WeCP Coding Questions

February 14, 2024

WeCP Practice Coding Questions for Freshers

WeCP is a technical skill assessment platform which helps the organization in recruiting students who are highly proficient in technical skills like hardware and software. One of the most important sections of WeCP exams is the coding round.

WeCP Practice Coding Questions for Freshers

Coding Round	Details
Number of Questions	2-3
Time Limit	2 hrs(approx)
Difficulty level	moderate
Package Offered	5 LPA – 7.5 LPA

Practice Questions for WeCP Coding Round

Question 1: Sentence counter

Problem statement-: We know sentences are terminated with certain punctuations like ‘.’,’!’,’?’, whereas there are several punctuations which don’t terminate the punctuations. Given for a paragraph, write a code to find out how many sentences are there.

Note that, ‘…’,’,’,’-‘ these don’t terminate a sentence.

Constraints:

Number of words in the paragraph<=10^8

Input Format:

Paragraph ended with an enter.

Output Format:

Single Integer denoting number of sentences.

Sample Input:

Hello! How are you? Last time I saw you… you were nervous.

Sample Output:

Explanation:

The three sentences are:

hello!

How are you?

Last time I saw you… you were nervous.

C++

Python

Java

C++

#include <bits/stdc++.h>
using namespace std;
 
int SentenceCount(string s)
{
    istringstream ss(s);
    int ans=0;
    while(ss)
    {
        string w;ss>>w;
        if(w=="") break;
        if(w[w.length()-1]=='!'||w[w.length()-1]=='?') ans++;
        else if(w==".") ans++;
        else if(w.length()>1)
        if(w[w.length()-1]=='.'&&w[w.length()-2]!='.') ans++;
    }
    return ans;
}
 
int main()
{
    string s;
    getline(cin,s);
    cout<<SentenceCount(s);
}

Python

def SentenceCount(L):
    ans=0
    for w in L:
        if w[len(w)-1]=='!' or w[len(w)-1]=='?':
            ans+=1
        elif(w=="."):
            ans+=1
        elif len(w)>1:
            if w[len(w)-1]=='.' and w[len(w)-2]!='.':
                ans+=1 
    return ans
            
    
def main():

    L=list(map(str,input().split()))
    
    result = SentenceCount(L);
    print(result)
if __name__ == "__main__":
    main()

Java

import java.util.Scanner;
public class Application12 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine();
		System.out.println(sentenceCount(s));
	}
	private static int sentenceCount(String s) {
		String[] w = s.trim().split("\\s+");
		int ans=0;
		for(int i=0;i<w.length;i++) {
			String k = w[i];
			int len = k.length();
			if(k.lastIndexOf('!')==(len-1) || k.lastIndexOf('?')==(len-1) ) {
				ans++;
			}else if(k.equals(".")) {
				ans++;
			}else if(len>1) {
				if(k.charAt(len-1)=='.'&&k.charAt(len-2)=='.') {
					ans++;
				}
			}else {	
			}
		}
		return ans;
	}
}

Question 2: Even Odd Even Odd

Problem Statement -: A number is even odd special if and only if there are even digits side by side, same goes for odd digits too. So the distributions must be in an alternating basis of even digits and odd digits. You are given a number N, and you have to say how many special numbers are possible <=N, and are greater than 0.

Constraints:

1<=N<=10^8

Input Format:

First Line containing a single integer N.

Output Format:

Single line containing a single integer denoting the possible number of special numbers.

Sample Input:

Output:

Explanation:

20 such special numbers are there :-

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 21, 23, 25, 27, 29, 30

C++

Python

C++

#include<bits/stdc++.h>
using namespace std;
int ans;
int a,l;
void func(string s)
{
    int S=stoi(s);
    if(a>=S)
    {
        cout<<s<<endl;
    ans++;}
    if(s.length()==l) {
     return;
    }
    if ((S%10)&1)
    for(int i=0;i<9;i+=2)
    func(s+to_string(i));
    else
    for(int i=1;i<=9;i+=2)
    func(s+to_string(i));
}
 
int main()
{
    string s;
    cin>>s;ans=0;
    l=s.length();
     a=stoi(s);
    for(int i=2;i<9;i+=2)
    func(to_string(i));
    for(int i=1;i<=9;i+=2)
    func(to_string(i));
    cout<<ans;
}

Python

a=input()
ans=0
l=len(a)
def func(s):
    global ans
    global a
    global l
    S=int(s)
    if int(a)>=S:
        ans+=1 
    if len(s)==l:
        return
    if (S&1):
        for i in range(0,9,2):
            func(s+str(i))
    else:
        for i in range(1,10,2):
            func(s+str(i))
    
    
    
for i in range(2,9,2):
    func(str(i))
for i in range(1,10,2):
    func(str(i))
print(ans)

Question 3: Profit balance

Problem Statement-: Anand and Brijesh got a bunch of profit in their business. Now it’s time to divide the profit between themselves. Anand being the one with the calculator was the one who could decide who will get how much. But as he was one hell of an honest guy, he can’t cheat on Brijesh to get some extra money. So he decided to divide them in such a way where they can get almost the same amount possible. Although, it is sometimes impossible to divide the profit into two, because the profits must be written in their production managed copy in a way where you cannot share the profit of a single thing, A single Profit will be held by a single person.

