Bosch Coding Questions

July 21, 2023

Bosch Coding Questions 2023

Bosch Coding Questions asked in Bosch Global Software Technologies Hiring Process which is an Computer Based Online Recruitment Test (AMCAT).

Before you go through the test process, we would like to tell you some of the guidelines for attempting the test and even about the further process. Please go through the page to know more about such points before you appear for Bosch Global Software Technologies for Hiring Process FY 22 – 23.

About Bosch Global Software Technologies

Bosch Global Software Technologies provide comprehensive engineering, information technology, and business solutions, making them one of the top global suppliers of technology and services.

Bosch Global Software Technologies are at the forefront of designing, developing, and implementing IoT ecosystems thanks to our global presence and presence in the US, Europe, Japan, China, and the Asia Pacific region. This is made possible by the comprehensive capability within the three IoT aspects of Sensors, Software, and Services.

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Recruitment Process at BGSW

Here is a complete process of recruitment at Bosch Global Software Technologies for the designation of Associate Software Engineer.

Online Assessment (AMCAT)
BGSW Day – Pre – Placement Talk
Technical Discussion
Final Online Assessment in lieu with HR Round

We have even tabulated some more information for your reference and understanding.

Bosch Global Software Technologies	Related Information
Batch	2023
Course	B.Tech (CSE and allied branches and ECE)
Role	Associate Software Engineer
Education	6.0 throughout all the semester No backlogs
Platform	AMCAT
Cost to Company (CTC)	5 LPA – 7.5 LPA
Work location	Any of the BGSW locations of Bangalore, Coimbatore, Pune or Hyderabad

Information Regarding Package Offered Package offered: 7 LPA for mobility requirements and 5 LPA for IT Services requirements.
(Consideration & Selection for Mobility & IT Services will depend on their online assessment scores and interview feedback)

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Bosch Coding Questions

Question 1 : Spiral Matrix

Problem Statement :

You will be given a 2d matrix. Write the code to traverse the matrix in a spiral format. Check the input and output for better understanding.

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

C++

JAVA

Python

C++

#include <bits/stdc++.h>

using namespace std;

void Spiral(vector < vector < int >> a) {
  if (a.size() == 0) return;
  int m = a.size(), n = a[0].size();
  int i, k = 0, l = 0;
  while (k < m && l < n) {
    for (i = l; i < n; ++i) cout << a[k][i] << " ";
    k++;
    for (i = k; i < m; ++i) cout << a[i][n - 1] << " ";
    n--;
    if (k < m) {
      for (i = n - 1; i >= l; --i) cout << a[m - 1][i] << " ";
      m--;
    }
    if (l < n) {
      for (i = m - 1; i >= k; --i) cout << a[i][l] << " ";
      l++;
    }
  }
}

int main() {
  int r, c;
  cin >> r >> c;
  vector < vector < int >> mat(r, vector < int > (c));
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++) cin >> mat[i][j];
  Spiral(mat);
}

JAVA

import java.util.*;
public class Main
{
    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 2 : Celsius to Fahrenheit

Problem Statement :

There are two major scales for measuring temperature, celsius & Fahrenheit.Given a floating point input for temperature in celsius scale, Write a program to convert celsius to fahrenheit, up till 2 decimal points.

Input Output
56 132.8

Explanation:
56 degrees celsius means132.8 degrees of Fahrenheit to be exact.

C++

JAVA

Python

C++

#include 

using namespace std;

int main() {
  float temperature;
  cin >> temperature;
  cout << (temperature * 9) / 5 + 32;
  return 0;
}

JAVA

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        float n = sc.nextFloat();
        float ans = ((n * 9.0 f / 5.0 f) + 32.0 f);
        System.out.println(ans);
    }
}

Python

def ctof(c):
    return (c * 9) / 5 + 32
temperature = float(input())
ans = ctof(temperature)
print(round(ans, 2))

Question 3 : Frequency of Array

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 <bits/stdc++.h>

using namespace std;

int main() {
  int n, k;
  cin >> n;
  map < int, int > m;
  while (n--) {
    cin >> k;
    m[k]++;
  }
  for (auto i: m)
    cout << i.first << " occurs " << i.second << " times" << endl;
}

JAVA

import java.util.Scanner;
public class Main {
    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 4 : Prime Number Sum

Problem Statement :

You will be given a left limit and a right limit. You need to Calculate the sum of all prime numbers between a given range x and y, both inclusive.

