Sopra Steria Coding Questions and Answers

April 28, 2023

Sopra Steria Coding Questions with Solutions

In this page, you will find out Sopra Steria Coding Questions and Answers asked in Online Assessments and Technical Interviews involved in the Recruitment Process of the Company.

Explore this page for getting more details like the Job Profile, CTC Offered, Service Agreement, Eligibility Criteria, etc of the company.

About Sopra Steria :

Sopra Steria is a leading European Tech company that is mainly popular for its Consulting Services, Software Development – Support Services, and other digital services. Their vision is to promote Digital Transformation of all Industries whether it is Large scale or small-scale ventures.

The company is also having its own subsidiary called Sopra Banking Software, which is known for its Financial Services and Solutions. They generally provide Banking Related Software, Support, and help many Financial Institutions to deploy an efficient Banking System.

About Sopra Steria Recruitment Process :

The Sopra Steria Recruitment Process consists of the following steps :

Online Assessment [ MCQs Based Aptitude + Technical ]
Technical Interview
HR Interview

We have mentioned further details of the Sopra Steria Recruitment Process in the following Tabular Form

Sopra Steria	Related Information
Position :	Engineer Trainee
Course :	B.E / B.Tech – CS, IT, or M.C.A. Eligible Batch – 2023
Eligibility Criteria / Academic Qualification Required :	Minimum 70 % or equivalent CGPA required in 10th / 12th. Minimum 60 % in Graduation No Current Backlogs.
Offered CTC :	₹ 6 L.P.A
Selection Process :	Online Aptitude & Technical Assessment Technical Interview HR Interview
Joining Location :	Noida Chennai Bangalore

ABOUT RECRUITMENT EXAM :

Exam Duration: 2 Hours

No. Of Questions: 100

FACTS ABOUT RECRUITMENT PROCESS :

The pattern of the Recruitment Process can be changed according to different colleges and Job Profile.

For the latest updates please check the official website of Sopra Steria.

FACTS ABOUT SERVICE AGREEMENT :

There will be 3 Years Service Agreement.

Bond Penalty: ₹ 2 Lakh

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Sample Sopra Steria Coding Questions with Solutions

Question 1: Consecutive Prime Sum

Problem Description

Question – : Some prime numbers can be expressed as a sum of other consecutive prime numbers.

For example
- 5 = 2 + 3,
- 17 = 2 + 3 + 5 + 7,
- 41 = 2 + 3 + 5 + 7 + 11 + 13.
  Your task is to find out how many prime numbers which satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.

Write code to find out the number of prime numbers that satisfy the above-mentioned property in a given range.

Input Format: First line contains a number N

Output Format: Print the total number of all such prime numbers which are less than or equal to N.

Constraints: 2<N<=12,000,000,000

C++

JAVA

Python

Run

#include <stdio.h>  
 
int prime(int b);
int main()
{
 int i,j,n,cnt,a[25],c,sum=0,count=0,k=0;
 printf("Enter a number : ");
 scanf("%d",&n);
 for(i=2;i<=n;i++)
 {
 	cnt=1;
 	for(j=2;j<=n/2;j++)
 	{
     	if(i%j==0)
     	cnt=0;
 	}
 	if(cnt==1)
 	{
    	a[k]=i;
    	k++;
    	}
 }
 for(i=0;i<k;i++)
 {
 	sum=sum+a[i];
	c= prime(sum);
	if(c==1)
	count++;
 }
 printf(" %d",count);
 return 0;
}

int prime(int b)
{
	int j,cnt;
	cnt=1;
 	for(j=2;j<=b/2;j++)
 	{
     	if(b%j==0)
     	cnt=0;
 	}
 	if(cnt==0)
 	return 1;
 	else
 	return 0;
}

