Darwinbox Coding Questions Round 1

September 20, 2023

Sample Darwinbox Coding Questions Round 1

Here, Darwinbox Coding Questions Round 1 page will help you to get Sample Darwinbox Coding Questions which will help you to prepare for Darwinbox Coding Test 1 included in Darwinbox Recruitment Process.

Go through this page to get all the sample Darwinbox Coding Questions for Round 1 as well as FAQs related to Darwinbox Recruitment Process at the end of this page.

Sample Darwinbox Coding Questions Round 1

Question 1: Order check

Problem Statement: In an IT company there are n number of Employees , they are asked to stand in ascending order according to their heights.But some employees are not currently standing in their correct position.

Your task is to find how many employees are there who are not standing in their correct positions.

Example

height=[1,2,1,3,3,4,3]

The 4 employees at indices 1,2,5 and 6 are not in the right positions. The correct positions are (1,1,2,3,3,3,4).Return 4.

Function Description

Complete the function countEmployees in the editor below.

count Employee has the following parameter(s):

int height[n]:an array of heights in the order the employees are standing

Returns:

Int : the number of employees not standing in the correct positions

Constraints

1<=n<=10^5
1<=height[i]<=10^9

Sample case 0

Sample input 0

7 ->height[] size n=7

1

2

1

3

3

4

3

Sample output 0:

4

Explanation

The four employees who are not standing in the correct positions are at indices [1,2,5,6] return 4. The correct positions are [1,1,2,3,3,3,4].

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Java

Python

C++

Java

import java.util.*;
public class Main
{
  public static int countEmployees (int arr[])
  {
    int n = arr.length;
    int temp[] = new int[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
      Arrays.sort (arr);
    int count = 0;
    for (int i = 0; i < n; i++)
      if (arr[i] != temp[i])
	  count++;
      return count;
  }
  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    int n = sc.nextInt ();
    int arr[] = new int[n];
    for (int i = 0; i < n; i++)
      arr[i] = sc.nextInt ();
    System.out.println (countEmployees (arr));
  }
}

Python

def countEmployees(n,arr):
    dup_arr=arr[:]
    dup_arr.sort()
    count=0
    for i in range(n):
        if arr[i]!=dup_arr[i]:
            count+=1 
        else:
            pass
    return count
n=int(input())
arr=[ ]
for i in range(n):
    arr.append(int(input()))
print(countEmployees(n,arr))

C++

#include<bits/stdc++.h>
using namespace std;

int countEmployees(int arr[], int n) {
    int temp[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
    
    sort(arr, arr + n);
    
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] != temp[i])
            count++;
    }
    return count;
}

int main() {
    int n;
    cin >> n;
    int arr[n];
    
    for (int i = 0; i < n; i++)
        cin >> arr[i];
    
    cout << countEmployees(arr, n) << endl;
    
    return 0;
}

Question 2 : Maximum Revenue

Problem statement : Amit is a salesman who wishes to know the maximum revenue received from a given item of the N products each day . Amit has a sales record of N products for M days.Helo amit to find the highest revenue received each day.

Input :

The first line of the input consists of two space-separated integers- day(M) and item(N) representing the days and the products in the sales record.
The next M lines consist of N space separated integers representing the sales revenue from each product each day.

Output:

Print m space-separated integers representing the maximum revenue received each day .

Example Input:

3 4
101 123 234 344
143 282 499 493
283 493 202 304

Output:

344 499 493

Explanation:

The maximum revenue received on the first day is 344 , followed by a maximum revenue of 499 on the second day and a maximum revenue of 493 on the third day.

C++

Java

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;

int main(){
    int m,n, i,j;
    cin >> m;
    cin >> n;
    int arr [m][n];
    for(i = 0;i < m;i++){
        for (j=0;j<n;j++){ 
            cin >> arr[i][j];
        }
    }
    for (i = 0;i < m;i++){
        int max = INT_MIN;
        for(j=0;j<n;j++){ 
            if(arr[i][j]>max){
                max = arr[i][j];
            }
        }
        cout  << max << " ";
    }
    return 0;
}

Java

import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    int m=sc.nextInt();
    int n=sc.nextInt();
    int arr[][]=new int[m][n];
    int i,j;
    for(i=0;i<m;i++){
        for (j=0;j<n;j++){
            arr[i][j]=sc.nextInt();
        }
    }
    for (i=0;i<m;i++){
        int max = Integer.MIN_VALUE;
        for(j=0;j<n;j++){
            if(arr[i][j]>max){
                max = arr[i][j];
            }
        }
        System.out.println(max);
    }
 }
}

Python

Tab Content

m = int(input())
n = int(input())
arr = []
for i in range(0, m):
   ar = []
   for j in range(0, n):
       ar.append(int(input()))
   arr.append(ar)

for i in arr:
   print(max(i), end=" ")

Question 3: Find the homeless

Problem Statement -: There are N Homeless people in the community and N houses in the community. It will be given in the array (people), the height of the person, and in the array house capacity of the house is given.