You are the one who is called to mitigate the problem. Given an array of profit, find out in the process, what can be the minimum difference between them in terms of income.

Constraints:

1<=N<=10^3
0<=Profits<=1000

Input Format:

First line contains an integer N, number of profits.
Next line, N space separated Integers denoting the i-th Profit.

Output:

A single integer denoting minimum possible difference between the total profits.

Sample Input:

1 2 3 4

Output:

Explanation:

He will take 1 and 4, so his profit will be 5, so the difference will be 0.

C++

Python

Java

C++

#include<bits/stdc++.h>
using namespace std;
vector<int> arr;
int n,s;
map<int,map<int,int>> m;
 
int MinDif(int i,int ss)
{
    if(m[i][ss]) return m[i][ss];
    if(i==n) return m[i][ss]= abs(s-2*ss);
    if(ss>=s/2) return m[i][ss]=abs(s-2*ss);
    return m[i][ss]=min(MinDif(i+1,ss+arr[i]),MinDif(i+1,ss));
}
 
int main()
{
    cin>>n;int a;
    s=0;
    for(int i=0;i<n;i++)
    {cin>>a;arr.push_back(a);s+=a;}
    cout<<MinDif(0,0);
}

Python

from collections import defaultdict
m = defaultdict(lambda: defaultdict(int))
def MinDif(i,ss,a,n,m):
    global s
    if m[i][ss] != 0:
        return m[i][ss]
    if i==n:
        m[i][ss]=abs(s-(2*ss))
        return m[i][ss]
    if ss>=s//2:
        m[i][ss]=abs(s-(2*ss))
        return m[i][ss]
    m[i][ss]=min(MinDif(i+1,ss+a[i],a,n,m),MinDif(i+1,ss,a,n,m))
    return m[i][ss]


n=int(input())
a=list(map(int,input().split()))
s=sum(a)
print(MinDif(0,0,a,n,m))

Java

import java.util.HashMap;
import java.util.Scanner;
public class Application14 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int s=0;
		int arr[]=new int[n];
		for(int i=0;i<n;i++) {
			arr[i] = sc.nextInt();
			s+=arr[i];
		}
		HashMap<Integer,HashMap<Integer,Integer>> m = new HashMap<>();
		System.out.println(minDiff(0,0,s,n,m,arr));
	}
	private static int minDiff(int i, int j, int s, int n, HashMap<Integer, HashMap<Integer, Integer>> m, int[] arr) {
		if(m.containsKey(i)) {
			if(m.get(i).containsKey(j))
			return m.get(i).get(j);
		}
		if(i==n) {
			HashMap<Integer,Integer> z = new HashMap<>();
			z.put(j,  Math.abs(s-2*j));
			m.put(i,z);
			return m.get(i).get(j);
		}
		if(j>=s/2) {
			HashMap<Integer,Integer> z = new HashMap<>();
			z.put(j,  Math.abs(s-2*j));
			m.put(i,z);
			return m.get(i).get(j);
		}
		int temp = Math.min(minDiff(i+1,j+arr[i],s,n,m,arr), minDiff(i+1,j,s,n,m,arr));
		HashMap<Integer,Integer> z = new HashMap<>();
		z.put(j, temp);
		m.put(i, z);
		return m.get(i).get(j);
	}
}

Question 4: Class Monitor

Problem Statement-: After JEE Mains, some students got admission into an engineering college. Now there is a class consisting of such n students, and the HOD came to say it is time to select the class monitor. But He never gets all of them at one time. So he brought a register, every time he gets someone with less rank than the previous time he cut the name and wrote the name of the student and the rank.

For a given number of ranks he gets each time, you have to predict how many names are cut in the list.

