89 / 224
In worst case Quick Sort has order
AO(n log n)
B\(O\left(n^{2}\right)\)
CO(log n)
D\(O\left(\frac{n^{2}}{4}\right)\)
Answer: Option B
Explanation:Here is no explanation for this answer
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90 / 224
Select the code snippet which returns the top of the stack.
I. public int top()
{
if(q1.size()>0)
{
return q1.poll();
}
else if(q2.size()>0)
{
return q2.poll();
}
return 0;
}
II. public int top()
{
if(q1.size()==0)
{
return q1.peek();
}
else if(q2.size()==0)
{
return q2.peek();
}
return 0;
}
III. public int top()
{
if(q1.size()>0)
{
return q1.peek();
}
else if(q2.size()>0)
{
return q2.peek();
}
return 0;
}
IV. public int top()
{
if(q1.size()>0)
{
return q2.peek();
}
else if(q2.size()>0)
{
return q1.peek();
}
return 0;
}
AI
BII
CIII
DIV
Answer: Option C
Explanation:Assuming its a push costly implementation, the top of the stack will be in the front end of the queue, note that peek() just returns the front element, while poll() removes the front element from the queue.
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