The government decided to give homes to people on the basis of the following conditions:

Priority is given to the people from left to right of the array
Each person is allotted to a house if and only if the capacity of the house is greater than or equal to the person’s height
Nearby empty Houses are allotted to the person( starting from the extreme left)

You need to find the number of homeless people who have not been allotted any home if the government follows the above conditions. So that government will have an idea of how many people they need to allot homes for next time.

Constraints:

1 <= N <= 10^3
1 <= people[i] <= 10^5
1 <= house[i] <= 10^5

Input Format for Custom Testing:

The first line contains an integer, N, denoting the number of people and number of houses.
Each line i of the N subsequent lines (where 0 <= i <= N) contains an integer describing peoplei.
Each line i of the N subsequent lines (where 0 <= i <= N) contains an integer describing housei.

Sample Test Cases

Sample Input 1
3
4
2
7
3
5
10
Sample Output 1
0
Explanation
people=[4,2,7]
house=[3,5,10]
People[0] has more priority , from left to right order in houses 5 is the nearest one which fits for people[0]
people[1]=2 will fit in 3 which is nearer from left
people[2]=7 will fit in remaining house of capacity of 10
So no homeless people left so return 0

Sample Input 2
3
3
8
5
1
9
4
Sample Output 2
2
Explanation
people=[3,8,5]
house=[1,9,4]
people[0]=3 can fit in 9 which is nearest from left in array house
people[1]=8 cannot fit in any home which is left (i.e, 1 and 4)
people[2]=5 cannot fit in any home which is left (i.e, 1 and 4)
So return 2,which is number of homeless people

C++

Java

Python

C++

Run

#include<bits/stdc++.h>
using namespace std;
int main(){  
    int n;
    cin>>n;    
    int people[n], house[n];
    for(int i=0; i<n; i++) cin>>people[i];       
    for(int i=0; i<n; i++) cin>>house[i];      
    int count = 0;
    for(int i=0; i<n; i++){      
        for(int j=0; j<n; j++){
    if(people[i]<house[j]){
            count+=1 ;
            house[j]=-1;
            break ;
            }
        }
    } 
    cout<<n-count;
  }

Java

Run

import java.util.*;
class Main
{ 
  public static void main(String args[])
  {
    Scanner sc = new Scanner (System.in);
    int n=sc.nextInt();
    int people[]= new int[n];
    int house[]= new int[n];
    for(int i=0; i<n; i++)
       people[i]=sc.nextInt();      
    for(int i=0; i<n; i++)
       house[i]=sc.nextInt();       
    int count = 0;
    for(int i=0; i<n; i++){       
        for(int j=0; j<n; j++){
         if(people[i]<house[j]){
            count+=1 ;
            house[j]=-1;
            break ;
            }
        }
    }
    System.out.println(n-count);
  }
 }

Python

Run

N=int(input())
people=[]
house=[]
#to read data of people and house arrays
for i in range(N): people.append(int(input()))
for i in range(N): house.append(int(input()))
count=0
for i in range(N):
    for j in range(N):
        if(people[i] < house[j]):
            count+=1 
            house[j]=-1
            break 
print(N-count)

Question 4 : Minimizing a string

Problem Statement :

Given a string, obtain the alphabetically smallest string possible by swapping either
adjacent ‘a’ and ‘b’ characters or adjacent ‘b’ and ‘c’ characters, any number of times.

Note: A string x is alphabetically smaller
than a string y if, for the first index i where x
and y differs, x[i] <y[i].

Example :
s=”abaacbac”

The alphabetically smallest possible string is
obtained by applying the following
operations:

‘c’at index 5 is swapped with ‘b’at index 6. So “abaacbac” becomes “abaabcac”
Then,’b’at index 2 is swapped with ‘a’at index 3. So “abaabcac” becomes “aababcac”.
Finally, ‘b’at index 3 is swapped with ‘a’at index 4 to obtain the final.
answer: “aaabbcac”.

