Wipro Elite NLTH Coding Questions | PrepInsta

January 14, 2021

Wipro NLTH Coding Questions and Answer

Wipro Elite NLTH Coding Questions and Answer are not like a general Programming Round Questions with Solutions it is all together different from C programming.

We have analyzed over 100+ Wipro NLTH Coding Questions. Below you will find Similar pattern based Wipro NLTH Coding Round Questions, Wipro NLTH Coding Questions that are asked constantly in Wipro NLTH Placement test. The languages that you can use in the test are –

C
C++
Java
Python

Total number of questions	2 questions
Total Time Duration	60 minutes
Type of Test	Adaptive
Negative Marking	No

WIPRO NLTH PROGRAMMING MODEL QUESTIONS AND ANSWERS

Here are some questions that are asked in Wipro for –

Wipro NLTH C Programming Questions for On Campus
Wipro NLTH Java Programming Questions for On Campus
Wipro NLTH C++ Programming Questions for On Campus

Wipro Programming Practice Questions and Coding Paper

Pattern Programming Dashboard
Most asked Questions
Pattern Making Questions – 1
Pattern Making Questions -2 Trapezium
Pattern Making Questions – 3
GCD of Two Numbers
Input Output Questions – 1
LCM of three numbers

Free Material

Program to calculate the GCD of two numbers – | C | C++ | Java | Python
Program to calculate the GCD of three numbers – | C | C++ | Java | Python
Program to check if the given number is Prime or not – | C | C++ | Java | Python
Program to Display the distinct elements of an array – | C | C++ | Java | Python
Program to Sort first half in ascending order and second half in descending order in an array – | C | C++ | Java | Python
Program for Matrix multiplication and finding transpose – | C | C++ | Java | Python
Program to Remove vowels from an input string – | C | C++ | Java | Python
Program to Print Sum of all Odd Numbers present in a larger number – | C | C++ | Java | Python
Program to Print the Transpose of a matrix – | C | C++ | Java | Python
Program to Display the largest element in each row in a 2-D Array – | C | C++ | Java | Python
Program to Check the balance of Parenthesis – | C | C++ | Java | Python

Some Coding Questions for Practice

Question 1

Wipro’s client eBay wants to run a campaign on their website, which will have the following parameters, eBay wants that on certain x products, they want to calculate the final price, for each product on eBay there will be a stock unit parameter, this parameter will denote, how many items are their in their fulfillment center

Now, while these numbers if are positive means product x is available in the fulfillment center and if not than the product is not available and cannot be shipped to the customer

Now the price on for each product varies based on the distance of the customer from the fulfillment center. Now, each product is in different fulfillment zone. Now, these values are 00’s kms for each centurion km. The price available would further increase by factor distance.

You’ve to find the maximum discount price for each product if the product can be shipped.

Following are the input/output parameters :

Input

The first line of the input will contain number of products.
The second line will contain price for each of these products.
The third line contains shipping distance in 00’s kms
The fourth line contains SKU’s

Output

It will contain the final price for each deliverable item in SKU’s

Example :

Input:

6
87 103 229 41 8 86
3 1 9 2 1 2
7 -21 30 0 -4 -3

Output

261 2061

Python

#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;
}

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

C++

JAVA

#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

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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

C++

JAVA

Python

#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++

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JAVA

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Python

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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

C++

JAVA

Python

#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++

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JAVA

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Python

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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

405

C++

JAVA

Python

/* 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++

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JAVA

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Python

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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.

C++

JAVA

Python

#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++

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JAVA

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Python

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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

C++

JAVA

Python

#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++

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JAVA

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Python

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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)

C++

JAVA

Python

/* 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++

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JAVA

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Python

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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

C++

JAVA

Python

#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

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Python

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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

C++

JAVA

Python

#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++

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JAVA

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Python

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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.

C++

JAVA

Python

#include <stdio.h>
int main()
{
    int num, target, count = 0, r;  
    scanf("%d",&num);
    scanf("%d",&target);
    while(num!=0)
    {
        r = num%10;
        if(r==target)
        {
            count = count + 1;
        }
        num = num/10;
    }
    printf("%d",count);   
}

C++

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JAVA

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Python

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36 comments on “Wipro Elite NLTH Coding Questions | PrepInsta”

Debanjan
Question 3 in python:

s = (input(“Enter string: “))
n = len(s.split())
print(“The number of words:”,n)

0

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Debanjan
Question 2 in python:
r, c = map(int, input(“Enter rows and columns: “).strip().split())
arr = []
for i in range(r):
a = []
for j in range(c):
a.append(int(input()))
arr.append(a)

s, temp, m = 0, 0, 0
sarr = []
for i in range(r):
for j in range(c):
s += arr[i][j]
print(“Row {} has sum = {}”.format(i, s))
sarr.append(s)
s = 0

n = len(sarr)
ct = 0
low = sarr[0]
for i in sarr:
if low < i:
low = i
ct += 1
print("Row {} has maximum: {}".format(ct, low))

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sonali
question 3(Python)
s=list(map(str,input().split()))
print(len(s))

0

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AKhil kumar
Q4-JAVA
import java.io.*;
import java.lang.*;
import java.util.*;
class PREPINSTA {
public static void main (String[] args) {
String in=”welcome world to the new world”;
String ot=”world”;
String[] ary = in.split(” “);
int k=0;
for(int i=0;i<ary.length;i++){
if(ary[i].equals(ot)){
k++;
}
}
System.out.println(k);
}
}

0

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Vikash
//problem no. 5

#include
using namespace std;

int main()
{
int m,n;
cin >> m >> n;
int s=m/15,sum=0;
if(m%15!=0)
s++;
while(15*s<=n)
{
sum=sum+(s*15);
s++;
}
cout << sum;
}

0

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Vikash
//problem no. 9

#include
using namespace std;

int main()
{
int c,r,sum,max=0,loc;
cin >> c >> r;
int arr[c][r];
for(int i=0;i<c;i++)
{
sum=0;
for(int j=0;j> arr[i][j];
sum += arr[i][j];
}
cout <<"Row "<<i+1<<" : "<<sum <max)
{
max=sum;
loc=i+1;
}
}
cout << "Row "<<loc<<" is having maximum sum : "<<max;
}

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Vikash
//problem no. 10

#include
using namespace std;

int main()
{
int n;
cin >> n;
int arr[n];
for(int i=0;i> arr[i];
sort(arr,arr+n);
cout << arr[0]+arr[n-1];
}

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Vikash
#include
using namespace std;

int main()
{
int num,even=0,odd=0;
cin >> num;
while(num!=0)
{
if((num%10)%2==0)
even += (num%10);
else
odd += (num%10);
num/=10;
}
cout << abs(even-odd);
}

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Arpan
Question 5 in python3:
m=int(input())
n=int(input())
def SumNumberDivisible(m,n):
sum=0
for i in range(m,n+1):
if((i%3==0) & (i%5==0)):
sum+=i
print(sum)
result=SumNumberDivisible(m,n)

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