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# Solved Placement Papers of Societe Generale

83.25K

## Total Qs: 224+

NA
SHSTTON
55
Solv. Corr.
214
Solv. In. Corr.
269
Attempted
0 M:0 S
Avg. Time

81 / 224

Choose the correct option.

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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NA
SHSTTON
194
Solv. Corr.
67
Solv. In. Corr.
261
Attempted
0 M:0 S
Avg. Time

82 / 224

Choose the correct option.

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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NA
SHSTTON
333
Solv. Corr.
474
Solv. In. Corr.
807
Attempted
0 M:41 S
Avg. Time

83 / 224

Choose the correct option.

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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NA
SHSTTON
254
Solv. Corr.
430
Solv. In. Corr.
684
Attempted
0 M:28 S
Avg. Time

84 / 224

Choose the correct option.

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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NA
SHSTTON
500
Solv. Corr.
227
Solv. In. Corr.
727
Attempted
0 M:14 S
Avg. Time

85 / 224

Choose the correct option.

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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NA
SHSTTON
110
Solv. Corr.
216
Solv. In. Corr.
326
Attempted
0 M:0 S
Avg. Time

86 / 224

Choose the correct option.

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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NA
SHSTTON
133
Solv. Corr.
161
Solv. In. Corr.
294
Attempted
0 M:0 S
Avg. Time

87 / 224

Choose the correct option.

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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NA
SHSTTON
434
Solv. Corr.
429
Solv. In. Corr.
863
Attempted
1 M:15 S
Avg. Time

88 / 224

Which one of the following statements about the function Process Array is CORRECT?
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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NA
SHSTTON
416
Solv. Corr.
349
Solv. In. Corr.
765
Attempted
0 M:0 S
Avg. Time

89 / 224

Choose the correct option.

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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NA
SHSTTON
166
Solv. Corr.
230
Solv. In. Corr.
396
Attempted
0 M:4 S
Avg. Time

90 / 224

Choose the correct option.

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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Here is the list of questions asked in Solved Placement Papers of Societe Generale Societe Generale aptitude questions. Practice Societe Generale Written Test Papers with Solutions and take Q4Interview Societe Generale Online Test Questions to crack Societe Generale written round test. Overall the level of the Societe Generale Online Assessment Test is moderate. Only those candidates who clear the written exam will qualify for the next round, so practic all the questions here and take all the free tests before going for final selection process of Societe Generale