Sopra Banking Coding Questions And Answers

September 19, 2023

Sopra Banking Coding Questions with Solution 2023

Sopra Banking Coding Questions and Answers from all the years that have been asked in Coding Round during hiring Process. This page also includes other details regarding hiring process like Job Description, Eligibility Criteria and Steps in Selection Process for specific Job-Profile in the company.

About Sopra Banking Software Ltd.

The parent company of Sopra Banking is Sopra Steria, which offers cloud-based products and solutions, specialised management systems, and support systems.

Sopra Banking Software Ltd. specialises in banking management systems. They provide banking software to control various tasks performed in banks. They also provide some specific API – Oriented softwares for various Financial organisations to control Banking and Money Related workflow in a specific way.

Sopra Banking Recruitment Process

The Online Assessment consists of following sections :

Online Coding Assessment [ 50 Min ]
Interview Round [ Technical + HR ]

For your reference, we have given further details for specific Job Profile offered by the company in Tabular Form :

Sopra Banking	Important Information
Position :	Software Engineer
Course :	Eligible Batch – 2023 B.Tech – CSE / IT
Eligibility Criteria / Academic Qualification Required :	Minimum 60 % throughout 10th and 12th B.Tech – Minimum 70 %
Cost to Company (CTC)	8.5 LPA
Selection Process :	Coding Test 1 on Hacker Earth. Test 2 Final Interview
Joining Location :	Noida Pune

Important Note :

There is a 2 Years service agreement with 2 Lakhs penalty in case of Bond Breach.

And the details given above are according to the On-Campus Hiring Notification.

Any details can be made for different colleges and Off-Campus Hiring.

Details of Sopra Banking Online Coding Assessment :

Points	Relevant Information
Coding Test	50 Minutes
Number Of Questions	2 – 3 Questions

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Sample Sopra Banking Practice Coding Questions and Answers

Question 1 : Borrow Number

Problem Statement :

You have two numbers number1 and number2, your job is to check the number of borrow operations needed for subtraction of number1 from number2. If the subtraction is not possible then return the string not possible.

Example :

754
658

Answer :
2

654
666

Answer :
Not possible

C++

JAVA

Python

C++

Run

#include &bits/stdc++.h>

using namespace std;

int main() {
    string s1, s2;
    int c = 0, f = 0;
    cin >> s1 >> s2;
    if (stoi(s1) < stoi(s2)) {
        cout << "Impossible";
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    for (int i = 0; i < s1.length(); i++)
        if (s1[i] < s2[i]) {
            f = 1;
            c++;
        }
    else if (s1[i] == s2[i]) {
        if (f == 1) {
            c++;
        }
        f = 0;
    } else f = 0;
    cout << c;
}

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int number1 = sc.nextInt();
        int number2 = sc.nextInt();
        int count = 0;
        if (number1 < number2) {
            System.out.println("Not possible");
        } else {
            boolean flag = false;
            while (number1 != 0 && number2 != 0) {
                int temp1 = 0;
                int temp2 = number2 % 10;
                if (flag)
                    temp1 = number1 % 10 - 1;
                else
                    temp1 = number1 % 10;
                if (temp1 < temp2) {
                    count++;
                    flag = true;
                } else
                    flag = false;
                number1 = number1 / 10;
                number2 = number2 / 10;
            }
            System.out.println(count);
        }

    }

}

Python

Run

number1 = int(input())
number2 = int(input())
count = 0
if number1 < number2:
    print("Not possible")
else:
    flag = 0
    while number1 != 0 and number2 != 0:
        temp1 = 0
        temp2 = number2 % 10
        if flag:
            temp1 = number1 % 10 - 1
        else:
            temp1 = number1 % 10
        if temp1 < temp2:
            count += 1
            flag = 1
        else:
            flag = 0
        number1 = number1 // 10
        number2 = number2 // 10
    print(count)

Question 2 : Iron Magnet

Problem Statement :

We know iron behaves like magnets when all the north and the south sides are placed accordingly in a balanced way while the north comes first and then the south. Now if you are given the distribution of the north and the south poles inside the iron metal piece, you have to say how many swaps is needed to turn it into an iron, if possible, otherwise print -1.

Input Format: A string consisting N and S as north and south poles.
Output Format:
An integer denoting the number of poles will be needed.

Sample Input:
SNNSN

Output:
3

Output Description:
After we balance the iron in the way NSNNSNSS, we will get a piece of metal that will be balanced as a magnet.