Function Description :
Complete the function smallestString in the
editor below.

smallestString has the following
parameter(s):

string s: the given string
Returns:
string: the lexicographically smallest string obtained after swapping

Constraints :

1 <_ length of s<_ 10^5 s only contains the characters ‘a’, ‘b’, and ‘c’.

C++

JAVA

Python

C++

#include <bits/stdc++.h>
using namespace std;
//a is the smaller char b is the bigger
void f(string & s, char a, char b) {
    int n = s.length();
    bool fg = false;
    int st = -1;
    for (int i = n; i >= 0; i--) {
        if (s[i] == b) {
            if (fg) {
                swap(s[i], s[st]);
                st--;
            } else {
                continue;
            }
        } else if (s[i] == a) {
            if (fg) {
                continue;
            } else {
                fg = true;
                st = i;
            }
        } else {
            fg = false;
        }
    }
}
int main() {
    string s;
    cin >> s;

    int n = s.length();

    f(s, 'b', 'c');
    f(s, 'a', 'b');
    cout << s << endl;
    return 0;
}

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        char arr[] = str.toCharArray();
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            count = 0;
            for (int j = 0; j < arr.length - 1; j++) {
                if ((arr[j] - arr[j + 1]) == 1) {
                    char temp = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = temp;
                    count++;
                }
            }
            if (count == 0)
                break;
        }
        System.out.println(new String(arr));
    }
}

Python

Run

s = list(input())
n = len(s)
af = False
fi = n + 1
for i in range(n - 1, -1, -1):
    if s[i] == "c":
        if af:
            s[i], s[fi] = s[fi], s[i]
            fi -= 1
            af = False
    elif s[i] == "b":
        if af:
            continue
        else:
            fi = i
            af = True
    else:
        af = False


af = False
fi = n + 1
for i in range(n - 1, -1, -1):
    if s[i] == "b":
        if af:
            s[i], s[fi] = s[fi], s[i]
            fi -= 1
            af = False
    elif s[i] == "a":
        if af:
            continue
        else:
            fi = i
            af = True
    else:
        af = False


print(*s, sep="")

Question 05: Password Creation

Problem Statement: A password manager wants to create new passwords using two strings given by the user, then combined to create a harder-to- guess combination. Given two strings,interleave the characters of the strings to create a new string. Beginning with an empty string, alternately append a character from string a and from string b. If one of the strings is exhausted before the other, append the remaining letters from the other
string all at once. The result is the new password.

Example :

If a = ‘hackerrank’ and b = ‘mountain’,
The result is hmaocuknetrariannk.

Function Description :

Complete the function newPassword in the editor below.

Parameter(s):

Str : string a
Str : string b
Returns:
Str: new password using two strings

Sample input:

abc → a=”abc”
def → b=”def”

Sample output 0:

Adbecf

JAVA

Python

C++

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str1 = sc.next();
        String str2 = sc.next();
        String password = "";
        int i, j;
        for (i = 0, j = 0; i < str1.length() && j < str2.length(); i++, j++) {
            password = password + str1.charAt(i) + str2.charAt(j);
        }
        if (i < str1.length())
            password = password + str1.substring(i, str1.length());
        if (j < str2.length())
            password = password + str2.substring(j, str2.length());
        System.out.println(password);
    }
}

Python

Run

def newPassword(a, b):
    ans = ""
    if len(a) > len(b):
        m = len(b)
        x = a[m:]
    if len(a) <= len(b):
        m = len(a)
        x = b[m:]
    for i in range(m):
        ans += a[i]
        ans += b[i]
    return ans + x

a = input()
b = input()
print(newPassword(a, b))

C++

#include<bits/stdc++.h>
using namespace std;

int main() {
    string str1, str2;
    cin >> str1 >> str2;
    string password = "";
    int i, j;
    for (i = 0, j = 0; i < str1.length() && j < str2.length(); i++, j++) {
        password = password + str1[i] + str2[j];
    }
    if (i < str1.length())
        password = password + str1.substr(i);
    if (j < str2.length())
        password = password + str2.substr(j);
    cout << password << endl;
    return 0;
}

Question 6 :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 7 : Seating Arrangement in Exam Hall

