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Kickdrum Coding Questions and Answers
Kickdrum Coding Questions with Solutions
Kickdrum is a group of enterprise-grade strategists, architects, and builders who collaborate with operations executives and technology investors to protect and improve enterprises. Kickdrum Coding Questions and Answers page will help you to get sample Coding Questions asked in the Online Selection Process of Kickdrum.
Go through this page to get all Sample Kickdrum Coding Questions for preparing for Online Assessments and Technical Interviews of the same.
Kickdrum Coding Questions with Answers
Question 1 : Guess the word
Problem Statement :
Kochouseph Chittilappilly went to Dhruv Zplanet , a gaming space, with his friends and played a game called “Guess the Word”.
Rules of games are –
Computer displays some strings on the screen and the player should pick one string / word if this word matches with the random word that the computer picks then the player is declared as Winner.
Kochouseph Chittilappilly’s friends played the game and no one won the game. This is Kochouseph Chittilappilly’s turn to play and he decided to must win the game.
What he observed from his friend’s game is that the computer is picking up the string whose length is odd and also that should be maximum. Due to system failure computers sometimes cannot generate odd length words. In such cases you will lose the game anyways and it displays “better luck next time”. He needs your help. Check below cases for better understand
Sample input 0:
5 → number of strings
Hello Good morning Welcome you
Sample output 0:
morning
Explanation:
Hello → 5
Good → 4
Morning → 7
Welcome → 7
You → 3
First word that is picked by computer is morning
Sample input 1:
3
Go to hell
Sample output 1:
Better luck next time
Explanation:
Here no word with odd length so computer confuses and gives better luck next time
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#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
string a, result = "";
while (n--) {
cin >> a;
if (a.length() & 1)
if (a.length() > result.length())
result = a;
}
if (result == "") cout << "Better luck next time";
else cout << result;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String arr[] = new String[n];
for (int i = 0; i < n; i++)
arr[i] = sc.next();
int len = 0;
ArrayList < String > oddLength = new ArrayList < String > ();
for (int i = 0; i < n; i++) {
len = arr[i].length();
if (len % 2 == 1)
oddLength.add(arr[i]);
}
if (oddLength.size() == 0)
System.out.println("Better luck next time");
else {
Iterator itr = oddLength.iterator();
int max = -1;
String res = "";
while (itr.hasNext()) {
String temp = (String) itr.next();
if (temp.length() > max) {
res = temp;
max = temp.length();
}
}
System.out.println(res);
}
}
}
n = int(input())
L = list(map(str, input().split()))
result = ""
for a in L:
if len(a) & 1:
if len(a) > len(result):
result = a
if result == "":
result = "Better luck next time"
print(result)
Question 2 : Stars Between Bars
Problem statement :
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 1.
- A Star is represented as an asterisk [*=ascii decimal 42]
- A Bar is represented as a Pipe [“|”=ascii decimal 124]
Example :
s=’|**|*|’
startIndex=[1,1]
endIndex=[5,6]
For the first pair of indices (1,5) the substrings is “|**|*” . There are 2 stars between a pair of bars
For the second pair of indices (1,6) 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 :
int [n];each element[i] answers the query of startIndex[i] to endindex[i]
Constraints
1<=n<=105
1<=StartInde[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. Each line i of the n subsequent lines contains an integer of startIndex.the next line contains an integer n , the no.of elements in endIndex. Each line i of the n subsequent lines contains an integer of endindex
>Sample Case 0
Sample Input for Custom Testing
*|*| → s=”*|*|”
1 → startindex[] size=1
1 → startindex= 1
1 → endindex[] size=1
3 → endindex=3
Sample output :
0
Explanation :
The substring from index =1 to index=3 is “*|*” . there is no consecutive pair of bars in this string.
