Tech Mahindra Coding Questions and Answers 2023 | Tech Mahindra LCM Coding Questions | Tech Mahindra Supercoder Questions | Tech Mahindra coding Questions in Python | Tech Mahindra Supercoder Questions and Answers
Tech Mahindra Coding Questions and Answers 2023 | Tech Mahindra coding Questions and Answers PDF: Professionals and recent grads alike might consider working at Tech Mahindra. The coding test is of medium complexity, which implies that although it is not impossible to pass, it is also not particularly easy. You would surely have an edge in the next Tech Mahindra coding test if you practiced some of the code Questions from last year’s test. Let’s examine a few C, C++, Java, Python, and Supercoder Test pattern, code issues, and their remedies. Prior recruiting campaigns included a heavy emphasis on arrays and other basic programming concepts in their coding tests. With enough preparation, one may easily pass the first round and even subsequent rounds. To answer the coding tasks, you must pick a language that you are familiar with, such as C, C++, Java, or even Python, and use your programming skills. To get more private jobs, click here
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Tech Mahindra coding Questions in Python
Questions: Andy wants to go on a vacation to de-stress himself. Therefore he decides to take a trip to an island. It is given that he has as many consecutive days as possible to rest, but he can only make one trip to the island. Suppose that the days are numbered from 1 to N. Andy has M obligations in his schedule, which he has already undertaken and which correspond to some specific days. This means that ith obligation is scheduled for day Di. Andy is willing to cancel at most k of his obligations in order to take more holidays.
Your task is to find out the maximum days of vacation Andy can take by canceling at most K of his obligations.
Input Format
- The first line contains an integer N, denoting the total number of days
- The next line contains an integer M denoting the total number of obligations.
- The next line contains an integer K denoting the largest number of obligations he could cancel
- Each line i of the M subsequent lines (where 0<=i<=M) contains an integer describing Di.
Constraints
- 1<=N<=10^6
- 1<=M<=2*10^6
- 1<=K<=2*10^6
- 1<=D[i]<=10^6
Sample Input 1:
10
5
2
6
9
3
2
7
Sample Output 1 :
5
Explanation:
Here he could cancel his 3rd and 4th obligation which makes vacation length 5.
Sample input 2:
7
2
0
3
4
Sample Output 2:
3
Explanation:
Here he could not cancel any obligation since K=0, so the vacation length is 3.
EXAMPLE PROGRAM
PYTHON PROGRAM
n = int(input()) m = int(input()) k = int(input()) arr = [0] * n for i in range(m): arr[i] = int(input()) ans = 0 arr.sort() if k > 0: for i in range(k + 1, m + 3, 1): ans = max(ans, arr[i] - arr[i - k - 1] - 1) else: j = 0 while arr[j] == 0: j = j + 1 count = 0 for i in range(1, n + 1, 1): count += 1 if j < n and (i == arr[j]): count = 0 j += 1 ans = max(count, ans) print(ans)
Q) Khaled has an array A of N elements. It is guaranteed that N is even. He wants to choose at most N/2 elements from array A. It is not necessary to choose consecutive elements. Khaled is interested in XOR of all the elements he chooses. Here, XOR denotes the bitwise XOR operation.
For example:
- If A=[2,4,6,8], then khaled can choose the subset [2,4,8] to achieve XOR=(2 XOR 4 XOR 8)=14.
Khaled wants to maximize the XOR of all the elements he chooses. Your task is to help khaled to find the max XOR of a subset that he can achieve by choosing at most N/2 elements?
Input format:
- The first line contains an integer, N, denoting the number of elements in A.
- Each line i of the N subsequent lines(where 0<=i<=N) contains an integer describing Ai.
Constraints
- 1<=N<=120
- 1<=A[i]<=10^6
Sample Input 1
2
1
2
Sample Output 1
2
Explanation:
N=2, A=[1,2] khaled can choose the subset[2]. The xor of the elements in the subset is 2. And the number of elements in the subset is 1 which is less than N/2.