Problem Statement :

Semester exams are going on for university students. Examiners noticed that a group of people are trying to cheat. They marked students of that group as ‘1’ and students of another group ( who are not cheating ) as ‘0’

We can reduce cheating by not allowing students from group 1 to sit together, means no two students from group 1 can sit together. Seatings are marked using above conditions. Your task is to give the seating placement of nth possibility Possibility order from 1 to 10 is given below

[1 10 100 101 1000 1001 1010 10000 10001 10010]

Sample input :

3 → number of test cases

4

6

9

Sample output :

101

1001

10001

Explanation :

4th possibility is 101

6th possibility is 1001

9th possibility is 10001

C++

JAVA

Python

C++

Run

#include<bits/stdc++.h>

using namespace std;

int main() {
    int n, m, Max = 0;
    cin >> n;
    vector < int > v(n);
    vector < string > arr;
    for (int i = 0; i < n; i++) { cin >> v[i];
        Max = max(Max, v[i]);
    }
    queue < string > q;
    q.push("1");
    int i = 1;
    arr.push_back("1");
    while (!q.empty()) { //cout<<"TEST"<<endl; string a = q.front(); q.pop(); q.push(a + "0"); arr.push_back(a + "0"); i++; if (a[a.length() - 1] == '0') { q.push(a + "1"); arr.push_back(a + "1"); i++; } if (i > Max) break;
    }
    for (int i = 0; i < n; i++) {
        cout << arr[v[i] - 1] << endl;
    }
}

JAVA

Run

import java.util.*;
class Main {
    public static void possibilities(int n) {
        int c = 0;
        String b = "";
        for (int i = 1; n != c; i++) {
            String s = Integer.toString(i, 2);
            if (!s.contains("11")) {
                c++;
                b = s;
            }
        }
        System.out.println(b);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        int[] a = new int[tc];
        for (int i = 0; i < tc; i++) {
            a[i] = sc.nextInt();
        }
        for (int i = 0; i < tc; i++) {
            possibilities(a[i]);
        }
    }
}

Python

Run

t = int(input())
a = []
m = -1
m1 = -1
for i in range(t):
    m1 = int(input())
    a.append(m1)
    m = max(m, m1)
k2 = 2
n = m + 1
k1 = ["0"] * (n)
k1[1] = "1"
a1 = 1
while k2 < (n): if k1[a1][-1] == "0": k1[k2] = k1[a1] + "0" k2 += 1 if k2 >= (n):
            break
        k1[k2] = k1[a1] + "1"
        k2 += 1
        if k2 >= (n):
            break
        a1 += 1
    elif k1[a1][-1] == "1":
        k1[k2] = k1[a1] + "0"
        k2 += 1
        if k2 >= (n):
            break
        a1 += 1
for i in a:
    print(k1[i])

Question 8 : Minimum Occurrence (R->Medium)

Problem Statement :
Given a sting , return the character that appears the minimum number of times in the string. The string will contain only ascii characters, from the ranges (“a”-”z”,”A”-”Z”,0-9), and case matters . If there is a tie in the minimum number of times a character appears in the string return the character that appears first in the string.

Input Format:
Single line with no space denoting the input string.

Output Format:
Single character denoting the least frequent character.

Constraints:
Length of string <=10^6

Sample Input:
cdadcda

Sample Output:
c

Explanation:
C and A both are with minimum frequency. So c is the answer because it comes first with less index.

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;
unordered_map < char,int > F;

int main()
{
    string s;
    int m=INT_MAX;
    cin >> s;
    for(auto i:s) F[i]++;
    for(auto i:F) m=min(m,i.second);
    for(auto i:s)
        if(F[i]==m)
            {
                cout << i;
                break;
            }
}

JAVA

Run

import java.util.*;
public class Main
{
  public static void main(String[] args)
{
   Scanner sc=new Scanner(System.in);
   String str=sc.next();
   char arr[]=str.toCharArray();
   int temp[]=new int[256];
    for(int i=0;i < arr.length;i++) { temp[arr[i]]++; } int min=Integer.MAX_VALUE; int index=0; for(int i=255;i >=  0;i--)
    {
       if(temp[i]==0)
        continue;
       min=Math.min(min,temp[i]);
      } 
    for(int i=0;i < arr.length;i++)
   {
       if(min==temp[arr[i]])
     {
        System.out.println(arr[i]);  
        break;
     }
  }
}
}

Python

Run

s=input()
ans=[]
for i in s:
    ans.append(s.count(i))
print(s[ans.index(min(ans))])

Question 9: Stars Between Bars

Given a string s consisting of stars “*” and bars “|” ,an array of starting indices starIndex, and an array of ending indices endIndex, determine the number of stars between any two bars within the substrings between the two indices inclusive . NOTE that in this problem indexing starts at 0.