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
int n;
cin >> n;
int st[n], ed[n];
for (int i = 0; i < n; i++) { cin >> st[i];
st[i]--;
}
cin >> n;
for (int i = 0; i < n; i++) { cin >> ed[i];
ed[i]--;
}
vector < int > bars;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '|')
bars.push_back(i);
}
for (int i = 0; i < n; i++) {
int idx = lower_bound(bars.begin(), bars.end(), st[i]) - bars.begin();
st[i] = bars[idx];
idx = lower_bound(bars.begin(), bars.end(), ed[i]) - bars.begin();
if (idx == 0 || bars[idx] == ed[i])
continue;
ed[i] = bars[idx - 1];
}
int sz = s.length();
int starCt[sz] = {
0
};
if (s[0] == '*')
starCt[0] = 1;
for (int i = 1; i < sz; i++) {
starCt[i] = starCt[i - 1] + (s[i] == '*');
}
for (int i = 0; i < n; i++) { if (st[i] >= ed[i]) {
cout << 0 << endl; } else { int ans = starCt[ed[i]]; if (st[i] > 0)
ans -= starCt[st[i] - 1];
cout << ans << endl;
}
}
return 0;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.next();
int n = sc.nextInt();
int startIndex[] = new int[n];
int endIndex[] = new int[n];
for (int i = 0; i < n; i++)
startIndex[i] = sc.nextInt();
for (int i = 0; i < n; i++)
endIndex[i] = sc.nextInt();
int count = 0;
for (int i = 0; i < n; i++) {
count = 0;
String sub = str.substring(startIndex[i] - 1, endIndex[i]);
int start = sub.indexOf("|");
int end = sub.lastIndexOf("|");
for (int j = start; j < end; j++)
if (sub.charAt(j) == '*')
count++;
System.out.print(count + " ");
}
s = input()
start = []
end = []
n = int(input())
for i in range(n):
start.append(int(input()))
for i in range(int(input())):
end.append(int(input()))
for i in range(n):
Str = s[start[i] - 1 : end[i]]
print(Str.strip("*").count("*"))
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 :
3
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
#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;
}
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);
}
}
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: Candies
Problem Description
Question:- Krishna loves candies a lot, so whenever he gets them, he stores them so that he can eat them later whenever he wants to.
He has recently received N boxes of candies each containing Ci candies where Ci represents the total number of candies in the ith box. Krishna wants to store them in a single box. The only constraint is that he can choose any two boxes and store their joint contents in an empty box only. Assume that there are an infinite number of empty boxes available.
At a time he can pick up any two boxes for transferring and if both the boxes contain X and Y number of candies respectively, then it takes him exactly X+Y seconds of time. As he is too eager to collect all of them he has approached you to tell him the minimum time in which all the candies can be collected.
Input Format:
- The first line of input is the number of test case T
- Each test case is comprised of two inputs
- The first input of a test case is the number of boxes N
- The second input is N integers delimited by whitespace denoting the number of candies in each box
Output Format: Print minimum time required, in seconds, for each of the test cases. Print each output on a new line.
Constraints:
- 1 < T < 10
- 1 < N< 10000
- 1 < [Candies in each box] < 100009
| S. No. | Input | Output |
|---|---|---|
| 1 | 1 4 1 2 3 4 | 19 |
| 2 | 1 5 1 2 3 4 5 | 34 |
Explanation for sample input-output 1:
4 boxes, each containing 1, 2, 3 and 4 candies respectively.Adding 1 + 2 in a new box takes 3 seconds.Adding 3 + 3 in a new box takes 6 seconds.Adding 4 + 6 in a new box takes 10 seconds.Hence total time taken is 19 seconds. There could be other combinations also, but overall time does not go below 19 seconds.
Explanation for sample input-output 2:
5 boxes, each containing 1, 2, 3, 4 and 5 candies respectively.Adding 1 + 2 in a new box takes 3 seconds.Adding 3 + 3 in a new box takes 6 seconds.Adding 4 + 6 in a new box takes 10 seconds.Adding 5 + 10 in a new box takes 15 seconds.Hence total time taken is 34 seconds. There could be other combinations also, but overall time does not go below 33 seconds.