Constraints:

Number of Visiting<=10^9
ranks <=10000

Input Format:

Number of Visiting N in their first line
N space separated ranks the HOD gets each time

Output Format:

Number of ranks cut in the list

Sample Input:

4 3 7 2 6 1

Sample Output:

C++

Python

Java

C++

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    int n,p,m=INT_MAX,ans=0;
    cin>>n;
    vector<int> a(n);
    for(int i=0;i<n;i++)
    {
        cin>>p;
        if(p<m) {m=p;ans++;}
    }
    cout<<ans-1;
}

Python

n=int(input())
a=list(map(int,input().split()))
m=a[0]
ans=0
for i in range(1,n):
    if a[i]<m:
        m=a[i]
        ans+=1
print(ans-1)

Java

import java.util.Scanner;

public class Application15 {

 public static void main(String[] args) {
  int n,p,ans=0,m=Integer.MAX_VALUE;
  Scanner sc = new Scanner(System.in);
  n = sc.nextInt();
  for(int i=0;i<n;i++) {
   p = sc.nextInt();
   if(p<m) {
    m=p;
    ans++;
   }
  }
  System.out.print(ans-1);
 }
}

Question 5: Corona for Computer

Problem Statement-: Every decimal number can be changed into its binary form. Suppose your computer has it’s own CoronaVirus, that eats binary digits from the right side of a number. Suppose a virus has 6 spikes, it will eat up 6 LSB binary digits in your numbers.

You will have a bunch of numbers, and your machine will have a virus with n spikes, you have to calculate what will be the final situation of the final numbers.

Input Format:

First line, a single Integer N
Second line N space separated integers of the bunch of values as array V
Third line a single integer n, the number of spikes in Corona for Computer

Output Format:

Single N space separated integers denoting the final situation with the array v.

Sample Input:

1 2 3 4 5

Sample Output:

0 0 0 1 1

Explanation:

5 is 101 in binary, when you cut the last two binary digits, its 1.

C++

Python

Java

C++

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    int N,n;cin>>N;
    vector<int> v(N);
    for(int i=0;i<N;i++) cin>>v[i];
    cin>>n;
    for(auto i:v)
    cout<<(i>>n)<<" ";
}

Python

N=int(input())
a=list(map(int,input().split()))
n=int(input())
s=""
for i in a:
    s+=str(i>>n)+" "
print(s)

Java

import java.util.Scanner;

public class Application16 {

 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();
  }
  int k = sc.nextInt();
  for(int i:arr) {
   System.out.print((i>>k));
   System.out.print(" ");
  }
 }

}

Question 6: Help of Prepsters

Problem Statement-: Arnab has given me a challenge. I have to calculate the number of numbers which are less than a certain value n, and have exactly k set bits in its binary form. As you are a Prepster like me, help me write a code that will take input for n and k and give the expected output.

Constraints :

1<=n<=10000
1<=k<=10

Input Format :

First line containing n and k space separated

Output Format :

Number of numbers present in a single line

Sample Input:

7 2

Sample Output:

Explanation:

011,110,101 -> These three numbers.

C++

Python

Java

C++

#include <bits/stdc++.h>
using namespace std;
int n,k,ans,l;
map<int,int> m;
 
void func(int i,string s,int L)
{
   // cout<<s<<endl;
    if(L>l) {return;}
    if(i==0 && m[stoull(s,0,2)]==0) {m[stoull(s,0,2)]++;ans++;}
    if(s!="") if(n<stoull(s,0,2)) {return;}
    func(i-1,s+"1",L+1);
    func(i,s+"0",L+1);
}
 
int main()
{
    cin>>n>>k;char res[10000];itoa(n,res,2);
    ans=0;l=strlen(res);
    func(k-1,"1",1);
    cout<<ans;
}

Python

ans=0
n,k=map(int,input().split())
l=len(bin(n)[2:])
def func(i,s,L):
    global l 
    global ans
    if(L>l):
        return
    if i==0 :
        ans+=1 
    if s!="":
        if n<int(s,2):
            return
    func(i-1,s+"1",L+1)
    func(i,s+"0",L+1)
    


func(k-1,"1",1)
print(ans)

Java

import java.util.Scanner;

public class Application17 {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int ans=0;
		for(int i=1;i<n;i++) {
			if(countSetBits(i)==k) {
				ans++;
			}
		}
		System.out.println(ans);
	}

	private static int countSetBits(int i) {
		if(i==0)
		return 0;
		else
			return 1 + countSetBits(i&(i-1));
	}

}

Question 7: Momentum LinkedList

Problem Statement -: Ratul made a linked list, a list made of n nodes, where every node has two variables, the velocity and the mass of a particle.

Since all the particles have the velocity in the same direction, find the total momentum of the entity made by the particles from the linked list.

Constraints :

1<=n<=10000
1<=m,v<=100

Input format:

First line containing n, number of nodes
Then n lines containing the mass and the velocity space separated.