A Star is represented as an asterisk [*=ascii decimal 42]
A Bar is represented as a Pipe [“|”=ascii decimal 124]

Example

s=’|**|*|’
n = 2
startIndex=[0,0]
endIndex=[4,5]

For the first pair of indices (0,4) the substrings is “|**|*” . There are 2 stars between a pair of bars
For the second pair of indices (0,5) the substring is “|**|*|” and there are 2+1=3 stars in between the bars.
Both of the answers are returned to the array [2,3].

Constraints

1 <= n <= 105
1 <= StartIndex[i] <= endIndex[i]
Each Character of s is either “*” or “|”

Input Format for Custom testing

First line contains a string S. The next line contains an integer n , the no.of elements in startIndex and endIndex. Each line i of the n subsequent lines contains an integer of startIndex. Each line i of the n subsequent lines contains an integer of endindex.

Sample Input

*|*| → s=”*|*|”
1 → size of startindex[] and endIndex[] is 1.
0 → startindex = 0
2 → endindex = 2

Sample output:

0

Explanation :

The substring from index = 0 to index = 2 is “*|*” . there is no consecutive pair of bars in this string.

C++

Java

Python

C++

Run

#include<bits/stdc++.h>
using namespace std;

int main() {
    string str;
    cin >> str;
    int n;
    cin >> n;
    vector startIndex(n);
    vector endIndex(n);
    for (int i = 0; i < n; i++) { cin >> startIndex[i];
    }
    for (int i = 0; i < n; i++) { cin >> endIndex[i];
    }
    int len = str.length();
    vector counter(len, -1);
    int count = 0;
    int lastIdx = -1;

    stack st;
    for (int i = 0; i < len; i++) {
        char ch = str[i];
        if (ch == '|') {
            while (!st.empty()) {
                int idx = st.top();
                st.pop();
                counter[idx] = i;
            }
        }
        st.push(i);
    }

    vector ansArr(n);
    for (int i = 0; i < n; i++) {
        int sIndex = startIndex[i];
        int eIndex = endIndex[i];
        int sCount = 0;
        if (str[sIndex] != '|')
            sIndex = counter[sIndex];
        if (sIndex == -1 || counter[sIndex] == -1) {
            ansArr[i] = 0;
            continue;
        }
        while (sIndex < eIndex) {
            int nextIdx = counter[sIndex];
            if ((nextIdx != -1) && (nextIdx <= eIndex))
            {
                sCount += nextIdx - sIndex - 1;
            }
            sIndex = nextIdx;
        }
        ansArr[i] = sCount;
    }

    for (int ele : ansArr) {
        cout << ele << " ";
    }
    cout << endl;

    return 0;
}

Java

Run

import java.util.*;

class Main
{
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        String str = scn.next();
        int n = scn.nextInt();
        int startIndex[] = new int[n];
        int endIndex[] = new int[n];
        for(int i = 0; i < n; i++)
        {
            startIndex[i] = scn.nextInt();
        }
        for(int i = 0; i < n; i++)
        {
            endIndex[i] = scn.nextInt();
        }
        int len = str.length();
        int counter[] = new int[len];
        int count = 0;
        int lastIdx = -1;
        
        Arrays.fill(counter, -1);
        Stack st = new Stack<>();
        for(int i = 0; i < len; i++)
        {
            char ch = str.charAt(i);
            if(ch == '|')
            {
                while(!st.empty())
                {
                    int idx = st.pop();
                    counter[idx] = i;
                }
            }
            st.push(i);
        }
        
        int ansArr[] = new int[n];
        for(int i = 0; i < n; i++)
        {
            int sIndex = startIndex[i];
            int eIndex = endIndex[i];
            int sCount = 0;
            if(str.charAt(sIndex) != '|')
                sIndex = counter[sIndex];
            if(sIndex == -1 || counter[sIndex] == -1)
            {
                ansArr[i] = 0;
                continue;
            }
            while(sIndex < eIndex)
            {
                int nextIdx = counter[sIndex];
                if((nextIdx != -1) && (nextIdx <= eIndex))
                {
                    sCount += nextIdx - sIndex - 1;
                }
                sIndex = nextIdx;
            }
            ansArr[i] = sCount;
        }
        
        for(int ele : ansArr)
        {
            System.out.print(ele + " ");
        }
    }
}