#include
#include
int main()
{
int i,j;
int num_box=0,k=0,sum=0,times=0,tst_case,temp=0;
long c[10000],s[10000];
printf("Enter the number of test case:");
scanf("%d",&tst_case);
printf("Enter the number of boxes:");
for(int l=0;l<tst_case;l++)
{
scanf("%d",&num_box);
}
printf("Enter the number of candies in each box:");
for(i=0;i<num_box;i++)
{
scanf("%ld",&c[i]);
}
for(i=0;i<num_box;i++)
{
for(j=i+1;j<num_box;j++) { if(c[i]>c[j])
{
temp=c[i];
c[i]=c[j];
c[j]=temp;
}
}
}
sum=0;
k=0;
for(i=0;i<num_box;i++)
{
sum=sum+c[i];
s[k]=sum;
k++;
}
times=0;
printf("Minimum time requried:");
for(i=1;i<k;i++)
times=times+s[i];
printf("%d\n",times);
return 0;
}
Output 1 4 1 2 3 4 19
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i,j;
int num_box=0,k=0,sum=0,times=0,tst_case,temp=0;
long c[10000],s[10000];
cout<<("Enter the number of test case:");
cin>>tst_case;
cout<<("Enter the number of boxes:");
for(int l=0;l<tst_case;l++) { cin>>num_box;
}
cout<<("Enter the number of candies in each box:");
for(i=0;i<num_box;i++) { cin>>c[i];
}
for(i=0;i<num_box;i++)
{
for(j=i+1;j<num_box;j++) { if(c[i]>c[j])
{
temp=c[i];
c[i]=c[j];
c[j]=temp;
}
}
}
sum=0;
k=0;
for(i=0;i<num_box;i++)
{
sum=sum+c[i];
s[k]=sum;
k++;
}
times=0;
cout<<("Minimum time requried:");
for(i=1;i<k;i++)
times=times+s[i];
cout<<times;
return 0;
}
Output 1 4 1 2 3 4 19
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n, i, k = 0, sum = 0, s1 = 0, t, temp = 0, j; long c[] = new long[1000009]; long s[] = new long[100009]; t = sc.nextInt(); for (int l = 0; l < t; l++) { n = sc.nextInt(); for (i = 0; i < n; i++) c[i] = sc.nextLong(); for (i = 0; i < n; i++) { for (j = i + 1; j < n; j++) { if (c[i] > c[j]) { temp = (int) c[i]; c[i] = c[j]; c[j] = temp; } } } sum = 0; k = 0; for (i = 0; i < n; i++) { sum = (int) (sum + c[i]); s[k] = sum; k++; } s1 = 0; for (i = 1; i < k; i++) s1 = (int) (s1 + s[i]); System.out.println(s1); } } }
Output 1 4 1 2 3 4 19
T = int(input())
arr1 = []
for _ in range(T):
N = int(input())
arr = list(map(int, input().split()))
arr.sort()
count = arr[0]
for i in range(1, len(arr)):
count += arr[i]
arr1.append(count)
print(sum(arr1))
Output 1 4 1 2 3 4 19
Question 5: Elements of Matrix
Problem Statement
You are required to input the size of the matrix then the elements of matrix, then you have to divide the main matrix in two sub matrices (even and odd) in such a way that element at 0 index will be considered as even and element at 1st index will be considered as odd and so on. then you have sort the even and odd matrices in ascending order then print the sum of second largest number from both the matrices
Example
- enter the size of array : 5
- enter element at 0 index : 3
- enter element at 1 index : 4
- enter element at 2 index : 1
- enter element at 3 index : 7
- enter element at 4 index : 9
Sorted even array : 1 3 9
Sorted odd array : 4 7
7
#include
int main() {
int arr[100];
int length, i, j, oddlen, evenlen, temp;
int odd[50], even[50];
printf("Enter the length of the array: ");
scanf("%d", &length);
for (i = 0; i < length; i++) {
printf("Enter element at index %d: ", i);
scanf("%d", &arr[i]);
}
if (length % 2 == 0) {
oddlen = length / 2;
evenlen = length / 2;
} else {
oddlen = length / 2;
evenlen = (length / 2) + 1;
}
for (i = 0; i < length; i++) {
if (i % 2 == 0) {
even[i / 2] = arr[i];
} else {
odd[i / 2] = arr[i];
}
}
for (i = 0; i < evenlen - 1; i++) {
for (j = i + 1; j < evenlen; j++) { if (even[i] > even[j]) {
temp = even[i];
even[i] = even[j];
even[j] = temp;
}
}
}
for (i = 0; i < oddlen - 1; i++) {
for (j = i + 1; j < oddlen; j++) { if (odd[i] > odd[j]) {
temp = odd[i];
odd[i] = odd[j];
odd[j] = temp;
}
}
}
printf("\nSorted even array: ");
for (i = 0; i < evenlen; i++) {