Output
Enter a number : 43
4

C++

Run

#include <iostream> 
using namespace std;

int prime(int b)
{
    int j,cnt;
    cnt=1;
     for(j=2;j<=b/2;j++)
     {
         if(b%j==0)
         cnt=0;
     }
     if(cnt==0)
     return 1;
     else
     return 0;
}
int main() 
{
 int i,j,n,cnt,a[25],c,sum=0,count=0,k=0;
 cout<<"Enter a number : "; 
cin>>n;
 for(i=2;i<=n;i++)
 {
     cnt=1;
     for(j=2;j<=n/2;j++)
     {
         if(i%j==0)
         cnt=0;
     }
     if(cnt==1)
     {
        a[k]=i;
        k++;
        }
 }
 for(i=0;i<k;i++)
 {
     sum=sum+a[i];
    c= prime(sum);
    if(c==1)
    count++;
 }
 cout<<count;
 return 0;
}

Output
Enter a number : 5
1

JAVA

Run

import java.util.Scanner;
 class Main {
 
    static int prime(int b) {
        int j,cnt;
        cnt=1;
        for (j = 2; j <= b/2; j++) {
            if(b%j==0)
                cnt=0;
        }
        if(cnt==0)
        return 1;
        else
            return 0;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int i,j,n=0,cnt,c=0,sum=0,count=0,k=0;
        Main t = new Main();
        int[] a = new int[25];
        System.out.println("Enter no");
        n = sc.nextInt();
        for (i = 2; i <=n ; i++) {
            cnt=1;
            for (j = 2; j <= n/2; j++) {
                if(i%j==0)
                    cnt=0;
            }
            if(cnt==1) {
                a[k]=i;
                k++;
            }
            
        }
        for (i = 0; i < k; i++) {
            sum=sum+a[i];
            c=t.prime(sum);
            if(c==1)
                count++;
        }
        System.out.println(count);
    }

}

Output
43
4

Python

Run

num = int(input())
arr = []
sum = 0
count = 0
if num > 1:
    for i in range(2, num + 2):
        for j in range(2, i):
            if i % j == 0:
                break
        else:
            arr.append(i)
def is_prime(sum):
    for i in range(2, (sum // 2) +2):
        if sum % i == 0:
            return False
        else:
            return True
for i in range(0, len(arr)):
    sum = sum + arr[i]
    if sum <= num:
        if is_prime(sum):
            count = count + 1

print(count)

Output
20
2

Question 2: Self Sufficient

Problem Statement – Abhijeet is one of those students who tries to get his own money by part time jobs in various places to fill up the expenses for buying books. He is not placed in one place, so what he does, he tries to allocate how much the book he needs will cost, and then work to earn that much money only. He works and then buys the book respectively. Sometimes he gets more money than he needs so the money is saved for the next book. Sometimes he doesn’t. In that time, if he has stored money from previous books, he can afford it, otherwise he needs money from his parents.

Now His parents go to work and he can’t contact them amid a day. You are his friend, and you have to find how much money minimum he can borrow from his parents so that he can buy all the books.

He can Buy the book in any order.

Function Description:

Complete the function with the following parameters:

Name	Type	Description
N	Integer	How many Books he has to buy that day.
EarnArray[ ]	Integer array	Array of his earnings for the ith book
CostArray[ ]	Integer array	Array of the actual cost of the ith book.

Constraints:

1 <= N <= 10^3
1 <= EarnArray[i] <= 10^3
1 <= CostArray[i] <= 10^3

Input Format:

First line contains N.
Second N lines contain The ith earning for the ith book.
After that N lines contain The cost of the ith book.

Output Format: The minimum money he needs to cover the total expense.

Sample Input 1:

[3 4 2]

[5 3 4]

Sample Output 1:

Explanation:

At first he buys the 2nd book, which costs 3 rupees, so he saves 1 rupee. Then he buys the 1st book, that takes 2 rupees more. So he spends his stored 1 rupee and hence he needs 1 rupee more. Then he buys the last book.

