Wipro Turbo Hiring Coding Questions 2025, Get Solved Questoions

Wipro Turbo Hiring Coding Questions 2025 | Wipro Turbo Coding Questions | Wipro Elite Coding Questions | Wipro Turbo Coding Questions & Answers 2025

Wipro Turbo Hiring Coding Questions 2025: The Company of Wipro is hiring the Wipro Turbo vacancies for eligible employees. Following of hiring, Wipro Turbo Hiring 2025/ Wipro Elite Hiring for capable and eligible candidates. The selection process of Wipro Turbo 2025 is Online Assessment Test, Interview Process, etc., Willing candidates eagerly prepare the interview process. Wipro Turbo Interview Questions & Answers and Wipro Turbo Coding Questions 2025 is available in this page. The Wipro will conduct many selections process. so, for candidates prepare common questions. Wipro Turbo Coding Questions 2025 contains mostly common and repeated asked questions. So, Candidates hopefully prepare the Wipro Turbo coding questions & answers.

The Wipro Company published some sample Wipro Elite coding questions and answers/ Wipro Turbo Coding questions in the official website. Interested candidates must prepare the Wipro Turbo Hiring Selection Process. When you will be qualified in the selection process, they will get Wipro Turbo Jobs. Candidates put your full efforts in the preparation of Wipro Elite Coding questions with Answers and Wipro Turbo Interview Coding questions. Continuous preparation of Wipro Turbo Hiring Coding questions and Answers, you definitely achieve the selection process. The Interview & test date will be intimated on mail. Keep check the mail.

Wipro Turbo Coding Interview Questions 

Wipro Turbo Coding Questions & Answers 

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 :

#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 PROGRAM 

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 PROGRAM 

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: 

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 PROGRAM

#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++ PROGRAM

#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;
}

JAVA PROGRAM

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: 

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

C PROGRAM 

/* 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++ PROGRAM 

#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;
}

JAVA PROGRAM

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 PROGRAM 

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 4: 

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 PROGRAM 

#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);
}

C++ PROGRAM 

#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;
}

JAVA PROGRAM 

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 PROGRAM 

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 5: 

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 PROGRAM 

#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;
}

JAVA PROGRAM

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 6: 

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 PROGRAM 

#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++ PROGRAM 

#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;
}

JAVA PROGRAM 

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 PROGRAM

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 7: 

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.

C PROGRAM 

#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);   
}

JAVA PROGRAM

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);
    }
}

Question 8: 

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 PROGRAM 

#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;
}

JAVA PROGRAM

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);
}
}

#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;
}

JAVA PROGRAM 

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: 

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 PROGRAM 

/* 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);
}

JAVA PROGRAM 

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));
}
}

The wipro Turbo Coding Questions and Answers/ Wipro Turbo Interview Coding Questions 2025 downloading easily and also prepare more conveniently in this page. Kindly watch dailyrecruitment.in site for more information updates about jobs and Education.

Find More Govt Jobs in

Railway Job

Government Exam Result

Competitive Exam

Upcoming Jobs

Defence job

Engineering Job