Wipro Elite NLTH Coding Questions | PrepInsta

#include <stdio.h>
int main()
{
    int noOfProducts;
    scanf("%d",&noOfProducts);
    
    int price[noOfProducts], distance[noOfProducts], sku[noOfProducts];
    for(int i=0;i<noOfProducts;i++)
    {
        scanf("%d",&price[i]);
    }
    for(int i=0;i<noOfProducts;i++)
    {
        scanf("%d",&distance[i]);
    }
    for(int i=0;i<noOfProducts;i++)
    {
        scanf("%d",&sku[i]);
    }
    
    int final_Price[noOfProducts];
    int count =0;
    for(int i=0;i<noOfProducts;i++)
    {
        if(sku[i]>0)
        {
            final_Price[count]= price[i] * distance[i];
            count++;
        }
    }
        for(int i=0;i<count;i++)
        {
            printf("%d ", final_Price[i]);
        }
    return 0;
}

Java

import java.util.*;
class Main 
{
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int noOfProducts=sc.nextInt();
        
        int price[]=new int[noOfProducts];
        int distance[]=new int[noOfProducts];
        int sku[]=new int[noOfProducts];
        
        for(int i=0;i<noOfProducts;i++)
            price[i]=sc.nextInt();
        
        for(int i=0;i<noOfProducts;i++)
            distance[i]=sc.nextInt();
        
        for(int i=0;i<noOfProducts;i++)
            sku[i]=sc.nextInt();
        
        int finalPrice[]=new int[noOfProducts];
        int count =0;
        
        for(int i=0;i<noOfProducts;i++)
        {
            if(sku[i]>0)
            {
                finalPrice[count]= price[i] * distance[i];
                count++;
            }
        }
        
        for(int i=0;i<count;i++)
        {
            System.out.print(finalPrice[i]+" ");
        }
    }
}

Python

number_of_products = int(input())
list_price = list(input().split(" "))
list_distance = list(input().split(" "))
list_sku = list(input().split(" "))
list_final=[]
for item in range (0, number_of_products):
    if int(list_sku[item]) >0 :
        temp_val = int(list_price[item]) * int(list_distance[item])
        list_final.append(temp_val)
for item in range(0, len(list_final)):
    print(list_final[item], sep=" ",end=" ")

Question : 2

(asked in Wipro On Campus, Sastra University, Aug, 2019)

Problem: in this first line you are required to take the value of m and n as the number of rows and columns of matrix, then you are required to take the input elements of array.
As an output you are required to print the sum of each row then the row having the maximum sum.
Test Case :
Input : 3 3
1 2 3
4 5 6
7 8 9
Output :
Row 1 : 6
Row 2 : 15
Row 3 : 24
Row 3 is having the maximum sum : 24

#include <stdio.h>
int main()
{
    int row, colm, i, j, temp, max = 1;
    int mat[100][100];
    int sum[100];
    printf("enter the number of rows : ");
    scanf("%d",&row);
    printf("enter the number of columms : ");
    scanf("%d",&colm);
    for(i=0; i<row; i++)
    {
        for(j=0; j<colm; j++)
        {
            printf("enter [%d %d] element : ",i+1,j+1);
            scanf("%d",&mat[i][j]);
        }
    }
    for(i=0; i<row; i++)
    {
        sum[i] = 0;
        for(j=0; j<colm; j++)
        {
            sum[i] = sum[i] + mat[i][j];
        }
        printf("\n");
    }
    for(i=0; i<row; i++)
    {
        printf("Row %d : %d\n",i+1,sum[i]);
    }
    for(i=0; i<row; i++)
    {
        if(sum[0]<sum[i+1])
        {
            temp = sum[0];
            sum [0] = sum[i+1];
            sum[i+1] = temp;
            max = max+1;
        }
    }
    printf("\nRow %d is having the maximum sum : %d",max,sum[0]);
    return 0;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
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];
        
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                arr[i][j]=sc.nextInt();
        
        int max=Integer.MIN_VALUE;
        int index=-1;
        for(int i=0;i<m;i++)
        {
            int sum=0;
            for(int j=0;j<n;j++)
            {
                sum=sum+arr[i][j];
            }
            System.out.println("Row "+(i+1)+" : "+sum);
            if(sum>max)
            {
                max=sum;
                index=i+1;
            }
        }
        