C++

Python

Java

C++

Run

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n, ans =0,sum=0;
    cin>>n;
    vector arr1(n),arr2(n);
    for(int i=0;i<n;i++) cin>>arr2[i];
    for(int i=0;i<n;i++) cin>>arr1[i];
    for(int i=0;i<n;i++) arr2[i]-=arr1[i];    sort(arr2.begin(),arr2.end(),greater());
    for(int i=0;i<n;i++)
    {
        sum+=arr2[i];
        if(sum<0)
        {ans+=abs(sum);sum=0;}
    }
    cout<<ans;
}

Python

Run

n=int(input())
L1=[0] *n
L2=[0]*n
for i in range(n):
    L2[i]=int(input())
for i in range(n):
    L1[i]=int(input())
    
for i in range(n):
    L2[i]=L2[i]-L1[i];

L2.sort()
su=0
ans=0
for i in range(n):
    su=su+L2[i]
    if su<0:
        ans = ans + abs(su)
        su=0
        
print(ans)

Java

Run

import java.util.*;
class Main
{ 
  public static void main(String args[])
  {
    Scanner sc = new Scanner (System.in);   
    int n=sc.nextInt();
    int arr1[]= new int[n];
    int arr2[]= new int[n];
    for(int i=0; i<n; i++)
       arr2[i]=sc.nextInt();       
    for(int i=0; i<n; i++)
       arr1[i]=sc.nextInt();       
    for(int i=0; i<n; i++) arr2[i] -= arr1[i]; Arrays.sort(arr2); int sum=0, ans=0; for(int i=n-1; i>=0; i--)
    {
        sum+=arr2[i];
        if(sum<0)
        {
         sum = -sum;
         ans= ans +sum ;
         sum=0;
        }
    }  
    System.out.println(ans);
  }
 }

Question : 3

Problem Statement : Fountains are installed at every position along a one-dimensional garden of length n. Array locations[] represents the coverage limit of these fountains. The ith fountain (where 1sisn) has a coverage limit of locations[i] that can range from the position max((i – locations[i]), 1) to min((i + locations[i]), n ). In other words, the h fountains do not reach outside the boundaries of the garden. In the beginning,all the fountains are switched off. Determine the minimum number of fountains that need to be activated to cover the n length garden by water.

Example :

n = 3
locations[] = {0, 2, 13, then
- For position 1: locations[1] = 0, max((1 – 0),
  - 1) to mini (1+0), 3) gives range = 1 to 1
- For position 2: locations[2] = 2, max((2-2),
  - 1) to min( (2+2), 3) gives range = 1 to 3
- For position 3: locations[3] = 1, max( (3-1),
  - 1) to min( (3+1), 3) gives range = 2 to 3

For the entire length of this garden to be covered, only the fountain at position 2 needs to be activated.

Function Description :

Complete the function fountainActivation in the editor below.

fountainActivation has the following Parameter:

int locations[n]: the fountain locations

Returns :

int: the minimum number of fountains that must be activated

Constraints :

1<_n<_ 10^5
O<_locations[i] <_ mini (n,100) (where 1 <_1<_10^5)

► Input Format For Custom Testing

Sample Case 0 :

Sample Input For Custom Testing :

3 ->locations[] size n = 3
1 ->locations[] [1, 1, 1
1 ->Sample Output

Sample Output :

Explanation :

Here, locations = {1, 1, 13

For position 1: locations[1] = 1, maxi (1 -1), 1) to min((1+1), 3) gives range = 1 to 2
For position 2: locations[2] = 1, max( (2 -1), 1) to min( (2+1), 3) gives range = 1 to 3
For position 3: locations[3] = 1, max((3 -1), 1) to min((3+1), 3) gyes range = 2 to 3

If the 2nd fountain is active, the range from position 7 to 3 will be covered. The total number of fountains needed is 1.