Sample Input 2
4
1
2
4
7
Sample Output 2
7
Explanation:
N=4, A=[1,2,4,7] Khaled can choose the subset [7]. The xor of the elements in the subset is 7, and the number of elements in the subset is 1 which is less than N/2.
EXAMPLE PROGRAMS
PYTHON PROGRAM
from itertools import combinations def fun(arr, N): sub = [] max_xor = max(arr) for i in range(1, N // 2): comb = combinations(arr, i + 1) for i in comb: sub.append(list(i)) for a in sub: z = 0 for b in a: z = z ^ b if z > max_xor: max_xor = z return max_xor N = int(input("Enter Length : ")) arr = [] for i in range(N): arr.append(int(input(f"Enter {i+1} Element : "))) if N < 3: print("Output :", max(arr)) else: print("Output :", fun(arr, N))
Q) Write a program to calculate and return the sum of absolute difference between the adjacent number in an array of positive integers from the position entered by the user.
You are expected to write code in the findTotalSum function only which receive three positional arguments:
1st: number of elements in the array
2nd: Array
3rd: position from where the sum is to be calculated
Example
Input
input 1 : 7
input 2 : 11 22 12 24 13 26 14
input 3 : 5
Output
25
Explanation
The first parameter 7 is the size of the array. Next is an array of integers and input 5 is the position from where you have to calculate the Total Sum. The output is 25 as per calculation below.
| 26-13 | = 13
| 14-26 | = 12
Total Sum = 13 + 12 = 25
Here is the code:
def findTotalSum(n,numbers,pos): total = 0 for i in range(pos-1,n-1): total+= abs(numbers[i]-numbers[i+1]) return total n = int(input()) numbers = list(map(int, input().split())) pos = int(input()) print(findTotalSum(n,numbers,pos))
Q) Write a program to find the difference between the elements at odd index and even index.
Note : You are expected to write code in the findDifference function only which receive the first parameter as the numbers of items in the array and second parameter as the array itself. You are not required to take the input from the console.
Example
Finding the maximum difference between adjacent items of a list of 5 numbers
Input
input 1 : 7
input 2 : 10 20 30 40 50 60 70
Output
40
Explanation
The first parameter 7 is the size of the array. Sum of element at even index of array is 10 + 30 + 50 + 70 = 160 and sum of elements at odd index of array is 20 + 40 + 60 = 120. The difference between both is 40
Here is the code:
def findDifference(n,values): total = 0 for i in range(n): if i%2 == 0: total+=values[i] else: total-=values[i] return total n = int(input()) values = list(map(int, input().split())) print(findDifference(n,values))
Q) Write a program to return the difference between the count of odd numbers and even numbers.
You are expected to write code in the countOddEvenDifference function only which will receive the first parameter as the number of items in the array and second parameter as the array itself. you are not required to take input from the console.
Example
Finding the difference between the count of odd and even numbers from a list of 5 number
Input
input 1 : 8
input 2 : 10 20 30 40 55 66 77 83
Output
-2
Explanation
The first paramter (8) is the szie of the array. Next is an array of integers. The calculation of difference between count sum of odd and even numbers is as follows:
3 (count of odd numbers) – 5 (count of even numbers) = -2
Here is the code:
def countOddEvenDifference(n,numbers): total=0 for i in numbers: if i%2 == 0: total = total - 1 else: total = total + 1 return total n = int(input()) numbers = list(map(int,input().split()))
Q) A Cloth merchant has some pieces of cloth of different lengths. He has an order of curtains of length of 12 feet. He has to find how many curtains can be made from these pieces. Length of pieces of cloth is recorded in feet.
Note : You are expected to write code in the findTotalCurtains function only which receive the first parameter as the number of items in the array and second parameter as the array itself. You are not required to take the input from the console.
Example
Finding the total curtains from a list of 5 cloth pieces.