Sample Test Case :
Input : Output :
30 68
40

Explanation :
31 and 37 are the two prime numbers in the window.

C++

JAVA

Python

C++

#include <bits/stdc++.h>

using namespace std;
unordered_map < int, bool > prime;
void SieveOfEratosthenes(int n) {
    for (int i = 0; i < n; i++) prime[i] = 1;
    for (int p = 2; p * p <= n; p++)
        if (prime[p] == true)
            for (int i = p * p; i <= n; i += p) prime[i] = false;
}
int main() {
    int n, m;
    int count = 0, sum = 0;
    cin >> n >> m;
    SieveOfEratosthenes(m + 1);
    for (int i = n; i <= m; i++)
        if (prime[i]) sum += i;
    cout << sum;
}

JAVA

import java.util.*;
public class Main {
    public static boolean isPrime(int no) {
        if (no <= 1)
            return false;
        if (no <= 3)
            return true;
        if (no % 2 == 0 || no % 3 == 0)
            return false;
        for (int i = 5; i * i <= no; i = i + 6) {
            if (no % i == 0 || no % (i + 2) == 0)
                return false;
        }
        return true;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int x = sc.nextInt();
        int y = sc.nextInt();
        int sum = 0;
        for (int i = x; i <= y; i++)
            if (isPrime(i))
                sum = sum + i;
        System.out.println(sum);
    }
}

Python

def isPrime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True


a = int(input())
b = int(input())
x = 0 if a > 2 else 2
step = 1 if (a % 2 == 0) else 0
for i in range(a + step, b + 1, 2):
    if isPrime(i):
        x = x + i
print(x)

Question 5 :Non Repeating Characters

Problem Statement :

You’re given a string, the characters in the string may get repeated in the string. You’re supposed to find the positions of the string elements that are not repeating in print them.
Note: case doesn’t matter (case insensitive)

Input : Output :
hghigiklblbmono 7 12 14

Explanation :
K, m and n do not repeat.

C++

JAVA

Python

C++

#include <bits/stdc++.h>

using namespace std;

int main() {
    string s;
    unordered_map < int, int > m;
    getline(cin, s);
    for (auto i: s) {
        m[i - 32]++;
        m[i]++;
        m[i + 32]++;
    }
    for (int i = 0; i < s.length(); i++)
        if (m[s[i]] == 1) cout << (i + 1) << " ";
}

JAVA

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        int arr[] = new int[26];
        String temp = str.toLowerCase();
        for (int i = 0; i < temp.length(); i++)
            arr[temp.charAt(i) - 97]++;
        for (int i = 0; i < str.length(); i++) {
            char ch = Character.toLowerCase(str.charAt(i));
            if (arr[ch - 97] == 1)
                System.out.print(i + 1 + " ");
        }
    }
}

Python

from collections import defaultdict

m = defaultdict(int)
s = input()
for i in s:
    m[ord(i)] += 1
    m[ord(i) - 32] += 1
    m[ord(i) + 32] += 1
for i in range(len(s)):
    if m[ord(s[i])] == 1:
        print(i + 1, end=" ")

FAQs on Bosch Coding Questions with Solutions

Question 1: What is the salary offered by Bosch?

Bosch offers a package of 5-7 LPA, depending on the performance on the recruitment rounds.

Question 2: What is the difficulty level of the Bosch exam?

Bosch exam is of medium to hard difficulty level. With adequate practice you can crack this exam.

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