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HCL Practice Coding Questions
As coding turns out to be the most crucial section of HCL recruitment examination, we believe that having a mindset for HCL Coding Questions & Answers is important. As the number of questions is only 2 but the time allotted is only 20 mins so it becomes very hard to pass all the test cases. Hence, this makes it obvious to understand the pattern of questions asked in HCL Placement Paper. This page will help you to know the solution of some coding questions.
About HCL Exam Pattern 2024
| Topics | Number of Questions | Time | Difficulty |
|---|---|---|---|
| Numerical Ability | 15 | 15 mins | Medium |
| Verbal Ability | 15 | 15 mins | Medium |
| Reasoning Ability | 15 | 15 mins | Medium |
| Computer Programming | 30 | 30 mins | High |
| Coding | 2 | 20 mins | High |
HCL Coding Questions : Detailed Analysis
Find the complete analysis for HCL Coding Section below in the tabular format for better understanding.
| HCL Section | No. of questions | Total Time |
|---|---|---|
| MCQ | 30 | 30 mins |
| Coding Challenge | 2 | 20 mins |
NOTE:
The HCL Recruitment and Placement Process has witnessed numerous changes in the paper pattern and syllabus.
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Some facts about Eligibility Criteria:-
The eligibility criteria for HCL is given below in the tabulated format.
| HCL | Requirements |
|---|---|
| Qualification | B.Tech |
| Eligible stream | (CSE,IT,ECE,EN,EI, ME) |
| Percentage Criteria | 75% throughout academics (10th, 12th / Diploma & UG) |
| Backlog | No Active Backlog |
| Service Agreement | Probation period for selected SE/GET will be of 12 Months. |
Practice Coding Question for HCL
Question 1 :
PrepInsta Name : Move Hash to Front
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
Run
#include<stdio.h>
#include<string.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++
Run
#include <bits/stdc++.h>
using namespace std;
string moveHash(string s)
{
string ans="";
for(auto i:s)
if(i=='#') ans='#'+ans;
else ans+=i;
return ans;
}
int main()
{
string s;
getline(cin,s);
cout<< moveHash(s);
}
JAVA
import java.util.*;
public class Main
{
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
Run
def moveHash(s):
x=s.count("#")
s=s.replace("#","")
return ("#"*x+s)
s=input()
print(moveHash(s))Question 2 :
PrepInsta Name : 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
C++
JAVA
Python
C
#include <stdio.h>
#include <string.h>
int main()
{
int number1,number2,count=0;
scanf("%d%d",&number1,&number2);
if(number1< number2)
printf("Not possible");
else
{
int flag=0;
int temp1,temp2;
while(number1!=0 && number2!=0)
{
temp1=0;
temp2=number2%10;
if(flag==1)
temp1=number1%10-1;
else
temp1=number1%10;
if(temp1< temp2)
{
count+=1;
flag=1;
}
else
flag=0;
number1=number1/10;
number2=number2/10;
}
printf("%d",count);
}
return 0;
}
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 3 :
PrepInsta Name : Capitalize/Decapitalize
Problem statement : You’re given a function that accepts the following, a string1, its length and a character c. Your job is to replace all the occurrences of character c in string1 and capitalize it or decapitalize it based on the character c.
Example 1
Example 2
Example 1
Input :
hello world
l
Output :
heLLo worLd
Example 2
Input :
prepinsta
p
Output :
PrePinsta
C
C++
JAVA
Python
C
Run
#include<stdio.h>
#include<string.h>
void changeCharacter (char str[], char chr, int len)
{
int i, j;
for (i = 0; i < len; i++)
if (chr == str[i])
{
if (str[i] >= 64 && str[i] <= 92)
str[i] += 32;
else if (str[i] >= 90 && str[i] <= 122)
str[i] -= 32;
}
printf ("%s", str);
}
int main ()
{
char str[1000], chr;
int len;
scanf ("%[^\n]s", str);
len = strlen (str);
scanf (" %c", &chr);
changeCharacter (str, chr, len);
}
C++
Run
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
getline(cin,s);
char k; cin>>k;
for(auto i:s)
if(i==k)
{
if(i>95)cout<<char(i-32);
else cout<<char(i+32);
}
else cout<<i;
}
JAVA
Run
import java.util.Scanner;
public class Main
{
public static void change(String str, char c, int len)
{
char[] ch = str.toCharArray();
for(int i=0; i<ch.length; i++)
{
if( c == ch[i] )
{
if(Character.isUpperCase(ch[i]))
{
ch[i]= Character.toLowerCase(ch[i]);
}
else if(Character.isLowerCase(ch[i]))
{
ch[i]= Character.toUpperCase(ch[i]);
}
}
}
System.out.print(new String(ch) );
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
char c = sc.next().charAt(0);
int len = str.length();
change(str, c, len);
}
}
Python
Run
s=input()
k=input()
if(k.isupper()):
s=s.replace(k,chr(ord(k)+32))
else:
s=s.replace(k,chr(ord(k)-32))
print(s)
Question 4 :
Problem statement : You’re given a string where multiple characters are repeated consecutively. You’re supposed to reduce the size of this string using mathematical logic given as in the example below :
Example 1
Example 2
Example 1
Input :
abbccccc
Output:
ab2c5
Example 2
Input :
aabbbbeeeeffggg
Output:
a2b4e4f2g3
C
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;
}
C++
Run
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
getline(cin,s);
map< char,int > m;
for(auto i:s) m[i]++;
for(auto i:m)cout<< i.first<< i.second;
}
JAVA
import java.util.*;
public class Main
{
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
s=input()
i=1
c=1
while i<len(s):
if s[i]==s[i-1]:
c+=1
else:
print(s[i-1],end="")
print(c,end="")
c=1
i+=1
print(s[i-1],end="")
print(c)
Question 5 :
PrepInsta Name : 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.
Example :
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
C++
Java
Python
C
Run
#include<stdio.h>
#include<conio.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 ",a[i][j]);
}
printf("\n");
}
while(rs< re && cs< ce)
{
for(i=cs;i< ce;i++)
{
printf("%d ",a[rs][i]);
}
rs++;
for(i=rs;i< re;i++)
{
printf("%d ",a[i][ce-1]);
}
ce--;
if(rs< re) { for(i=ce-1; i>=cs; --i)
{
printf("%d ", a[re - 1][i]);
}
re--;
}
if(cs< ce) { for(i=re-1; i>=rs; --i)
{
printf("%d ", a[i][cs]);
}
cs++;
}
}
return 0;
}
C++
Run
#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
Run
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
Run
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))
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