        System.out.println("Row "+index+" is having the maximum sum : "+max);
    }
}

Question : 3

(asked in Wipro On Campus, PESIT Bangalore, Oct, 2019)

Problem: You are required to count the number of words in a sentence.
You are required to pass all the test cases

Test Cases :
Test Case : 1
Input : welcome to the world
Output : 4

Test Case : 2
Input : [space] say hello
Output : 2

Test Case : 3
Input : To get pass you need to study hard [space] [space]
Output : 8

#include <stdio.h>
#include <string.h>
int main()
{
    char a[100];
    int i=0,count;
    printf("Enter the string : ");
    scanf("%[^\n]s",a);
    if(a[0]==' ')
    {
        count = 0;
    }
    else
    {
        count = 1;
    }
    while(a[i]!='\0')
    {
        if(a[i]==' ' && a[i+1]!=' ' && a[i+1]!='\0')
        {
            count = count + 1;
        }
        i++;
    }
    printf("%d",count);
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        String str=sc.nextLine();
        str=str.trim();
        String arr[]=str.split(" ");
        int count=0;
        for(int i=0;i<arr.length;i++)
        {
            arr[i]=arr[i].trim();
            if(arr[i].length()>0)
                count++;
        }
        System.out.println(count);
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question : 4

(asked in Wipro On Campus, VIT University, Aug, 2019)

Problem: First of all you need to input a whole sentence in the string, then you need to enter the target
word.
As an output you need to find the number of times that particular target word is repeated in the
sentence.

Test Cases :

Test Case : 1
Input : welcome world to the new world
world
Output : 2

Test Case : 2
Input : working hard and working smart both are different ways of working
working
Output : 3

#include <stdio.h>
#include <string.h>
int main()
{
    char str[100];
    char toSearch[100];
    int count;
    int i, j, found;
    int stringLen, searchLen;
    
    scanf("%[^\n]s",str);
    scanf("%s",toSearch);
    
    stringLen = strlen(str);      
    searchLen = strlen(toSearch); 
    count = 0;
    for(i=0; i <= stringLen-searchLen; i++)
    {
        found = 1;
        for(j=0; j<searchLen; j++)
        {
            if(str[i + j] != toSearch[j])
            {
                found = 0;
                break;
            }
        }
        if(found == 1)
        {
            count++;
        }
    }
    printf("%d",count);
    return 0;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        String str=sc.nextLine();
        String target=sc.next();
        str=str.trim();
        String arr[]=str.split(" ");
        int count=0;
        for(int i=0;i<arr.length;i++)
        {
            arr[i]=arr[i].trim();
            if(arr[i].equals(target))
                count++;
        }
        System.out.println(count);
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question : 5

(asked in Wipro On Campus, MIET Meerut, Aug, 2019)

Problem Statement

You are required to implement the following function:
Int SumNumberDivisible(int m, int n);

The function accepts 2 positive integer ‘m’ and ‘n’ as its arguments.
You are required to calculate the sum of numbers divisible both by 3 and 5, between ‘m’ and ‘n’ both inclusive and return the same.

Note
0 < m <= n

Example
Input:
m : 12
n : 50
Output 90

Explanation:
The numbers divisible by both 3 and 5, between 12 and 50 both inclusive are
{15, 30, 45} and their sum is 90.

Sample Input

m : 100
n : 160

Sample Output

510

/* Programming Question */
#include <stdio.h>
int Calculate(int, int);
int main()
{
    int m, n, result;
    // Getting Input 
    printf("Enter the value of m : ");
    scanf("%d",&m);
    printf("Enter the value of n : ");
    scanf("%d",&n);
    result = Calculate(n,m);
    // Getting Output
    printf("%d",result);
    return 0;
}
/* Write your code below . . . */
int Calculate(int n, int m)
{
// Write your code here
    int i, sum = 0;
    for(i=m;i<=n;i++)
    {
        if((i%3==0)&&(i%5==0))
        {
            sum = sum + i;
        }
    }
return sum;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
    public static int sumNumberDivisible(int m,int n)
    {
        int sum=0;
        for(int i=m;i<=n;i++)
        {
            if((i%3==0)&&(i%5==0))
            {
                sum=sum+i;
            }
        }
        return sum;
    }
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int m=sc.nextInt();
        int n=sc.nextInt();
        int res=sumNumberDivisible(m,n);
        System.out.println(res);
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question : 6

(asked in Wipro On Campus, Galgotia University Noida, Aug, 2019)

Input
The first line of input consists of three space separated integers. Num, start and end representing the size of list [N], the starting value of the range and the ending value of the range respectively.
The second line of input consists N space separated integers representing the distances of the employee from the company.