Input
input 1 : 5
input 2 : 3 42 60 6 14
Output
9
Explanation
The first parameter 5 is the size of the array. Next is an array of measurements in feet. The total number of curtains is 5 which is calculated as under
3 -> 0
42 -> 3
60 -> 5
6 -> 0
14 -> 1
total = 9
Here is the code:
def findTotalCurtains(n,numbers): total = 0 for i in numbers: total += i//12 return total n = int(input()) numbers = list(map(int, input().split())) print(findTotalCurtains(n,numbers))
Tech Mahindra Coding Questions in C++
Q) Calculate the total interest on loans for an array of amounts. However, till the amount of 2000, there is no interest applicable, but, there is a 20% interest applicable for the remaining amounts in the array.
In this case, we will be writing a calculate TotalInterest function that receives the first input as the number of amounts in the array and the second input as the array of amounts.
For instance, if there are 5 amounts with the first input as 5 and the amounts being 2000, 4000, 6000, 8000 and 10000.
The total sum would be = 20% of 4000 + 20% of 6000 + 20% of 8000 + 20% of 10000 = 5600 (as the 2000 will not be considered for interest)
Here is the code:
#include<iostream>
using namespace std;
int calculateTotalInterest(int n, int arr[])
{
int tax=0;
for(int i=0;i<n;i++)
{
tax=tax+(int)((arr[i]-1000)*0.2);
}
return tax;
}
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)cin>>arr[i];
cout<<calculateTotalInterest(n,arr);
return 0;
}
Q) Write code for returning the differences between the sums of even and odd numbers in an array of positive integers.
In this case, we will be writing a calculateOddEvenDifference function that receives the first input as the number of positive integers in the array and the second input as the array of positive integers.
For instance, if there are 4 numbers as the first input and the numbers are 13, 8, 9 and 4, the difference of the sum of odd numbers and the sum of even numbers would be = (13+9) – (8+4) = 22 – 12 = 10
Here is the code:
#include<iostream>
using namespace std;
int calculateOddEvenDifference(int n, int arr[])
{
int odd=0,even=0;
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
even+=arr[i];
else
odd+=arr[i];
}
return odd-even;
}
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)cin>>arr[i];
cout<<calculateOddEvenDifference(n,arr);
return 0;
}
Tech Mahindra coding Questions in Java
Questions: While playing an RPG game, you were assigned to complete one of the hardest quests in this game. There are n monsters you’ll need to defeat in this quest. Each monster i is described with two integer numbers – poweri and bonusi. To defeat this monster, you’ll need at least poweri experience points. If you try fighting this monster without having enough experience points, you lose immediately. You will also gain bonusi experience points if you defeat this monster. You can defeat monsters in any order.
The quest turned out to be very hard – you try to defeat the monsters but keep losing repeatedly. Your friend told you that this quest is impossible to complete. Knowing that, you’re interested, what is the maximum possible number of monsters you can defeat?
Input:
- The first line contains an integer, n, denoting the number of monsters. The next line contains an integer, e, denoting your initial experience.
- Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, poweri, which represents power of the corresponding monster.
- Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, bonusi, which represents bonus for defeating the corresponding monster.
Output
- 2
Example Programs:
JAVA PROGRAM
import java.util.Arrays; import java.util.Comparator; import java.util.Scanner; class Main { public static void main(String[] args) { Scanner s = new Scanner(System.in); int n = s.nextInt(); int exp = s.nextInt(); int monst[] = new int[n]; int bonus[] = new int[n]; for (int i = 0; i < n; i++) { monst[i] = s.nextInt(); } for (int i = 0; i < n; i++) { bonus[i] = s.nextInt(); } class Monster { private final int power, bonus; public Monster(int power, int bonus){ this.power = power; this.bonus = bonus; } } Monster[] monsters = new Monster[n]; for(int i = 0; i < n; i++) monsters[i] = new Monster(monst[i], bonus[i]); Arrays.sort(monsters, Comparator.comparingInt(m -> m.power)); int count = 0; for(Monster m: monsters){ if(exp < m.power) break; exp += m.bonus; ++count; } System.out.println(count); } }
Questions: One of the first lessons IT students learn is the representation of natural numbers in the binary number system (base 2) This system uses only two digits, 0 and 1. In everyday life we use for convenience the decimal system (base 10) which uses ten digits, from 0 to 9. In general, we could use any numbering system.
