Capgemini Exceller Coding Questions 2023

October 17, 2023

Capgemini Exceller Coding Questions and Answers 2023

Capgemini Exceller Coding Questions has been included as a new round in Capgemini Exceller selection process for 2023 graduates. This round will be conducted through CoCubes, containing 2 coding questions and you will have a total of 45 mins to solve. This round has been added to check the candidates programming, logical and problem solving techniques. Below we have given some previous year sample based questions for Capgemini Exceller Coding Round, make sure you practice all of them

Details for Capgemini Exceller Coding Round

Coding Round	Important Information
Total no. of question	2
Allotted Time	45 mins
Section Property	Mandatory
Difficulty	High

Capgemini Exceller has started asking Coding Questions, but it is asked in a special hiring where the package being offered is more than the normal hiring. The package offered for this post is 7.5LPA, and since the package being offered is comparatively higher than the normal hiring process, the selection process and test difficulties level are also higher than normal.

Practice Questions for Capgemini Exceller Coding Round

Question 1

Problem Statement –

You have write a function that accepts, a string which length is “len”, the string has some “#”, in it you have to move all the hashes to the front of the string and return the whole string back and print it.

char* moveHash(char str[],int n);

example :-

Sample Test Case

Input:

Move#Hash#to#Front

Output:

###MoveHashtoFront

C

C++

Java

Python

C

#include<string.h>
#include<stdio.h>
char *moveHash(char str[],int n)
{
char str1[100],str2[100];
int i,j=0,k=0;
for(i=0;str[i];i++)
{
if(str[i]=='#')
str1[j++]=str[i];
else
str2[k++]=str[i];
}
strcat(str1,str2);
printf("%s",str1);
}
int main()
{
char a[100], len;
scanf("%[^\n]s",a);
len = strlen(a);
moveHash(a, len);
return 0;
}

C++

#include<bits/stdc++.h>
using namespace std;
char *moveHash(char str[],int n)
{
    char str1[100],str2[100];
    int i,j=0,k=0;
    for(i=0;str[i];i++)
    {
        if(str[i]=='#')
        str1[j++]=str[i];
        else
        str2[k++]=str[i];
    }
    strcat(str1,str2);
    cout<<str1;
}
int main()
{
    char a[100], len;
    cin>>a;
    len = strlen(a);
    moveHash(a, len);
    return 0;
}

Java

import java.util.*;
public class MoveHash 
{
    public static void moveHash(String str ,int n)
    {
        String str1= new String("");
        String str2= new String("");
        int i=0;
        for(i=0;i<n;i++)
        {
            if(str.charAt(i)=='#')
                str1=str1 + str.charAt(i);
            else
                str2 = str2 + str.charAt(i);
        }
        String result = str1.concat(str2);
        System.out.println(result);
    }
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        String a = sc.nextLine();
        int len= a.length();
        moveHash(a, len);
    }
}

Python

s=input()
x=s.count("#")
s=s.replace("#","")
print("#"*x+s)

Question 2

Problem Statement –

Capgemini in its online written test have a coding question, wherein the students are given a string with multiple characters that are repeated consecutively. You’re supposed to reduce the size of this string using mathematical logic given as in the example below :

Input :

aabbbbeeeeffggg

Output:

a2b4e4f2g3

Input :

abbccccc

Output:

ab2c5

C

Java

Python

C

#include<stdio.h>
int main()
{
    char str[100];
    scanf("%[^\n]s",str);
    int i, j, k=0, count = 0;
    char str1[100];
    for(i=0; str[i]!='\0'; i++)
    {
        count = 0;
        for(j=0; j<=i; j++)
        {
            if(str[i]==str[j])
            {
                count++;
            }
        }
        if(count==1)
        {
            str1[k] = str[i];
            k++;
        }
    }
    for(i=0; i<k; i++)
    {
        count = 0;
        for(j=0; str[j]!='\0'; j++)
        {
            if(str1[i]==str[j])
            {
                count++;
            }
        }
        if(count==1)
        {
            printf("%c",str1[i]);
        }
        else
        {
            printf("%c%d",str1[i],count);
        }
    }
    return 0;
}

