Tech Mahindra 2nd Round 13th April Exam Shifts All Coding Question with Solution
Tech Mahindra 2nd Round 12th to 15th April Exam Shifts All Coding Question with Solution | Non-SDE Tech Technical & NonTechnical Solved by Pappu Career Guide
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TECH MAHINDRA Non-SDE ALL Technical & Non-Technical All Repeated Questions SOLUTIONS :
1. Gp ( Done)
2. Bitwise operation (Done)
3. Frequency Count (Done)
4. Caesar Cipher (Done)
5. Number of Decoding (Done)
6. Longest Common Subsequence
7. Numbers Puzzle
8. Longest Increasing Subsequence
9. Moving Apples
10. infix to postfix
11. Penalty
12. Euler’s Totient Function
13. Next Number Generator
14. Next Number Element
15. Maximum Subarray
16. Minimise a String
17. Dubai Airport
18. Possible Decodings
19. Longest Decreasing Subsequence
20. Longest Palindromic subsequence
21. Module 11 code
22. Roots of the Quadratic Equation
23. Maximum Subarray
24. Number of Selective Arrangement
25. The Cuckoo Sequent
26. Character Count
1 . Geometric Progression Question with Solution: Geometric Progression In C++
#include <bits/stdc++.h>
using namespace std;
int main() {
double second_term;
double third_term;
int n;
cin>>second_term>>third_term>>n;
double r = third_term/second_term; //9/3=3
double a = second_term/r;
double nth_term = a * pow( r, n-1);
cout<<nth_term;
return 0;
}
In Python
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In C
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2. BitWise Operation Question with Solution: In C++
// solved by Pappu Kumar Guide
#include <bits/stdc++.h>
using namespace std;
int main() {
// a#b#c
// ex
// 10 - binary- 1 0 1 0
// output 2#1#3
int n;
cin>>n;
int a=0; // most
int b=-1; // least
int c=-1;
int i=0;
// solved by Pappu Career Guide
// performing bit marking
while(n>0)
{
if(n&1)
{
a++;
if(b==-1)
{
b=i;
}
c=i;
}
i++;
n=n>>1; //right shift
}
cout<< to_string(a)<<"#" << to_string(b) <<"#" << to_string(c);
return 0;
}
n=int(input())
bi=bin(n)
bi=int(bi[2:])
x=[]
// Solved By Pappu Career Guide
while(bi>0):
x.append(bi%10)
bi//=10
a=x.count(1)
b=x.index(1)
x.reverse()
c=len(x)-x.index(1)-1
print(a,b,c,sep="#")
In C
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3. Frequency Count Question with Solution: Frequency Count In C++
#include <bits/stdc++.h>
using namespace std;
string helper(string s)
// Solved By Pappu Kumar (Coder Guy)
{
int arr[26]={0}; // there are 26 alphabets
in the english char
for(int i=0;i<s.length();i++) // loop through
the input string
{
arr[s[i]-97]++; //
to come back with the 0 position
}
string result="";
for(int i=0;i<26;i++) // for character
count with loop
{
if(arr[i]>0) // if char is present
{
char ch = 97+i; // add the value of
ASCII value of i = 1 - taking b
result+=ch;
result+=to_string(arr[i]);
}
}
return result;
}
int main() {
string s;
cin>>s;
cout<<helper(s);
return 0;
}
In Python
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In C
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4. Caesar Cipher Question with Solution: Caesar Cipher In C++
#include <bits/stdc++.h>
using namespace std;
// Solved By Pappu Kumar (Coder Guy)
string helper(string s)
{
string result = ""; // answer store string
for(int i=0;i<s.length();i++) // loop through
the input string
{
char ch = char((s[i] + 3 - 97)%26 + 97);
result+=ch;
}
return result;
}
int main() {
string s;
cin>>s;
cout<<helper(s);
return 0;
}
In C
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In Python
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https://t.me/placemate
5. Number of Decodings
The solution in Python:
def numDecodings(s):
if not s:
return 0
dp = [0 for x in range(len(s) + 1)]
# base case initialization
dp[0] = 1
dp[1] = 0 if s[0] == "0" else 1 #(1)
for i in range(2, len(s) + 1):
# One step jump
if 0 < int(s[i-1:i]) <= 9: #(2)
dp[i] += dp[i - 1]
# Two step jump
if 10 <= int(s[i-2:i]) <= 26: #(3)
dp[i] += dp[i - 2]
return dp[len(s)]
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s=input()
print(numDecodings(s))
In C
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In c++
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6. Longest Common Subsequence
// Solved By Pappu Career Guide
#include <bits/stdc++.h>
using namespace std;
string helper(string s1 , string s2)
{
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int dp[s1.length()+1][s2.length()+1];
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for(int i=0;i<=s1.length();i++)
{
for(int j=0;j<s2.length();j++)
{
dp[i][j]=0;
}
}
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for(int i=1;i<=s1.length();i++)
{
for(int j=1;j<=s2.length();j++)
{
if(s1[i-1]==s2[j-1])
{
dp[i][j]=1+dp[i-1][j-1];
}
else
{
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
}
return dp[s1.length()][s2.length()];
}
int main() {
string s1;
string s2;
cin>>s1>>s2;