C++

Java

C++

Run

#include<bits/stdc++.h>

#define ll long long

using namespace std;

bool compare(pair < int, int > A, pair < int, int > B)

{

  if (A.first = B.first)

    return A.second < B.second;

  return A.first < B.first;

}

int solve(int location[], int n)

{

  pair < int, int > range[n];

  for (int i = 0; i < n; i++)

  {

    int id = i + 1;

    range[i].first = max(1, id - location[i]);

    range[i].second = min(n, id + location[i]);

  }

  sort(range, range + n, compare);

  int i = 0;

  int ans = 0;

  while (i < n)

  {

    pair < int, int > p = range[i];

    ans++;

    while (i + 1 < n && range[i].first == range[i + 1].first)

    {

      p.second = max(p.second, range[i + 1].second);

      i++;

    }

    //cout<<p.second<<" "<<i<<endl;

    while (i < n && range[i].second <= p.second)

      i++;

    //cout<<p.second<<" "<<i<<endl; } return ans; } int main() { int n; cin >> n;

  int location[n];

  for (int i = 0; i < n; i++) cin >> location[i];

  cout << solve(location, n) << endl;

  return 0;

}

Java

Run

import java.util.*;
class Pair {
  Integer a;
  Integer b;
  Pair() {}

  Pair(Integer a, Integer b) {
    this.a = a;
    this.b = b;
  }

  public Integer getA() {
    return a;
  }

  public void setA(Integer a) {
    this.a = a;
  }

  public Integer getB() {
    return b;
  }

  public void setB(Integer b) {
    this.b = b;
  }
}

class SortingPair implements Comparator {

  @Override
  public int compare(Pair o1, Pair o2) {
    if (o1.getA() == o2.getA()) {
      if (o1.getB() < o2.getB()) {
        return 1;
      } else {
        return 0;
      }
    }

    if (o1.getA() < o2.getA()) {
      return 1;
    } else {
      return 0;
    }
  }
}

public class Solution {

  public static void main(String args[]) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int location[] = new int[n];

    for (int i = 0; i < n; i++) {
      location[i] = sc.nextInt();
    }
    System.out.println(solve(location, n));
  }
  private static int solve(int[] location, int n) {
    Pair range[] = new Pair[n];
    for (int i = 0; i < n; i++) {
      int id = i + 1;
      range[i] = new Pair();
      range[i].setA(Math.max(1, id - location[i]));
      range[i].setB(Math.min(n, id + location[i]));
    }
    Arrays.sort(range, new SortingPair());
    int i = 0, ans = 0;
    while (i < n) {
      Pair p = range[i];
      ans++;

      while (i + 1 < n && range[i].getA() == range[i + 1].getA()) {
        p.b = Math.max(p.getB(), range[i + 1].getB());
        i++;
      }
      while (i < n && range[i].getB() <= p.getB()) {
        i++;
      }
    }
    return ans;
  }
}

Question 4

Problem Statement : Street Lights are installed at every position along a 1-D road of length n.

Locations[] (an array) represents the coverage limit of these lights. The ith light has a coverage limit of locations[i] that can range from the position max((i – locations[i]), 1) to min((i + locations[i]), n ) (Closed intervals). Initially all the lights are switched off. Find the minimum number of fountains that must be switched on to cover the road.

Example :

n = 3

locations[] = {0, 2, 13}then

For position 1: locations[1] = 0, max((1 – 0),

1) to mini (1+0), 3) gives range = 1 to 1

For position 2: locations[2] = 2, max((2-2),

1) to min( (2+2), 3) gives range = 1 to 3

For position 3: locations[3] = 1, max( (3-1),

1) to min( (3+1), 3) gives range = 2 to 3

For the entire length of this road to be covered, only the light at position 2 needs to be activated.