printf("%d ", even[i]);
}
printf("\n");
printf("Sorted odd array: ");
for (i = 0; i < oddlen; i++) {
printf("%d ", odd[i]);
}
printf("\n");
printf("%d", even[evenlen - 2] + odd[oddlen - 2]);
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int main ()
{
int arr[100];
int length, i, j, oddlen, evenlen, temp, c, d;
int odd[50], even[50];
cout << "enter the length of array : "; cin >> length;
for (i = 0; i < length; i++)
{
cout << "Enter element at " << i << " index : "; cin >> arr[i];
}
if (length % 2 == 0)
{
oddlen = length / 2;
evenlen = length / 2;
}
else
{
oddlen = length / 2;
evenlen = (length / 2) + 1;
}
for (i = 0; i < length; i++) // seperation of even and odd array
{
if (i % 2 == 0)
{
even[i / 2] = arr[i];
}
else
{
odd[i / 2] = arr[i];
}
}
for (i = 0; i < evenlen - 1; i++) // sorting of even array
{
for (j = i + 1; j < evenlen; j++) { temp = 0; if (even[i] > even[j])
{
temp = even[i];
even[i] = even[j];
even[j] = temp;
}
}
}
for (i = 0; i < oddlen - 1; i++) // sorting of odd array
{
for (j = i + 1; j < oddlen; j++) { temp = 0; if (odd[i] > odd[j])
{
temp = odd[i];
odd[i] = odd[j];
odd[j] = temp;
}
}
}
cout << "\nSorted even array : "; // printing even array
for (i = 0; i < evenlen; i++)
{
cout << even[i] << " ";
}
cout << "\n";
cout << "Sorted odd array : "; // printing odd array
for (i = 0; i < oddlen; i++)
{
cout << odd[i] << " ";
}
cout << endl;
cout << even[evenlen - 2] + odd[oddlen - 2]; // printing final result
}
import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;
public class Main
{
public static void main (String[]args)
{
Scanner sc = new Scanner (System.in);
System.out.print ("Enter size of array : ");
int arrsize = sc.nextInt ();
int[] main = new int[arrsize];
ArrayList < Integer > even = new < Integer > ArrayList ();
ArrayList < Integer > odd = new < Integer > ArrayList ();
System.out.println ("Enter " + arrsize + " Elements");
for (int i = 0; i < arrsize; i++)
main[i] = sc.nextInt ();
for (int i = 0; i < arrsize; i++)
{
if (i % 2 == 0)
even.add (main[i]);
else
odd.add (main[i]);
}
Collections.sort (even);
Collections.sort (odd);
System.out.println ("Sorted even array ");
for (int e:even)
System.out.print (e + " ");
System.out.println ();
System.out.println ("sorted odd array ");
for (int e:odd)
System.out.print (e + " ");
System.out.println ();
int evensec = even.get (even.size () - 2);
int oddsec = odd.get (odd.size () - 2);
System.out.println ("Second Largest Element in Even List is:" + evensec);
System.out.println ("Second Largest Element in Odd List is:" + oddsec);
System.
out.println ("Sum Of Second Largest Element Of Odd and Even List:" +
(evensec + oddsec));
}
}
array = []
evenArr = []
oddArr = []
n = int(input("Enter the size of the array:"))
for i in range(0,n):
number = int(input("Enter Element at {} index:".format(i)))
array.append(number)
if i % 2 == 0:
evenArr.append(array[i])
else:
oddArr.append(array[i])
evenArr = sorted(evenArr)
print("Sorted Even Array:", evenArr[0:len(evenArr)])
oddArr = sorted(oddArr)
print("Sorted Odd Array:", oddArr[0:len(oddArr)])
print(evenArr[1] + oddArr[1])
evenArr = sorted(evenArr)
print("Sorted Even Array:", evenArr[0:len(evenArr)])
oddArr = sorted(oddArr)
print("Sorted Odd Array:", oddArr[0:len(oddArr)])
print(evenArr[-2] + oddArr[-2])
Question 6 : Duplicates
Problem statement : The principal has a problem with repetitions. Everytime someone sends the same email twice he becomes angry and starts yelling. His personal assistant filters the mails so that all the unique mails are sent only once, and if there is someone sending the same mail again and again, he deletes them. Write a program which will see the list of roll numbers of the student and find how many emails are to be deleted.