Computer scientists often use systems based on 8 or 16. The numbering system based on K uses K digits with a value from 0 to K-1. Suppose a natural number M is given, written in the decimal system To convert it to the corresponding writing in the system based on K, we successively divide M by K until we reach a quotient that is less than K
The representation of M in the system based on K is formed by the final quotient (as first digit) and is followed by the remainder of the previous divisionsFor example :
- If M=122 and K=8, 122 in base 10= 172 in base 8 This means that the number
- In decimal system = 172 in octal system.
- 172 in base 8 = 1*8^2 + 7*8 + 2 = 122
You made the following observation in applying the above rule of converting natural numbers to another numbering system
- In some cases in the new representation all the digits of the number are the same. For example 63 in base 10= 333 in base 4
Given a number M in its decimal representation, your task is find the minimum base B such that in the representation of M at base B all digits are the same.
Input Format
- The first line contains an integer, M, denoting the number given
Constraints
- 1 <= M = 10^12
Sample Input 1 :
41
Sample Output 1 :
40
Explanation :
Here 41 in base 40. will be 11 so it has all digits the same, and there is no smaller base satisfying the requirements
Sample Input 2 :
34430
Sample Output 2 :
312
Explanation :
Here 34430 in base 312 will have all digits the same and there is no smaller base satisfying the requirements
Example Programs:
JAVA PROGRAM
import java.util.*; class Main { public static boolean convertBase (int m, int base) { int rem = m % base; m = m / base; while (m >= base && (m % base == rem)) m = m / base; if (m == rem) return true; return false; } public static void main (String[]args) { Scanner sc = new Scanner (System.in); int m = sc.nextInt (); int base = 2; while (convertBase (m, base) != true) base++; System.out.println (base); } }
Q) You have been given a string S of length N. The given string is a binary string that consists of only 0’s and ‘1’s. Ugliness of a string is defined as the decimal number that this binary string represents.
Example:
- “101” represents 5.
- “0000” represents 0.
- “01010” represents 10.
There are two types of operations that can be performed on the given string.
- Swap any two characters by paying a cost of A coins.
- Flip any character by paying a cost of B coins
- flipping a character means converting a ‘1’to a ‘0’or converting a ‘0’ to a ‘1’.
Initially, you have been given coins equal to the value defined in CASH. Your task is to minimize the ugliness of the string by performing the above mentioned operations on it. Since the answer can be very large, return the answer modulo 10^9+7.
Note:
- You can perform an operation only if you have enough number of coins to perform it.
- After every operation the number of coins get deducted by the cost for that operation.
Input Format
- The first line contains an integer, N, denoting the number of character in the string
- The next line contains a string, S, denoting the the binary string
- The next line contains an integer, CASH, denoting the total number of coins present initially
- Next will contains an integer, A, denoting the cost to swap two characters.
- Then the next line contains an integer, B, denoting the cost to flip a character.
Constraints
- 1 <= N <= 10^5
- 1< len(S)<= 10^5
- 1<=CASH <=10^5
- 1<=A<=10^5
- 1<=B<=10^5
Sample Input 1 :
4
1111
7
1
2
Sample Output 1 :
1
Explanation:
3 flips can be used to create “0001” which represents 1.
Sample Input 2:
6
111011
7
1
3
Sample Output 2:
7
Explanation:
First swap 0 with the most significant 1, then use flip twice first on index one and then on index two “111011”=>”0111111″=>”001111″=>”000111″ the value represented is 7.