Function Description

Returns

int: the minimum number of street lights that must be activated

Constraints

1<_n<_ 10^5
O<_locations[i] <_ mini (n,100) (where 1 <_1<_

10^5)

► Input Format For Custom Testing

Sample Case 0

Sample Input For Custom Testing

3 ->locations[] size n = 3

1 ->locations[] [1, 1, 1]

1 ->Sample Output

Sample Output

C++

Run

#include 
#define ll long long
using namespace std;
bool compare(pair A, pair B)
{
    if (A.first = B.first)
        return A.second < B.second;
    return A.first < B.first;
}
int solve(int location[], int n)
{
 
    pair range[n];
    for (int i = 0; i < n; i++)
    {
        int id = i + 1;
        range[i].first = max(1, id - location[i]);
        range[i].second = min(n, id + location[i]);
    }
 
    sort(range, range + n, compare);
 
    int i = 0;
    int ans = 0;
    while (i < n)
    {
        pair p = range[i];
        ans++;
        while (i + 1 < n && range[i].first == range[i + 1].first)
        {
            p.second = max(p.second, range[i + 1].second);
            i++;
        }
        //cout<> n;
    int location[n];
    for (int i = 0; i < n; i++)
        cin >> location[i];
 
    cout << solve(location, n) << endl;
    return 0;
}

Question 5

Problem Statement : You just received another bill which you cannot pay because you lack the money.

Unfortunately, this is not the first time to happen, and now you decide to investigate the cause of your constant monetary shortness. The reason is quite obvious: the lion’s share of your money routinely disappears at the entrance of party localities.

You make up your mind to solve the problem where it arises, namely at the parties themselves. You introduce a limit for your party budget and try to have the most possible fun with regard to this limit.

You inquire beforehand about the entrance fee to each party and estimate how much fun you might have there. The list is readily compiled, but how do you actually pick the parties that give you the most fun and do not exceed your budget?

Write a program which finds this optimal set of parties that offer the most fun. Keep in mind that your budget need not necessarily be reached exactly. Achieve the highest possible fun level, and do not spend more money than is absolutely necessary.

Input

The first line of the input specifies your party budget and the number n of parties.
The following n lines contain two numbers each. The first number indicates the entrance fee of each party. Parties cost between 5 and 25 francs. The second number indicates the amount of fun of each party, given as an integer number ranging from 0 to 10.
The budget will not exceed 500 and there will be at most 100 parties. All numbers are separated by a single space.
There are many test cases. Input ends with 0 0.

Output

For each test case your program must output the sum of the entrance fees and the sum of all fun values of an optimal solution. Both numbers must be separated by a single space.

Example

Sample input:
50 10
12 3
5 8
16 9
16 6
10 2
21 9
18 4
12 4
17 8
18 9
50 10
13 8
19 10
16 8
12 9
10 2
12 8
13 5
15 5
11 7
16 2
0 0

Sample output:
50 29
48 32

Python

Java

C++

Python

Run

def knapSack(W, wt, val, n): 
    K = [[0 for x in range(W + 1)] for x in range(n + 1)] 
    
     
    for i in range(n + 1): 
        for w in range(W + 1): 
        	if i == 0 or w == 0: 
    		    K[i][w] = 0
        	elif wt[i-1] <= w: 
        		K[i][w] = max(val[i-1] 
        				+ K[i-1][w-wt[i-1]], 
        					K[i-1][w]) 
        	else: 
        		K[i][w] = K[i-1][w] 

    res = K[n][W] 
    res1=res 
    x=0  
    w = W 
    for i in range(W+1):
        if(K[n][i]==res):
            x=i 
            break
        
    print(x , res1)

b,n=map(int,input().split())
fun=[]
cost=[]
for i in range(n):
    x,y=map(int,input().split())
    fun.append(y)
    cost.append(x) 
    
(knapSack(b,cost,fun,n))