Output
Print space separated integers representing the ID’s of the employee whose distance liew within the given range else return -1.

Example
Input
6 30 50
29 38 12 48 39 55

Output
1 3 4

Explanation :
There are 3 employees with id 1, 3, 4 whose distance from the office lies within the given range.

#include <stdio.h>
int main()
{
    int start, end, a[50], num, i, flag;
    scanf("%d %d %d",&num,&start,&end);
    for(i=0; i<num; i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<num;i++)
    {
        if(a[i]>start && a[i]<end)
        {
            flag = 1;
        }
        else
        {
            flag = -1;
        }
        if(flag == 1)
        {
            printf("%d ",i);
        }
    }
    return 0;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
   
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int num=sc.nextInt();
        int start=sc.nextInt();
        int end=sc.nextInt();
        int list[]=new int[num];
        
        for(int i=0;i<num;i++)
            list[i]=sc.nextInt();
        
        boolean flag=false;
        for(int i=0;i<num;i++)
        {
            if(list[i]>start && list[i]<end)
            {
                flag=true;
                System.out.print(i+" ");
            }
        }
        
        if(flag==false)
            System.out.println("-1");
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question : 7

(asked in Wipro On Campus, IIIT Hyderabad, Aug, 2019)

Write a code for the following functions
Function 1 : sortArray
It’s function is to sort the passed array with a particular length, there are 2 parameters passed into it first is the length of array and next is the elements of list.
Function 2 : findMaxElement
It’s function is to get the 2 array with their own particular elements and length, and find the maximum element from both the array and print it.

Input
12
2 5 1 3 9 8 4 6 5 2 3 11
11
11 13 2 4 15 17 67 44 2 100 23

Output
100

#include <stdio.h>
int* sortArray(int len, int* arr)
{
    int i=0, j=0, temp = 0;
    for(i=0; i<len; i++)
    {
        for(j=i+1; j<len; j++)
        {
            if(arr[i]>arr[j])
            {
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
    return arr;
}
int findMaxElement(int len1, int* arr1, int len2, int* arr2)
{
    sortArray(len1, arr1);
    sortArray(len2, arr2);
    if(arr1[len1-1]>arr2[len2-1])
        return arr1[len1-1];
    else
        return arr2[len2-1];
}
int main()
{
    int len1, len2, arr1[20], arr2[20], i;
    int ans;
    scanf("%d",&len1);
    for(i=0; i<len1; i++)
    {
        scanf("%d",&arr1[i]);
    }
    scanf("%d",&len2);
    for(i=0; i<len2; i++)
    {
        scanf("%d",&arr2[i]);
    }
    ans = findMaxElement(len1, arr1, len2, arr2);
    printf("%d",ans);
    return 0;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
    public static void sortArray(int len,int arr[])
    {
        int i=0,j=0,temp=0;
        for(i=0;i<len;i++)
        {
            int countOfSwap=0;
            for(j=0;j<len-1-i;j++)
            {
                if(arr[j]>arr[j+1])
                {
                    temp=arr[j];
                    arr[j]=arr[j+1];
                    arr[j+1]=temp;
                    countOfSwap++;
                }
            }
            if(countOfSwap==0)
                break;
        }
    }
    public static int findMaxElement(int len1, int arr1[],int len2,int arr2[]) 
    {
        sortArray(len1,arr1);
        sortArray(len2,arr2);
        if(arr1[len1-1]>arr2[len2-1])
            return arr1[len1-1];
        else 
            return arr2[len2-1];
    }
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int len1=sc.nextInt();
        int arr1[]=new int[len1];
        for(int i=0;i<len1;i++)
            arr1[i]=sc.nextInt();
        
        int len2=sc.nextInt();
        int arr2[]=new int[len2];
        for(int i=0;i<len2;i++)
            arr2[i]=sc.nextInt();
        
        int ans=findMaxElement(len1,arr1,len2,arr2);
        System.out.println(ans);
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question 8

(asked in Wipro On Campus, HBTI Kanpur, Sep, 2019)

Problem: you are given a number, and you have to extract the key by finding the difference between the sum of the even and odd numbers of the input.