Sample Input 3:
6
111011
7
3
2
Sample Output 3:
3
Explanation:
Flip the 3 most significant characters to get “000011” : the value represented by this string is 3.N
EXAMPLE PROGRAMS
JAVA PROGRAM
import java.util.*; class Main { static String str; static int cash, n, a, b; static void swapf () { char s[] = str.toCharArray (); int i = 0; for (int a = 0; a < s.length; a++) if (s[a] == '1') { i = a; break; } int j = s.length - 1; while (j > i) { if (cash < a) break; if (s[j] == '0') { if (s[i] == '0') i++; else { char temp = s[i]; s[i] = s[j]; s[j] = temp; cash -= a; j--; } } else j--; } str = new String (s); } static void flipf () { char s[] = str.toCharArray (); int i = 0; for (int a = 0; a < s.length; a++) if (s[a] == '1') { i = a; break; } while (cash >= b) { if (i == s.length) break; if (s[i] == '1') { s[i] = '0'; i++; cash -= b; } } str = new String (s); } public static void main (String[]args) { Scanner sc = new Scanner (System.in); n = sc.nextInt (); str = sc.next (); cash = sc.nextInt (); a = sc.nextInt (); b = sc.nextInt (); if (a < b) { swapf (); flipf (); } else { flipf (); swapf (); } System.out.println (Integer.parseInt (str, 2)); } }
Tech Mahindra Coding Questions in C
Q) Here are some Tech Mahindra Coding Questions from 2020 which are solved in C: Build a program for calculating and returning the sums of absolute differences between adjacent numbers in arrays of positive integers. This must be calculated from the position determined by the current user.
In the case of this coding problem, you use three positional arguments through a findTotalSum function. The three inputs you would require are the number of elements inside the array, the elements in the array and the position from where this function will take place.
For example, suppose the total number of elements is 5 and these are the elements:
1 2 3 6 4
Then, if we decide to start from the 3rd position or enter 3 as input, the function will occur from ‘3’, the 3rd number in the array.
Hence, the sum would be a total of (6-3)+(4-6)= 5
Here is the code:
#include <stdio.h>
#include <stdlib.h>
int findTotalSum(int n, int arr[], int start)
{
int difference, sum=0;
for(int i=start-1; i<n-1; i++)
{
difference = abs(arr[i]-arr[i+1]);
sum = sum + difference;
}
return sum;
}
int main()
{
int n;
int start;
scanf("%d",&n);
int array[n];
for(int i=0; i<n; i++)
{
scanf("%d",&array[i]);
}
scanf("%d",&start);
int result = findTotalSum(n, array, start);
printf("\n%d",result);
return 0;
}
Q) Build a program that will allow you to find out how many clothing pieces in total of a certain length can be extracted from a particular number of cloth pieces. We can take the required length for each clothing piece as 10 feet.
In this case, we must first decide upon the length unit as feet and determine the inputs we need. For this function, we will need two inputs, first the number of pieces ( in the array) and the size of each piece in feet inside the array.
A cloth merchant has some pieces of cloth of different lengths. He has an order of curtains of length 12 feet. He has to find how many curtains can be made from these pieces. Length of pieces of cloth is recorded in feet.
For example, suppose the total number of elements is 3 and these are the elements:
0 10 40
Then, the first input is 3 followed by the second input of 0, 10 and 40.
Hence, the sum would be a total of 0 + (10/10) + (40/10) = 5
Thus, there could be 5 pieces of clothing extracted from these 3 pieces of cloth of variable sizes.
Here is the code:
#include <stdio.h>
int findTotalPieces(int n, int arr[])
{
int feet, total = 0;
for(int i=0; i<n; i++)
{
feet = arr[i] / 10;
total = total + feet;
}
return total;
}
int main()
{
int n;
scanf("%d",&n);
int array[n];
for(int i=0; i<n; i++)
{
scanf("%d",&array[i]);
}
int result = findTotalPieces(n, array);
printf("%d",result);
return 0;
}
Q) A Cloth merchant has some pieces of cloth of different lengths. He has an order of curtains of length of 12 feet. He has to find how many curtains can be made from these pieces. Length of pieces of cloth is recorded in feet.
Note : You are expected to write code in the findTotalCurtains function only which receive the first parameter as the number of items in the array and second parameter as the array itself. You are not required to take the input from the console.
Example
Finding the total curtains from a list of 5 cloth pieces.