Java

import java.util.*;
public class CharacterCount 
{
    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        int i, j, k = 0, count = 0;
        String uni = new String("");
        for(i=0; i<str.length(); i++)
        {
            count = 0;
            for(j=0; j<=i; j++)
            {
                if(str.charAt(i)==str.charAt(j))
                {
                    count++;
                }
            }
            if(count==1)
            {
                uni = uni + str.charAt(i);
            }
        }
        for(i=0; i<uni.length(); i++)
        {
            count = 0;
            for(j=0; j<str.length(); j++)
            {
                if(uni.charAt(i)==str.charAt(j))
                {
                    count++;
                }
            }
            if(count==1)
            {
                System.out.printf("%c",uni.charAt(i));
            }
            else
            {
                System.out.printf("%c%d",uni.charAt(i),count);
            }
        }
    }    
}

Python

def solve(s):
    ans=""
    c=1
    for i in range(len(s)-1):
        if(s[i]==s[i+1]):
            c+=1
        else:
            if(c==1):
                ans+=s[i]
            else:
                ans+=s[i]+str(c)
            c=1 
    if(c==1):
        ans+=s[i+1]
    else:
        ans+=s[i+1]+str(c)
    return ans
s=input()
print(solve(s))

Question 3

Problem Statement –

Write the code to traverse a matrix in a spiral format.

Sample Input

Input

5 4

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

17 18 19 20

Output

1 2 3 4 8 12 16 20 19 18 17 13 9 5 6 7 11 15 12 14 10

C

Java

Python

C

#include<stdio.h>
int main()
{
int a[5][4] = {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16},{17,18,19,20}};
int rs = 0, re = 5, cs = 0, ce = 4;
int i, j, k=0;

for(i=0;i<5;i++)
{
for(j=0;j<4;j++)
{
printf("%d\t",a[i][j]);
}
printf("\n");
}

printf("\n");

while(rs {
for(i=cs;i<ce;i++)
{
printf("%d\t",a[rs][i]);
}
rs++;

for(i=rs;i<re;i++)
{
printf("%d\t",a[i][ce-1]);
}
ce--;

if(rs<re)
{
for(i=ce-1; i>=cs; --i)
{
printf("%d\t", a[re - 1][i]);
}
re--;
}

if(cs<ce)
{
for(i=re-1; i>=rs; --i)
{
printf("%d\t", a[i][cs]);
}
cs++;
}
}
return 0;
}

Java

import java.util.*;
public class Solution
{
    public static List<Integer> solve(int [][]matrix,int row,int col)
   {
      List<Integer> res=new ArrayList<Integer>();
      boolean[][] temp=new boolean[row][col];
      int []arr1={0,1,0,-1};
      int []arr2={1,0,-1,0};
      int di=0,r=0,c=0;
      for(int i=0;i<row*col;i++)
      {
             res.add(matrix[r][c]);
             temp[r][c]=true;
             int count1=r+arr1[di];
             int count2=c+arr2[di];
            if(count1>=0 && row>count1 && count2>=0 && col>count2 && !temp[count1][count2]){
	r=count1;
                  c=count2; 
            } 
            else 
            {
                 di=(di+1)%4;
                 r+=arr1[di];
                 c+=arr2[di];
             }
      }  
     return res;
    }
    public static void main(String[] args)
   {
         Scanner sc=new Scanner(System.in);
         int m=sc.nextInt();
         int n=sc.nextInt();
         int matrix[][]=new int[m][n];
         for(int i=0;i<m;i++)
         {
                 for(int j=0;j<n;j++)
                      matrix[i][j]=sc.nextInt();
         }
       System.out.println(solve(matrix,m,n));
  }
}

Python

def spiralOrder(arr):
    ans=[]
    while arr:
        ans+=arr.pop(0)
        arr= (list(zip(*arr)))[::-1]
    return ans
arr=[]
n,m=map(int,input().split())
for i in range(n):
    arr.append(list(map(int,input().split())))
print(*spiralOrder(arr))

Question 4

Problem Statement –

You’re given an array of integers, print the number of times each integer has occurred in the array.