Sample Input:
6
1
3
3
4
3
3
Sample Output:
3
#include<bits/stdc++.h>
using namespace std;
map<int,int> m;
int main()
{
int n,a,ans=0;
cin>>n;
while(n--)
{
cin>>a;
m[a]++;
if(m[a]>1) ans++;
}
cout<<ans;
}
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
TreeSet list=new TreeSet<>();
for(int i=0;i<n;i++)
list.add(sc.nextInt());
System.out.println(Math.abs(n-list.size()));
}
}
def countDuplicate(numbers):
c=0
for i in set(numbers):
if numbers.count(i)>1:
c+=numbers.count(i)-1
return c
n=int(input())
numbers=[]
for i in range(n):
numbers.append(int(input()))
print(countDuplicate(numbers))
Question 7 : Set Bit
Problem Statement :
You are given an integer, N. You have to turn it into the binary representation of it, and find out how many set bits are there in the binary representation.
Input Format:
The first line contains the integer.
Output Format:
One line containing an integer denoting the number of setbits.
Constraints:
1<=N<=10^9
Sample Input:
8
Output:
1
Output Description:
8 in binary is 1000.
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, ans = 0;
cin >> n;
while (n) {
n &= (n - 1);
ans++;
}
cout << ans;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int count = 0;
while (n != 0) {
if ((n & 1) == 1)
count++;
n = n >> 1;
}
System.out.println(count);
}
}
n = int(input())
ans = 0
while n:
n &= n - 1
ans += 1
print(ans)
Question 8 : 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.
#include
using namespace std;
int main() {
float temperature;
cin >> temperature;
cout << (temperature * 9) / 5 + 32;
return 0;
}
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);
}
}
def ctof(c):
return (c * 9) / 5 + 32
temperature = float(input())
ans = ctof(temperature)
print(round(ans, 2))
Question 9 :Number with 2
Problem Statement :
Suppose you are in a number system, where if the number doesn’t contain 2 in the unit digit then the number is not valid. So the first number of the number system is 2, the second number is 12, and the third is 22.
for a given integer n, you have to print the nth element of the number system.
Input Format:
First line, containing n denoting the number of test cases.
then n number of lines for the query.
Output Format:
Print the consecutive number in the number system for each query.
Sample Input:
3
Sample Output:
22
Explanation:
1st number will be 2 , 2nd number will be 12 and third number will be 32
#include
using namespace std;
int main() {
int n;
cin >> n;
cout << (n - 1) * 10 + 2;
return 0;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
System.out.println((n - 1) * 10 + 2);
}
}
n = int(input()) print((n - 1) * 10 + 2)
Question 10 :Jack’s Text
Problem Statement :
Jack is learning to type english from the beginning and he is making an error of repeating the same words in his texts over whatsapp. Write a function that will take input for his text sent to you and then keep only the unique texts.
Note that, the uniqueness is about being word specific not position, there are nothing but alphabets in the sentences and words are separated only with white space.
Constraints:
Words in the line<=10^5
Alphabets in the words<=20
Sample Input:
Send send the image send to to to me
Output:
Send the mage to me
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, ans = 0;
cin >> n;
while (n) {
n &= (n - 1);
ans++;
}
cout << ans;
}
import java.util.*;
class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
String arr[] = str.split(" ");
String first = arr[0];
arr[0] = arr[0].toLowerCase();
LinkedHashMap < String, Boolean > map = new LinkedHashMap < String, Boolean > ();
for (int i = 0; i < arr.length; i++)
map.put(arr[i], true);
Set s = map.entrySet();
Iterator itr = s.iterator();
System.out.print(first + " ");
int j = 0;
while (itr.hasNext()) {
Map.Entry m = (Map.Entry) itr.next();
if ((Boolean) m.getValue() == true && j == 1)
System.out.print(m.getKey() + " ");
j = 1;
}
}
}
arr = list(input().split())
dup = []
ans = ""
for i in arr:
if i not in dup:
dup.append(i)
ans += i + " "
print(ans)
FAQs on Kickdrum Coding Questions and Answers
Question 1: What is the job role offered in Kickdrum?
The Job Role Offered is of Software Developer Intern in which a student needs to develop complex web application with responsive design techniques and industry leading UX.
Question 2: What is the salary offered by Kickdrum during Internship?
The stipend during internship is INR 35, 000 PM and post completion of internship it comes with a full time role in which the package offered is INR 16 Lacs
Question 3: What is the eligibility criteria to apply for this role?
As per placement policy the eligibility criteria is
- BTech – CSE, IT
- CGPA 8 and above
- 80% & above in X, XII. No active backlog(s).
Question 4: What is the selection process ?
The selection process consist of
- Online Assessment
- Video Screening
- Technical Round – (virtual interviews) and Culture Round.
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