Input
input 1 : 5
input 2 : 3 42 60 6 14
Output
9
Explanation
The first parameter 5 is the size of the array. Next is an array of measurements in feet. The total number of curtains is 5 which is calculated as under
3 -> 0
42 -> 3
60 -> 5
6 -> 0
14 -> 1
total = 9
Here is the code:
#include <stdio.h> int findTotalCurtains(int n, int arr[]) { int feet, total = 0; for(int i=0; i<n; i++) { feet = arr[i] / 12; total = total + feet; } return total; } int main() { int n; scanf("%d",&n); int array[n]; for(int i=0; i<n; i++) { scanf("%d",&array[i]); } int result = findTotalCurtains(n, array); printf("%d",result); return 0; }
Q) Write a program to find the difference between the elements at odd index and even index.
Note : You are expected to write code in the findDifference function only which receive the first parameter as the numbers of items in the array and second parameter as the array itself. You are not required to take the input from the console.
Example
Finding the maximum difference between adjacent items of a list of 5 numbers
Input
input 1 : 7
input 2 : 10 20 30 40 50 60 70
Output
40
Explanation
The first parameter 7 is the size of the array. Sum of element at even index of array is 10 + 30 + 50 + 70 = 160 and sum of elements at odd index of array is 20 + 40 + 60 = 120. The difference between both is 40
Here is the code:
#include <stdio.h> int findDifference(int n, int arr[]) { int odd=0, even=0; for(int i=0; i<n; i++) { if(i%2==0) { even = even + arr[i]; } else { odd = odd + arr[i]; } } return even-odd; } int main() { int n; scanf("%d",&n); int array[n]; for(int i=0; i<n; i++) { scanf("%d",&array[i]); } int result = findDifference(n, array); printf("%d",result); return 0; }
Q) Write a program to calculate and return the sum of absolute difference between the adjacent number in an array of positive integers from the position entered by the user.
Note : You are expected to write code in the findTotalSum function only which receive three positional arguments:
1st : number of elements in the array
2nd : array
3rd : position from where the sum is to be calculated
Example
Input
input 1 : 7
input 2 : 11 22 12 24 13 26 14
input 3 : 5
Output
25
Explanation
The first parameter 7 is the size of the array. Next is an array of integers and input 5 is the position from where you have to calculate the Total Sum. The output is 25 as per calculation below.
| 26-13 | = 13
| 14-26 | = 12
Total Sum = 13 + 12 = 25
Here is the code:
#include <stdio.h> #include <stdlib.h> int findTotalSum(int n, int arr[], int start) { int difference, sum=0; for(int i=start-1; i<n-1; i++) { difference = abs(arr[i]-arr[i+1]); sum = sum + difference; } return sum; } int main() { int n; int start; scanf("%d",&n); int array[n]; for(int i=0; i<n; i++) { scanf("%d",&array[i]); } scanf("%d",&start); int result = findTotalSum(n, array, start); printf("\n%d",result); return 0; }
Q) Write a program to return the difference between the count of odd numbers and even numbers.
You are expected to write code in the countOddEvenDifference function only which will receive the first parameter as the number of items in the array and second parameter as the array itself. you are not required to take input from the console.
Example
Finding the difference between the count of odd and even numbers from a list of 5 number
Input
input 1 : 8
input 2 : 10 20 30 40 55 66 77 83
Output
-2
Explanation
The first paramter (8) is the szie of the array. Next is an array of integers. The calculation of difference between count sum of odd and even numbers is as follows:
3 (count of odd numbers) – 5 (count of even numbers) = -2
Here is the code:
#include <stdio.h> int countOddEvenDifference(int n, int arr[]) { int odd = 0, even = 0; for(int i=0; i<n; i++) { if(arr[i]%2==0) { even = even+1; } else { odd = odd+1; } } return odd - even; } int main() { int n; scanf("%d",&n); int array[n]; for(int i=0; i<n; i++) { scanf("%d",&array[i]); } int result = countOddEvenDifference(n, array); printf("%d",result); return 0; }
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