Example

Input :

10

1 2 3 3 4 1 4 5 1 2

Output :

1 occurs 3 times

2 occurs 2 times

3 occurs 2 times

4 occurs 2 times

5 occurs 1 times

C

Java

Python

C

#include<stdio.h> 
int main()
{
    int n;
    scanf("%d",&n);
    int arr[n];
    int count;
    int k=0;
    for(int i=0; i<n; i++)
    {
        scanf("%d",&arr[i]);        //input of arrray
    }
    int newarr[n];
    //step 1
    for(int i=0; i<n; i++)
    {
        count = 0;
        for(int j=0; j<=i; j++)
        {
            if(arr[i]==arr[j])
            {
                count++;
            }
        }
        if(count==1)
        {
            newarr[k] = arr[i];
            k++;
        }
    }
    //step 2
    for(int i=0; i<k; i++)
    {
        count = 0;
        for(int j=0; j<n; j++)
        {
            if(newarr[i]==arr[j])
            {
                count++;
            }
        }
        printf("%d occurs %d times\n",newarr[i],count);
    }
    return 0;
}

Java

import java.util.Scanner;
public class ArrayFrequency
{
    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int array[] = new int[n];       // taking value of integer
        int count = 0;
        int k = 0;
        for(int i=0; i<n; i++)
        {
            array[i] = sc.nextInt();    //elements of array
        }
        //step - 1
        int newarray[] = new int[n];
        for(int i=0; i<n; i++)
        {
            count = 0;
            for(int j=0; j<=i; j++)
            {
                if(array[i]==array[j])
                {
                    count++;
                }
            }
            if(count == 1)
            {
                newarray[k] = array[i];
                k++;
            }
        }
        //step 2;
        for(int i=0; i<k; i++)
        {
            count  = 0;
            for(int j=0; j<n; j++)
            {
                if(newarray[i] == array[j])
                {
                    count++;
                }
            }
            System.out.printf("%d occurs %d times\n",newarray[i],count);
        }
    }
}

Python

n=int(input())
arr=list(map(int,input().split()))
dup=[]
for i in arr:
    if(i not in dup):
        dup.append(i)
        print("{} occurs {} times".format(i,arr.count(i)))

Question 5

Problem Statement –

Write a function to solve the following equation a³ + a²b + 2a²b + 2ab² + ab² + b³.

Write a program to accept three values in order of a, b and c and get the result of the above equation.

C

C++

Python

C

#include<stdio.h> 
int main()
{
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    int sum;
    sum=(a+b)*(a+b)*(a+b);
    printf("%d",sum);
}

C++

#include<iostream.h> 
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    int sum;
    sum=(a+b)*(a+b)*(a+b);
    cout<<sum;
}

Python

def spiralOrder(arr):
    ans=[]
    while arr:
        ans+=arr.pop(0)
        arr= (list(zip(*arr)))[::-1]
    return ans
arr=[]
n,m=map(int,input().split())
for i in range(n):
    arr.append(list(map(int,input().split())))
print(*spiralOrder(arr))

Question 6

Problem Statement –

A function is there which tells how many dealerships there are and the total number of cars in each dealership.

Your job is to calculate how many tyres would be there in each dealership.