cout<<helper(s1,s2);
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return 0;
}
7. Numbers Puzzles Solution in C
// Solved By Pappu Career Guide
#include<stdio.h>
#include <stdlib.h>
void swap(int *x, int *y){
int temp;
temp = *x;
*x = *y;
*y = temp;
}
void BubbleSort( int arr[], int n ){
int i, j;
for(i=0;i<n-1;i++){
for (j = 0; j < n-i-1; j++){
if (arr[j] > arr[j+1]){
swap(&arr[j], &arr[j+1]);
}
}
}
}
int Display(int arr[], int n){
int i;
for(i=0;i<n;i++){
printf("%d \n", arr[i]);
}
}
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int AbsValue(int arr[],int n){
int sum=0;
for(int i=n-1;i>0;--i){
sum += abs(arr[i]-arr[i-1]);
}
printf("%d",sum);
}
int main(){
int arr[] = {3,2,1};
int n = sizeof(arr) / sizeof(arr[0]);
BubbleSort(arr, n);
//Display(arr,n);
AbsValue(arr,n);
}
One More Solution of Number of Puzzle
8. Longest Increasing Subsequence Solution in Python
// Solved By Pappu Career Guide
https://t.me/freshersoffcampjobs 9. Moving Apples
https://t.me/techmahindra2021crack 10. inFixToPostFix Solution in Python
def inFixToPostFix():
inFix = '3*(x+1)-2/2'
postFix = ''
s = Stack()
for c in inFix:
# if elif chain for anything that c can be
if c in "0123456789x":
postFix += c
elif c in "+-":
if s.isEmpty():
s.push(c)
elif s.top() =='(':
s.push(c)
elif c in "*/":
if s.isEmpty():
s.push(c)
elif s.top() in "+-(":
s.push(c)
elif c == "(":
s.push(c)
elif c == ")":
while s.top() is not '(':
postFix += s.pop()
s.pop()
else:
print("Error")
print(postFix)
return postFix
https://t.me/techmahindra2021crack
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11. Penalty Solution in C++
// Solved By Pappu Career Guide
#include <bits/stdc++.h>
using namespace std;
int helper(int n, int arr[])
{
int penalty = 0;
sort(arr, arr+n); // inbuild method foro sort the array
for(int i=1;i<n;i++) // look through all the element in the array
{
penalty+=abs(arr[i]-arr[i-1]); // adding the adjecent element in a penalty variables
}
return penalty;
}
int main() {
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout<<helper(n, arr);
return 0;
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}
12. Euler’s Totient Function in C++
// Solved By Pappu Career Guide
#include <stdio.h>
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
int phi(unsigned int n)
{
unsigned int result = 1;
for (int i = 2; i < n; i++)
if (gcd(i, n) == 1)
result++;
return result;
}
int main()
{
int n;
for (n = 1; n <= 10; n++)
printf("phi(%d) = %d\n", n, phi(n));
return 0;
}
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13. Next Number Generator in C++
// Solved By Pappu Career Guide
14. Next Number Element in C++
// Solved By Pappu Career Guide
15. Maximu Subarray in C++
// Solved By Pappu Career Guide
16. Minimise a String in C++
// Solved By Pappu Career Guide
17. Dubai Airport in C++
// Solved By Pappu Career Guide
18. Possible Decodings in C++
// Solved By Pappu Career Guide
19. Longest Decreasing Subsequence in C++
// Solved By Pappu Career Guide
20. Longest Palindromic subsequence in C++
// Solved By Pappu Career Guide
21. module 11 in Python
// Solved By Pappu Career Guide
def remainder(st) :
ln = len(st)
rem = 0
for i in range(0, ln) :
num = rem * 10 + (int)(st[i])
rem = num % 11
return rem
st = "3435346456547566345436457867978"
print(remainder(st))
22. Roots of the Quadratic Equation Solution import static java.lang.Math.*;
public class Main
{
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static void calculateRoots(int a, int b, int c)
{
if (a == 0)
{
System.out.println("The value of a cannot be 0.");
return;
}
int d = b * b - 4 * a * c;
double sqrtval = sqrt(abs(d));
if (d > 0)
{
System.out.println("The roots of the equation are real and different. \n");
System.out.println((double)(-b + sqrtval) / (2 * a) + "\n"+ (double)(-b - sqrtval) / (2 * a));
}
else if (d == 0)
{
System.out.println("The roots of the equation are real and same. \n");
System.out.println(-(double)b / (2 * a) + "\n"+ -(double)b / (2 * a));
}
else
{
System.out.println("The roots of the equation are complex and different. \n");
System.out.println(-(double)b / (2 * a) + " + i"+ sqrtval + "\n"+ -(double)b / (2 * a)+ " - i" + sqrtval);
}
}
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public static void main(String args[])
{
int a = 1, b = -2, c = -3;
calculateRoots(a, b, c);
}
}
23. Maximum Subarray Solution class Solution(object): def maxSubArray(self, nums): """ :type nums: List[int] :rtype: int """ dp = [0 for i in range(len(nums))] dp[0] = nums[0] for i in range(1,len(nums)): dp[i] = max(dp[i-1]+nums[i],nums[i]) #print(dp) return max(dp) nums = [-2,1,-3,7,-2,2,1,-5,4] ob1 = Solution() print(ob1.maxSubArray(nums)) 24. Number of Selective Arrangement
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25. The Cuckoo Sequent Solution
26. Character Count Solution
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