Output Format:

Single integer denoting the momentum

Sample Input:

1 3

2 4

2 3

4 5

Sample Output:

C++

Python

Java

C++

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    int n,s=0,m,v;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>m>>v;
        s+=(m*v);
    }
    cout<<s;
}

Python

n=int(input())
s=0
for i in range(n):
    m,v=map(int,input().split())
    s+=(m*v)
print(s)

Java

import java.util.Scanner;

public class Application18 {

 public static void main(String[] args) {
  Scanner sc = new Scanner(System.in);
  int n = sc.nextInt();
  int s = 0;
  for(int i=0;i<n;i++) {
   int m=sc.nextInt();
   int v= sc.nextInt();
   s+=(m*v);
  }
  System.out.println(s);
 }

}

Question 8: Lazy String

Problem Statement-: Anish is the laziest person you can ever see. He is tasked to write the name of the winner in a game where two people take part. And he just writes the longest common subsequence over there, so that with minimum chane or no backspace he can edit the name to the winner’s name.

For two given names, you have to predict what Anish will write in his computer before the start of the name. If there are more than two longest subsequences possible,write the one with less lexicographic value.

Input Format:

Two lines including two strings of name(All with capital letters)

Output Format:

A single line with the lexicographically smallest possible longest common subsequence.

Sample Input:

ABCD
BACD

Sample Output:

ACD

Explanation:

ACD and BCD these are the two possible biggest substring

C++

Python

Java

C++

#include <bits/stdc++.h>
using namespace std;
string s1,s2,s3;
 
int fun(int i,int j,int k,string s)
{
    if(i==s1.length()||j==s2.length())
    {
        if(s3.length()==s.length()) s3=min(s,s3);
        else if(s3.length()<s.length()) s3=s;
        return k;
    }
    if(s1[i]==s2[j]) return fun(i+1,j+1,k+1,s+s1[i]);
    return max(fun(i+1,j,k,s),fun(i,j+1,k,s));
}
 
int main()
{
    cin>>s1>>s2;
    fun(0,0,0,"");
    cout<<s3;
}

Python

s1=input()
s2=input()
s3=""
def fun(i,j,k,s):
    global s1
    global s2
    global s3
    if i==len(s1) or j==len(s2):
        if len(s3)==len(s):
            s3=min(s,s3)
        elif len(s3)<len(s):
            s3=s
        return k 
    if s1[i]==s2[j]:
        return fun(i+1,j+1,k+1,s+s1[i])
    return max(fun(i+1,j,k,s),fun(i,j+1,k,s))
    
fun(0,0,0,"")
print(s3)

Java

import java.util.Scanner;
public class Application19 {
	public static String s1,s2,s3="";
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		 s1 = sc.nextLine();
		 s2 = sc.nextLine();
		 fun(0,0,0,"");
		 System.out.println(s3);
	}
	private static int fun(int i, int j, int k, String s) {
		// TODO Auto-generated method stub
		if(i==s1.length() || j==s2.length()) {
			if(s3.length()==s.length()) {
				if(s3.compareTo(s)>0) {
					s3=s;
				}
			}
			else if(s3.length()<s.length()) {
				s3=s;
			}
			return k;	
		}
		if(s1.charAt(i)==s2.charAt(j)) {
			return fun(i+1,j+1,k+1,s+s1.charAt(i));
		}
		return Math.max(fun(i+1,j,k,s), fun(i,j+1,k,s));
	}

}

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WeCP Coding Questions

WeCP Practice Coding Questions for Freshers

WeCP Practice Coding Questions for Freshers

More About WeCP Companies based Coding Questions

Pattern for WeCP Companies based Coding Questions

Practice Questions for WeCP Coding Round

Question 1: Sentence counter

Question 2: Even Odd Even Odd

Question 3: Profit balance

Question 4: Class Monitor

Question 5: Corona for Computer

Question 6: Help of Prepsters

Question 7: Momentum LinkedList

Question 8: Lazy String

30+ Companies are Hiring

Unlock this article for Free, by logging in

WeCP Coding Questions

WeCP Practice Coding Questions for Freshers

WeCP Practice Coding Questions for Freshers

More About WeCP Companies based Coding Questions

Pattern for WeCP Companies based Coding Questions

Practice Questions for WeCP Coding Round

Question 1: Sentence counter

Question 2: Even Odd Even Odd

Question 3: Profit balance

Question 4: Class Monitor

Question 5: Corona for Computer

Question 6: Help of Prepsters

Question 7: Momentum LinkedList

Question 8: Lazy String

30+ Companies are Hiring

Unlock this article for Free,
by logging in