Input

3

4 2

4 0

1 2

Output

20

16

8

There are total 3 dealerships

dealerships1 contains 4 cars and 2 bikes

dealerships2 contains 4 cars and 0 bikes

dealerships3 contains 1 cars and 2 bikes

Total number of tyres in dealerships1 is (4 x 4) + (2 x 2) = 20

Total number of tyres in dealerships2 is (4 x 4) + (0 x 2) = 16

Total number of tyres in dealerships3 is (1 x 4) + (2 x 2) = 8

C

C++

Java

Python

C

#include<stdio.h> 
int main()
{
    int t,cars,bikes;
    scanf("%d",&t);
    for(int i=0;i<t;i++)
    {
        scanf("%d%d",&cars,&bikes);
        printf("%d\n",cars*4+bikes*2);
    }
}

C++

#include<iostream.h> 
using namespace std;
int main()
{
    int t,cars,bikes;
    cin>>t;
    for(int i=0;i<t;i++)
    {
        cin>>cars>>bikes;
        cout<<cars*4+bikes*2;
        cout<<"\n";
    }
}

Java

import java.util.*;
public class Solution
{
    public static void main(String[] args)
   {
        Scanner sc=new Scanner(System.in);
        int dealership=sc.nextInt();
        while(dealership-->0)
       {
            int cars=sc.nextInt();
            int bikes=sc.nextInt();
            System.out.println(cars*4+bikes*2);           
       }
   }
}

Python

for i in range(int(input())):
    cars,bikes=map(int,input().split())
    print(cars*4+bikes*2)

30 comments on “Capgemini Exceller Coding Questions 2023”

Asif Basheer
Question 4 (Using LinkedHashMap)
import java.util.*;
class Asif {
public static void main(String[] args) {

int s[] = {1, 2, 3, 3, 4, 1, 4, 5, 1, 2};
LinkedHashMap mapp = new LinkedHashMap();
for(int i = 0; i ~~System.out.println(key + ” occures “+value+” times”));~~

}
}

0

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kapilsingh
import java.util.*;
public class MyClass {
public static void main(String args[]) {
String s = “Move#Hash#to#Front”;
String v = “”;
String h = “”;
for(int i=0;i<s.length();i++){
if(s.charAt(i)=='#'){
v += s.charAt(i);
}else{
h +=s.charAt(i);
}
}
System.out.println(v +""+h);

0

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swapnil
problem statement 2(in java ): public class demo3 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s = sc.next();
char arr[]=s.toCharArray();
HashMap c = new HashMap();
for(int i=0;i<s.length();i++){
if(c.containsKey(arr[i])){
c.put(arr[i],c.get(arr[i])+1);
}
else{
c.put(arr[i],1);
}
}
for (Map.Entry entry: c.entrySet()){
System.out.print(entry.getKey()+""+entry.getValue());
}

}
}

0

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NISHANT
Q. 1.)
Solution :-

import java.util.*;

public class Main
{

public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String s=sc.next();
String s1[]=s.split(“#”);
List ll = new ArrayList();

for (String i:s1
) {
ll.add(i);
}
LinkedHashMap lhm = new LinkedHashMap();
for (int i=0;i<s.length();i++)
{
if (lhm.containsKey(s.charAt(i)))
{
lhm.put(s.charAt(i),lhm.get(s.charAt(i))+1);
}
else
{
lhm.put(s.charAt(i),1);
}
}
int count=0;
for (Map.Entry entry:lhm.entrySet()
) {
if (entry.getKey()==’#’)
{
count=entry.getValue();
}

}
for (int i=0;i<count;i++)
{
ll.add(0,"#");
}
for (String i:ll
) {
System.out.print(i);
}

}
}

0

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NISHANT
Here is one solution :-

import java.lang.reflect.Array;
import java.util.*;
import java.io.*;

public class Main
{

public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n= sc.nextInt();
ArrayList al1 = new ArrayList();
ArrayList al2 = new ArrayList();
while (n>0)
{
al1.add(sc.nextInt());
al2.add(sc.nextInt());
n–;

}
ArrayList al = new ArrayList();
ListIterator li1 = al1.listIterator();
ListIterator li2 = al2.listIterator();
while (li1.hasNext()&& li2.hasNext())
{
al.add(li1.next()*4+ li2.next()*2);
}
for (Integer i:al
) {
System.out.println(i);
}