Python

Run

str = input()
n = int(input())
startIndex = []
endIndex = []
for i in range(n):
    startIndex.append(int(input()))
for i in range(n):
    endIndex.append(int(input()))

len_str = len(str)
counter = [-1] * len_str
count = 0
lastIdx = -1

st = []
for i in range(len_str):
    ch = str[i]
    if ch == '|':
        while st:
            idx = st.pop()
            counter[idx] = i
    st.append(i)

ansArr = []
for i in range(n):
    sIndex = startIndex[i]
    eIndex = endIndex[i]
    sCount = 0
    if str[sIndex] != '|':
        sIndex = counter[sIndex]
    if sIndex == -1 or counter[sIndex] == -1:
        ansArr.append(0)
        continue
    while sIndex < eIndex:
        nextIdx = counter[sIndex]
        if nextIdx != -1 and nextIdx <= eIndex:
            sCount += nextIdx - sIndex - 1
        sIndex = nextIdx
    ansArr.append(sCount)

for ele in ansArr:
    print(ele, end=" ")

Question 10 : Death Note

Problem Statement :

Ryuk, the Shinigami (God of death) had allowed Light Yagami, a school student, to kill as many people as he can by using a death note. But writing the names barely will allow other people to watch them. So he encrypts the names using digits, where a means 1, b means 2 and so on upto z is 26. Now if he gives numbers, there is a communication error because a number string can be decrypted by the death note in various ways and eventually killing them all. If everyone in the world has a unique name, for a given number, how many people can die?
NOTE THAT: There is every possible name in the world with the 26 letters, and capital or small letters is not a problem.

Input format:
A number stream denoting the first name’s encrypted version

Output Format:
Number of people dying by this.

Constraints:
1<=stream length<=10^8

Sample Input: 1267
Sample Output: 3
Output Specification:Two people of the name azg and lfg die.

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;
string s;
int ans;
void func(int i, int n) {
    if (i == n - 1 || i == n) {
        ans++;
        return;
    }

    if (s[i] == '1')
        func(i + 2, n);
    else if (s[i] == '2' && s[i + 1] < '7') func(i + 2, n); func(i + 1, n); } int main() { ans = 0; cin >> s;
    func(0, s.length());
    cout << ans;
}

JAVA

Run

import java.util.*;
class Main {
    static String s;
    static int ans;
    public static void func(int i, int n) {
        if (i == n - 1 || i == n) {
            ans++;
            return;
        }
        if (s.charAt(i) == '1')
            func(i + 2, n);
        else if (s.charAt(i) == '2' && s.charAt(i + 1) < '7')
            func(i + 2, n);
        func(i + 1, n);

    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        s = sc.next();
        func(0, s.length());
        System.out.println(ans);
    }
}

Python

Run

def solve(s, n):
    if n == 0 or n == 1:
        return 1
    ans = 0
    if s[n - 1] > "0":
        ans = solve(s, n - 1)
    if s[n - 2] == "1" or (s[n - 2] == "2" and s[n - 1] < "7"):
        ans += solve(s, n - 2)
    return ans
s = input()
print(solve(s, len(s)))

FAQs on Darwinbox Coding Questions

Question 1: How many coding rounds are conducted for Darwinbox Recruitment Drive?

There are 2 Rounds of Coding Questions conducted by Darwinbox for their Recruitment Process.

Coding Test 1 – Time Duration – 60 Mins Approx.
Coding Test 2 – Time Duration – 90 Mins Approx.

Question 2: What are the roles offered by Darwinbox for freshers ?

Darwinbox usually offers the Job Profile for Freshers like:

Product Development Intern
Software Development Engineer
Fullstack Developer Intern, etc.

Question 3: What is the eligibility criteria for Darwinbox Recruitment Process ?

Basic eligibility criteria for Darwinbox Recruitment Process for Hiring freshers:

Branch – B.tech from CS/IT.
Batch – 2024
Score Required: Minimum 60% or Equivalent CGPA throughout the Academics.

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