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;

int main() {
    string s;
    int ans = 0, k = 0;
    cin >> s;
    for (auto i: s) {
        if (i == 'N') k++;
        else k--;
        if (k < 0) {
            k++;
            ans++;
        }
    }
    ans += k;
    cout << ans;
}

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        int res = 0, j = 0;
        for (int i = 0; i < str.length(); i++) {
            if (str.charAt(i) == 'N')
                j++;
            else
                j--;
            if (j < 0) {
                j++;
                res++;
            }
        }
        res = res + j;
        System.out.println(res);
    }
}

Python

Run

s = input()
ans = 0
if s[0] != "N":
    ans += 1
    s = "N" + s
print(ans + abs(s.count("N") - s.count("S")))

Question 3 : Array Subarray

Problem Statement: You are given an array, You have to choose a contiguous subarray of length ‘k’, and find the minimum of that segment, return the maximum of those minimums.

Sample input :

1 → Length of segment x =1
5 → size of space n = 5
1 → space = [ 1,2,3,1,2]
2
3
1
2

Sample output :

Explanation :

The subarrays of size x = 1 are [1],[2],[3],[1], and [2],Because each subarray only contains 1 element, each value is minimal with respect to the subarray it is in. The maximum of these values is 3. Therefore, the answer is 3

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;
vector < int > arr;
int prevmin=-1;
int flag=0;
int x,n,q;

int sorting(int start,int end)
{
    if(start+1==n) {start=0;end=end-n;}
    if(start==end) return arr[start];
    return min(arr[start],sorting(start+1,end));
}

int func(int start,int end)
{
    if(flag==0) {flag++;return prevmin=sorting(start,end);}
    if(arr[start-1]==prevmin) return prevmin;
    return prevmin=(arr[end] <= prevmin)?prevmin:sorting(start,end); } int main() { cin >> x >> n;
    int ans=0;
    for(int i=0;i < n;i++) {cin >> q;arr.push_back(q);}
    for(int i=0;i < n;i++)
    {
       ans=max(ans,func(i,i+x-1));
    }
    cout << ans;
}

JAVA

Run

import java.util.*;
public class DiskSpace
{
  public static void main(String[] args)
{
  Scanner sc=new Scanner(System.in);
  int x=sc.nextInt();
  int n=sc.nextInt();
  int arr[]=new int[n];
  for(int i=0;i < n;i++)
   arr[i]=sc.nextInt();

  int min=Integer.MAX_VALUE;
  int max=Integer.MIN_VALUE;

  for(int i=0;i <= n-x;i++)
 {
    min=Integer.MAX_VALUE;
    for(int j=i;j < (i+x);j++)
     min=Math.min(min,arr[j]);
   max=Math.max(min,max);     
 }
 System.out.println(max);
}
}

Python

Run

n=int(input())
arr=[]
for i in range(n):
    arr.append(int(input()))
s,a=0,0
for i in arr:
    s=s+i   
    if(s < 1):
        a=a+(-1*s) 
        s=0 
print(a)

Question 4 : Maximum Toys

In a toy shop there are a number of toys presented with several various – priced toys in a specific order. You have a limited budget and would like to select the greatest number of consecutive toys that fit within the budget. Given prices of the toys and your budget, what is the maximum number of toys that can be purchased for your child?

Example:

prices=[1,4,5,3,2,1,6]

money=6

All sub arrays that sum to less than or equal to 6 .

length 1: [1] [4] [5] [3] [2] [1] [6]

length 2: [1,4] [3,2] [2,1]

length 3: [3,2,1]

The longest of these or the maximum number of toys that can be purchased is 3.

Function description

Complete the function getMaxToys in the editor below

getMaxToys has the following parameters:

int prices[n] : the prices of the various toys

int money: the amount of money you can spend on toys

Returns

Int the maximum number of toys you can purchase

Constraints

1<=n<=10^5

1<=price[i]<=100

1<=money<=10^6

Sample case

Sample input 0

7->n=7
1-> price[]=[1,4,5,3,2,1,6]
4
5
3
2
1
6
6 ->money=6

Sample Output

Java

Python

Java

Run

import java.util.*;
public class Main
{
  public static int getMaxToys (int prices[], int money)
  {
    int sum = 0;
    int count = 0;
    int max = 0;
    for (int i = 0; i < prices.length; i++)
      {
	sum = count = 0;
	for (int j = i; j < prices.length; j++)
	  {
	    if ((sum + prices[j]) <= money)
	      {
		sum = sum + prices[j];
		count = count + 1;
		max = Math.max (count, max);
	      }
	    else
	      {
		count = 0;
		sum = 0;
		break;
	      }
	  }
      }
    return max;
  }
  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    int n = sc.nextInt ();
    int prices[] = new int[n];
    for (int i = 0; i < n; i++)
      prices[i] = sc.nextInt ();
    int money = sc.nextInt ();
    System.out.println (getMaxToys (prices, money));
  }
}