}
}

0

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Maths
Q-2
from collections import Counter
d=Counter()
s=input()
d.update(s)
l={}
for i in d.elements():
if d[i]==1:
l.update({i:(str(”))})
else:
l.update({i:(str(d[i]))})

new_dict= ”.join(map(”.join, l.items()))
print(new_dict)

0

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rajkanyam02
Problem Statement 1 :
str = int(input(“String :: “))
def hashfun(str):
str1=””
str2=””
for i in str:
if i == ‘#’:
str1 +=i
else:
str2 += i
str1 += str2
return str1
print(hashfun(str))

0

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saragada
code for 6th question in python
l=[]
t= int(input())
for i in range(t):
a=[int(b) for b in input().split()]
l.append(a)
#print(l)
for i in l:
print(i[0]*4+i[1]*2)

1

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Santosh
import java.util.Arrays;
import java.util.Scanner;

public class dealear_ship_car {
static int car(int car) {
int cartyre = 4;
return (car * cartyre);
}
static int bike(int bike){
int biketyre = 2;
return (bike * biketyre);
}
static int totaltyres(int noofcars ,int noofbikes){
int total = car(noofcars) + bike(noofbikes);
return total;
}

public static void main(String[] args) {
System.out.println(“enter total no of dealerships : “);
Scanner in = new Scanner(System.in);
int dealers = in.nextInt();
int[] arr = new int[dealers];
int noofcars = 0, noofbikes = 0;
for (int i = 0; i < dealers; i++) {
noofcars = in.nextInt();
noofbikes = in.nextInt();
arr [i] = totaltyres(noofcars,noofbikes);
}
System.out.println(Arrays.toString(arr));
}
}

1

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Asif
import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner t=new Scanner(System.in);
int n=t.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
a[i]=t.nextInt();
int visited[]=new int[100];
for(int i=0;i<visited.length;i++)
visited[i] = 0;
int count=0;
for(int i=0;i<n;i++)
{
count=0;
if(visited[a[i]]==0)
{
for(int j=i;j<n;j++)
{
if(a[i]==a[j])
{
count++;
visited[a[i]]=1;
}
}
System.out.println(a[i]+" occurs "+count+" times");
}
}
}
}

0

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Reddy
What is the 5th question solution ?

4

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Vaibhav Jain
The equation is being simplified in order to find out the optimum solution

1

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Ravindra
print((a+b)**3)

0

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Asif
import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner t=new Scanner(System.in);
int n=t.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
a[i]=t.nextInt();
int visited[]=new int[100];
for(int i=0;i<visited.length;i++)
visited[i] = 0;
int count=0;
for(int i=0;i<n;i++)
{
count=0;
if(visited[a[i]]==0)
{
for(int j=i;j<n;j++)
{
if(a[i]==a[j])
{
count++;
visited[a[i]]=1;
}
}
System.out.println(a[i]+" occurs "+count+" times");
}
}
}
}

0

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Santosh
public static void main(String[] args) {
System.out.println(“enter value of a and b : “);
Scanner in = new Scanner(System.in);
int a = in.nextInt();
int b = in.nextInt();
System.out.println(equation(a,b));
}
}

0

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Santosh
import java.util.Scanner;

public class Solving_an_equation {
static int equation(int x,int y){
return ((x+y)*(x+y)*(x+y));
}

public static void main(String[] args) {
System.out.println(“enter value of a and b : “);
Scanner in = new Scanner(System.in);
int a = in.nextInt();
int b = in.nextInt();
System.out.println(equation(a,b));
}
}

2

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saragada
question 4 answer:
w= int(input())
a=[int(i) for i in input().split()]

l=[]
for i in a:
c=a.count(i)
l.append(c)

d=(dict(zip(a,l)))
#print(d)
for i,j in d.items():
print(i,”occurs”,j,”items”)

1

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Sainithin
where are the coding questions i have paid 700 for that there are only 6 questions over here

5

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Vaibhav Jain
The coding material is provided free of cost, the mock contains the questions of rest of the sections

2

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