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Wipro NTH Coding Questions and Answer
Wipro Elite NTH 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 NTH Coding Questions. Below you will find Similar pattern based Wipro NTH Coding Round Questions, Wipro NTH Coding Questions that are asked constantly in Wipro NTH Placement test. The languages that you can use in the test are C , C++ , Java and Python.
| Total number of questions | 2 questions |
| Total Time Duration | 60 minutes |
| Type of Test | Adaptive |
| Negative Marking | No |
WIPRO NTH PROGRAMMING MODEL QUESTIONS AND ANSWERS
In this page you will find some previous years coding questions asked during the campus placement:
- Wipro NTH C Programming Questions for On Campus
- Wipro NTH Java Programming Questions for On Campus
- Wipro NTH C++ Programming Questions for On Campus
Free Material
- Program to calculate the GCD of two numbers –
- Program to calculate the GCD of three numbers –
- Program to check if the given number is Prime or not –
- Program to Display the distinct elements of an array –
- Program to Sort first half in ascending order and second half in descending order in an array –
- Program for Matrix multiplication –
- Program to Remove vowels from an input string –
- Program to Print Sum of all Odd Numbers present in a larger number –
- Program to Print the Transpose of a matrix –
- Program to Display the largest element in each row in a 2-D Array –
- Program to Check the balance of Parenthesis –
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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
#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; }
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]+" ");
}
}
}
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; }
#include <iostream>
int main() {
int row, colm, i, j, temp, max = 1;
int mat[100][100];
int sum[100];
std::cout << "Enter the number of rows: "; std::cin >> row;
std::cout << "Enter the number of columns: "; std::cin >> colm;
for(i = 0; i < row; i++) {
for(j = 0; j < colm; j++) {
std::cout << "Enter [" << i + 1 << "][" << j + 1 << "] element: "; std::cin >> mat[i][j];
}
}
for(i = 0; i < row; i++) {
sum[i] = 0;
for(j = 0; j < colm; j++) {
sum[i] += mat[i][j];
}
std::cout << std::endl;
}
for(i = 0; i < row; i++) {
std::cout << "Row " << i + 1 << ": " << sum[i] << std::endl;
}
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;
}
}
std::cout << std::endl;
std::cout << "Row " << max << " is having the maximum sum: " << sum[0] << std::endl;
return 0;
}
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 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); }
#include <iostream>
#include <string>
#include <sstream>
int main() {
std::string str;
std::getline(std::cin, str);
str = str.substr(0, str.find_last_not_of(" \t") + 1);
std::stringstream ss(str);
std::string word;
int count = 0;
while (ss >> word) {
count++;
}
std::cout << count << std::endl;
return 0;
}
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);
}
}
str = input()
str = str.strip()
arr = str.split()
count = 0
for word in arr:
word = word.strip()
if len(word) > 0:
count += 1
print(count)
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; }
#include <iostream>
#include <cstring>
int main() {
char str[100];
char toSearch[100];
int count;
int i, j, found;
int stringLen, searchLen;
std::cin.getline(str, 100);
std::cin >> 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++;
}
}
std::cout << count;
return 0;
}
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);
}
}
str = input()
toSearch = input()
stringLen = len(str)
searchLen = len(toSearch)
count = 0
for i in range(stringLen - searchLen + 1):
found = True
for j in range(searchLen):
if str[i + j] != toSearch[j]:
found = False
break
if found:
count += 1
print(count)
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; }
#include <iostream
#include <cstring>
int main() {
char str[100];
char toSearch[100];
int count;
int i, j, found;
int stringLen, searchLen;
std::cin.getline(str, 100);
std::cin >> 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++;
}
}
std::cout << count;
return 0;
}
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);
}
}
str = input()
toSearch = input()
stringLen = len(str)
searchLen = len(toSearch)
count = 0
for i in range(stringLen - searchLen + 1):
found = True
for j in range(searchLen):
if str[i + j] != toSearch[j]:
found = False
break
if found:
count += 1
print(count)
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; }
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");
}
}
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; }
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);
}
}
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); }
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));
}
}
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; }
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 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; }
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);
}
}
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
3
Output :
2
Explanation :
3 occured 2 times in the integer.
#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); }
import java.util.*;
class Main
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int num=sc.nextInt();
int target=sc.nextInt();
int count=0;
while(num>0)
{
int rem=num%10;
if(rem==target)
count++;
num=num/10;
}
System.out.println(count);
}
}
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code for 8th question:
num=int(input())
even=0
odd=0
while(num>0):
rem=num%10
if(rem%2==0):
even+=rem
num//=10
else:
odd+=rem
num//=10
print(odd-even)
Question number 3 in python:-
str1=input()
ans=str1.split()
print(len(ans))
Question number 4 in python:-
str2=input()
target=input()
l1=str2.split()
count=0
for i in l1:
if i==target:
count+=1
else:
continue
print(count)
Question number 5 in python:-
def SumNumberDivisible(a,b):
ans=[]
sum=0
for i in range(a,b+1):
if i%3==0 and i%5==0:
ans.append(i)
for i in ans:
sum=sum+i
return sum
a=int(input())
b=int(input())
SumNumberDivisible(a,b)
Question number 6 in python:-
n,start,end=map(int,input().split())
list_dist=list(map(int,input().split()))
for i in range(n):
if list_dist[i]start:
print(i,end=” “)
Question number 7 in python:-
m=int(input())
l1=list(map(int,input().split()))
n=int(input())
l2=list(map(int,input().split()))
max1=max(l1)
max2=max(l2)
if max1>max2:
print(max1)
else:
print(max2)
Question number 8 in python:-
inp=int(input())
ans=str(inp)
odd=[int(x) for x in ans if int(x)%2==1]
even=[int(x) for x in ans if int(x)%2==0]
sum_odd=0
sum_even=0
for i in odd:
sum_odd+=i
for i in even:
sum_even+=i
final=sum_odd-sum_even
print(final)
Question number 10 in python:-
n=int(input())
arr=list(map(int,input().split()))
max1=max(arr)
min1=min(arr)
ans=max1+min1
print(ans)
Question number 11 in python:-
a=int(input())
b=str(a)
tar=int(input())
target=str(tar)
count=0
for i in b:
if i==target:
count+=1
print(count)
Question number 6 in python :-
num,start,end = map(int,input(“->”).split())
L = list(map(int,input(“->”).split()[:num]))
for j in range(0,len(L)):
for i in range(start,end+1):
if L[j] == i:
print(j,end=” “)
Question number 5 in python :-
m = int(input(“enter value of m:”))
n = int(input(“enter value of n:”))
Sum = 0
L = []
for i in range(m,n+1):
if i % 3 == 0 and i % 5 == 0:
Sum = Sum + i
print(Sum)
Question number 3 in python:-
input1= list(map(str,input(“enter your string:”).split()))
print(len(input1))