Java

Run

import java.util.*;
class Solution 
{
    public static void main (String[]args) 
    {
        Scanner sc = new Scanner (System.in);
        while (true)
        {
            int budget = sc.nextInt ();
            int n = sc.nextInt ();
        
            if (budget == 0 && n == 0)
                break;
        
            int cost[] = new int[n + 1];
            int fun[] = new int[n + 1];
            int arr[][] = new int[n + 1][budget + 1];

            for (int i = 0; i < n; i++)
        	{
                cost[i] = sc.nextInt ();
                fun[i] = sc.nextInt ();
            } 
            for (int i = 0; i <= n; i++)
                for (int j = 0; j <= budget; j++)
                    arr[i][j] = 0;
        
            for (int i = 1; i <= n; i++)
    	    {
                for (int j = 1; j <= budget; j++)
                    if (cost[i - 1] <= j)
                        arr[i][j] =Math.max (fun[i - 1] + arr[i - 1][j - cost[i - 1]],arr[i - 1][j]);
        	        else
                        arr[i][j] = arr[i - 1][j];
            }
            int c = 0;
            for (int i = 0; i <= budget; i++)
        	{       
                if (arr[n][i] == arr[n][budget])
    	        {
                    c = i;
                    break;
                }       
            }
            System.out.println (c + " " + arr[n][budget]);
        }
    }
}

C++

Run

#include
using namespace std;

int main() {
  int w, n;

  x: cin >> w >> n;

  if (w == 0 and n == 0)
    goto r;

  else {

    int ct[n], val[n];

    for (int i = 0; i < n; i++) {
      cin >> ct[i] >> val[i];
    }

    int t[n + 1][w + 1];

    for (int i = 0; i <= n; i++) {

      for (int j = 0; j <= w; j++)
        t[i][j] = 0;
    }

    for (int i = 1; i <= n; i++) {

      for (int j = 1; j <= w; j++) {

        if (ct[i - 1] <= j)
          t[i][j] = max(val[i - 1] + t[i - 1][j - ct[i - 1]], t[i - 1][j]);

        else
          t[i][j] = t[i - 1][j];
      }
    }

    int cost = 0;

    for (int i = 0; i <= w; i++) {

      if (t[n][i] == t[n][w]) {
        cost = i;
        break;
      }
    }

    cout << cost << " " << t[n][w] << endl;

    goto x;
  }

  r: return 0;

}

FAQs on Sopra Steria Coding Questions

Question 1: Is Coding questions asked in Sopra Steria Recruitment Process?

Yes, coding questions were asked in the Sopra Steria’s 1st Round of recruitment process (the online assessment) and second round (the technical interview).

Question 2: What kind of skills does Sopra Steria look for in job applicants in India?

Sopra Steria typically looks for candidates with technical skills and qualifications in areas such as software development, cloud computing, cybersecurity, and data analytics. They also value soft skills such as teamwork, communication, and problem-solving abilities.

Question 3: In what field does Sopra Steria works?

Sopra Steria works in a variety of fields and industries, including banking and financial services, insurance, healthcare, government and public sector, energy and utilities, aerospace and defense, and transportation. The company also provides business process services, outsourcing solutions, and enterprise resource planning (ERP) services.

Question 4: How long does the recruitment process take at Sopra Steria?

The recruitment process at Sopra Steria in India can vary depending on the role and the number of candidates being considered. On average, the process can take anywhere from a few weeks to several months.

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Sopra Steria Coding Questions and Answers

Sopra Steria Coding Questions with Solutions

About Sopra Steria :

About Sopra Steria Recruitment Process :

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Sample Sopra Steria Coding Questions with Solutions

Question 1: Consecutive Prime Sum

Question 2: Self Sufficient

Question : 3

FAQs on Sopra Steria Coding Questions

Question 1: Is Coding questions asked in Sopra Steria Recruitment Process?

Question 2: What kind of skills does Sopra Steria look for in job applicants in India?

Question 3: In what field does Sopra Steria works?

Question 4: How long does the recruitment process take at Sopra Steria?

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