Test Case :

Input : 24319587

Output : 11

Explanation : odd terms : 3 + 1 + 9 + 5 + 7 = 25

even terms : 2 + 4 + 8 = 14

output : 11 (25-14)

/* Program to find key */
#include <stdio.h>
int main()
{
    int n;
    int r, odd=0, even=0, key;
    scanf("%d",&n);
    while(n!=0)
    {
        r = n%10;
        if(r%2==0)
        {
            even = even + r;
        }
        else
        {
            odd = odd + r;
        }
        n =  n/10;
    }
    if(odd>even)
    {
        key = odd - even;
    }
    else
    {
        key = even - odd;
    }
    printf("%d",key);
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int num=sc.nextInt();
        int odd=0,even=0;
        while(num>0)
        {
            int rem=num%10;
            if(rem%2==0)
                even=even+rem;
            else 
                odd=odd+rem;
            num=num/10;
        }
        
        System.out.println(Math.abs(even-odd));
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question 9

(asked in Wipro NLTH 2019)

Problem: in this first line you are required to take the value of m and n as the number of rows and columns of matrix, then you are required to take the input elements of array.

As an output you are required to print the sum of each row then the row having the maximum sum.

Test Case :

Input : 3 3

1 2 3

4 5 6

7 8 9

Output :

Row 1 : 6
Row 2 : 15
Row 3 : 24

Row 3 is having the maximum sum : 24

#include <stdio.h>
int main()
{
    int m, n, i, j, temp, max = 1;
    int mat[100][100];
    int sum[100];
    scanf("%d %d",&m, &n);
    for(i=0; i<m; i++)
    {
        for(j=0; j<n; j++)
        {
            scanf("%d",&mat[i][j]);
        }
    }
    
    for(i=0; i<m; i++)
    {
        sum[i] = 0;
        for(j=0; j<n; j++)
        {
            sum[i] = sum[i] + mat[i][j];
        }
        printf("\n");
    }
    
    for(i=0; i<m; i++)
    {
        printf("Row %d : %d\n",i+1,sum[i]);
    }
    
    for(i=0; i<m; i++)
    {
        if(sum[0]<sum[i+1])
        {
            temp = sum[0];
            sum [0] = sum[i+1];
            sum[i+1] = temp;
            max = max+1;
        }
    }
    
    printf("\nRow %d is having the maximum sum : %d",max,sum[0]);
    return 0;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
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];
        
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                arr[i][j]=sc.nextInt();
        
        int max=Integer.MIN_VALUE;
        int index=-1;
        for(int i=0;i<m;i++)
        {
            int sum=0;
            for(int j=0;j<n;j++)
            {
                sum=sum+arr[i][j];
            }
            System.out.println("Row "+(i+1)+" : "+sum);
            if(sum>max)
            {
                max=sum;
                index=i+1;
            }
        }
        
        System.out.println("Row "+index+" is having the maximum sum : "+max);
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question 10

(asked in Wipro NLTH 2019)

Problem: in this first line you are required to take the input that set the number of elements to be inserted, the next line takes the elements as input.

You are required to print the sum of maximum and minimum element of the list

Test Case :

Input : 6

55 87 46 21 34 79

Output :

108

Explanation :

21 + 87 = 108

#include <stdio.h>
int main()
{
    int a[100];
    int n, sum, temp;
    int i, j;
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%d",&a[i]); 
    }
    for(i=0; i<n; i++)
    {
        for(j=i+1; j<n; j++)
        {
            if(a[i]>a[j])
            {
                temp = a[i];
                a[i] = a[j];
                a[j] = temp;
            }
        }
    }
    sum = a[0]+a[n-1];
    printf("%d",sum);
    return 0;
}

C++

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

JAVA

import java.util.*;
class Main 
{
    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 max=Integer.MIN_VALUE,min=Integer.MAX_VALUE;
        for(int i=0;i<n;i++)
        {
            min=Math.min(min,arr[i]);
            max=Math.max(max,arr[i]);
        }
        System.out.println(max+min);
    }
}

Python

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

Question 11

(asked in Wipro NLTH 2019)

Problem: in this first line you are required to take the input of an integer number, and in the second line you are required to input the target element

You are required to print the number of time target element occured in the integer

Test Case :

Input : 734139

Output :

Explanation :

3 occured 2 times in the integer.