Python

Run

def getMaxToys(n,arr,money):
    L,H=0,n-1
    while(L<=H):
        if(sum(arr[L:H+1])<=money): break else: if arr[L]>arr[H]:
                L+=1 
            else:
                H-=1 
    return (H-L+1)
n=int(input())
arr=[]
for i in range(n):
    arr.append(int(input()))
money=int(input())
print(getMaxToys(n,arr,money))

Question 5 : Maximize Earnings (R->Hard)

Problem Statement :

A company has a list of jobs to perform. Each job has a start time, end time and profit value. The manager has asked his employee Anirudh to pick jobs of his choice. Anirudh being greedy wants to select jobs for him in such a way that would maximize his earnings.
Given a list of jobs how many jobs and total earning are left for other employees once Anirudh
Picks jobs of his choice.
Note: Anirudh can perform only one job at a time.

Input format:

Each Job has 3 pieces of info – Start Time,End Time and Profit
The first line contains the number of Jobs for the day. Say ‘n’. So there will be ‘3n lines
following as each job has 3 lines.
Each of the next ‘3n’ lines contains jobs in the following format:
start_time
end-time
Profit
start-time and end-time are in HHMM 24HRS format
i.e. 9am is 0900 and 9PM is 2100

Constraints :

The number of jobs in the day i.e’ is less.
than 10000
0<_n<_10000
start-time is always less than end time.

Output format :

Program should return an array of 2 integers where
1st one is number of jobs left and earnings of other employees

Sample Input 1 :
4
0200
0300
10
0400
0700
20
0300
0800
30
0900
1000
50

Sample Output 1:
1
20

Sample Explanation 1

CHooses 1st, 3rd and 4th job cause they don’t overlap. So only second job is remaining.

Sample Input 2:

5
0805
0830
100
0835
0900
100
0905
0930
100
0935
1000
100
1005
1030
100

Sample output 2:

0
0

Sample Explanation 2:
Anirudh can work on all appointments as there are none overlapping.
Hence 0 appointments and 0 earnings for other employees.

C++

JAVA

C++

Run

#include <bits/stdc++.h>

using namespace std;
class job {
    public: int st,
    ed,
    cost;
};
int getTime(string s) {
    int hr = (s[0] - '0') * 10 + (s[1] - '0');
    int min = (s[2] - '0') * 10 + (s[3] - '0');

    return hr * 60 + min;
}
bool compare(job A, job B) {
    return A.ed < B.ed;
}
int searchJob(job arr[], int st, int ed, int key) {
    int ans = -1;
    while (st <= ed) {
        int mid = (st + ed) / 2;
        if (arr[mid].ed <= key) {
            ans = mid;
            st = mid + 1;
        } else {
            ed = mid - 1;
        }
    }
    return ans;
}
pair < int, int > solve(job arr[], int n) {
    int dp[n] = {
        0
    };
    int numOfJobs[n] = {
        0
    };

    dp[0] = arr[0].cost;
    numOfJobs[0] = 1;

    for (int i = 1; i < n; i++) { int cur = arr[i].cost; int num = 1; int idx = searchJob(arr, 0, i - 1, arr[i].st); if (idx != cur) { cur += dp[idx]; num += numOfJobs[idx]; } if (cur > dp[i - 1]) {
            dp[i] = cur;
            numOfJobs[i] = num;
        } else {
            dp[i] = dp[i - 1];
            numOfJobs[i] = numOfJobs[i - 1];
        }
    }
    return {
        numOfJobs[n - 1],
        dp[n - 1]
    };
}
int main() {
    int n;
    cin >> n;

    job arr[n];
    int cost;
    string st, ed;
    int total = 0;

    for (int i = 0; i < n; i++) { cin >> st >> ed >> cost;
        arr[i].st = getTime(st);
        arr[i].ed = getTime(ed);
        arr[i].cost = cost;
        total += cost;
    }
    sort(arr, arr + n, compare);

    pair < int, int > res = solve(arr, n);

    cout << n - res.first << endl;
    cout << total - res.second << endl;

    return 0;
}

JAVA

Run

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
class Job {
    public Integer st;
    public Integer ed;
    public Integer cost;

    public Job() {
        super();
    }

    public Job(Integer st, Integer ed, Integer cost) {
        super();
        this.st = st;
        this.ed = ed;
        this.cost = cost;
    }
}
class Pair {
    public Integer first;
    public Integer second;
    public Pair() {
        super();
    }
    public Pair(Integer first, Integer second) {
        super();
        this.first = first;
        this.second = second;
    }
}
class SortingJobs implements Comparator < Job > {

    @Override
    public int compare(Job o1, Job o2) {
        if (o1.ed < o2.ed) {
            return 1;
        } else {
            return 0;
        }
    }
}
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        Job[] arr = new Job[n];
        int cost;
        String st, ed;
        int total = 0;
        for (int i = 0; i < n; i++) {
            st = sc.next();
            ed = sc.next();
            if (st.length() < 4) {
                while (st.length() != 4) {
                    st += "0";
                }
            }
            if (ed.length() < 4) {
                while (ed.length() != 4) {
                    ed += "0";
                }
            }
            cost = sc.nextInt();
            arr[i] = new Job();
            arr[i].st = getTime(st);
            arr[i].ed = getTime(ed);
            arr[i].cost = cost;
            total += cost;
        }
        Arrays.sort(arr, new SortingJobs());
        Pair res = new Pair();
        res = solve(arr, n);
        if (res == null) {
            System.out.println(0);
        } else {
            System.out.println(n - res.first);
            System.out.println(total - res.second);
        }

    }
    private static Pair solve(Job[] arr, int n) {
        if (n == 0) {
            return null;
        }
        int dp[] = new int[n];
        int numOfJobs[] = new int[n];
        for (int i = 0; i < n; i++) {
            dp[i] = 0;
            numOfJobs[i] = 0;
        }
        dp[0] = arr[0].cost;
        numOfJobs[0] = 1;
        for (int i = 1; i < n; i++) { int curr = arr[i].cost; int num = 1; int idx = searchJob(arr, 0, i - 1, arr[i].st); if (idx != curr && idx != -1) { curr += dp[idx]; num += numOfJobs[idx]; } if (curr > dp[i - 1]) {
                dp[i] = curr;
                numOfJobs[i] = num;
            } else {
                dp[i] = dp[i - 1];
                numOfJobs[i] = numOfJobs[i - 1];
            }
        }
        return new Pair(numOfJobs[n - 1], dp[n - 1]);
    }
    private static int searchJob(Job[] arr, int st, int ed, int key) {
        int ans = -1;
        while (st <= ed) {
            int mid = (st + ed) / 2;
            if (arr[mid].ed <= key) {
                ans = mid;
                st = mid + 1;
            } else {
                ed = mid - 1;
            }
        }
        return ans;
    }
    private static int getTime(String st) {
        int hr = (st.charAt(0) - '0') * 10 + (st.charAt(1) - '0');
        int min = (st.charAt(2) - '0') * 10 + (st.charAt(3) - '0');
        return hr * 60 + min;
    }
}

FAQs on Sopra Banking Coding Questions

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

Yes, Coding Questions were asked in Online Assessment and Technical Interview of Sopra Banking Recruitment Process.

Question 2: In what field does Sopra Banking works?

Sopra Banking works in the field of banking and financial services to help there customers to operate efficiently on their financial systems and services.

Question 3: What is the minimum salary of Sopra Banking?

The salary offered by the company for Engineer Trainee is a minimum ₹6.0 Lakhs per year, for Software Engineers the minimum salary is ₹8.5 L.P.A and so on.

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Sopra Banking Coding Questions And Answers

Sopra Banking Coding Questions with Solution 2023

About Sopra Banking Software Ltd.

Sopra Banking Recruitment Process

Details of Sopra Banking Online Coding Assessment :

Prime Course Trailer

Related Banners

Sample Sopra Banking Practice Coding Questions and Answers

Question 1 : Borrow Number

Question 2 : Iron Magnet

Question 3 : Array Subarray

Sample input :

Sample output :

Explanation :

Question 4 : Maximum Toys

Question 5 : Maximize Earnings (R->Hard)

FAQs on Sopra Banking Coding Questions

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

Question 2: In what field does Sopra Banking works?

Question 3: What is the minimum salary of Sopra Banking?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

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30+ Companies are Hiring

Unlock this article for Free, by logging in

Sopra Banking Coding Questions And Answers

Sopra Banking Coding Questions with Solution 2023

About Sopra Banking Software Ltd.

Sopra Banking Recruitment Process

Details of Sopra Banking Online Coding Assessment :

Prime Course Trailer

Related Banners

Sample Sopra Banking Practice Coding Questions and Answers

Question 1 : Borrow Number

Question 2 : Iron Magnet

Question 3 : Array Subarray

Sample input :

Sample output :

Explanation :

Question 4 : Maximum Toys

Question 5 : Maximize Earnings (R->Hard)

FAQs on Sopra Banking Coding Questions

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

Question 2: In what field does Sopra Banking works?

Question 3: What is the minimum salary of Sopra Banking?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

30+ Companies are Hiring

Unlock this article for Free,
by logging in