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Tech Mahindra 2nd Round 12th April Exam Shifts All Sections Question with Solution | Non-SDE Tech & Psychometric Assessment (1A/1B/1C)

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Tech Mahindra 2nd Round 12th April Exam Shifts All Sections Question with Solution
Tech Mahindra 2nd Round 12th April Exam Shifts All Sections Question with Solution


Tech Mahindra 2nd Round 12th April Exam Shifts All Sections Question with Solution | Non-SDE Tech & Psychometric Assessment (1A/1B/1C) Solved by Pappu Career Guide

Tech Mahindra 2nd Round 13th April Exam Shifts Non-SDE Tech All Sections Question with Solution

TECH MAHINDRA Non-SDE ALL CODING SOLUTIONS:

1. Geometric Progression Question with Solution:

Geometric Progression In C++

// Solved By Pappu Career Guide

#include <bits/stdc++.h>

using namespace std;
// Solved By Pappu Kumar (Coder Guy)
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

Coming Soon.....

In C

Coming Soon.....

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);
// solved by Pappu Kumar Guide
return 0;

}

In Python

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

Coming Soon.....

https://t.me/techmahindra2021crack

3. Frequency Count Question with Solution:

Frequency Count In C++
// Solved By Pappu Career Guide

#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

Coming Soon.....


In C

Coming Soon.....


4. Caesar Cipher Question with Solution:

Caesar Cipher In C++
// Solved By Pappu Career Guide

#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

Coming Soon.....

In Python

Coming Soon.....

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)]

https://t.me/techmahindra2021crack

s=input()

print(numDecodings(s))

In C

Coming Soon.....

In c++

Coming Soon.....

How To for Download All Section Non-SDE Tech & Psychometric Assessment (1A/1B/1C) and Coding solution Complete in One PDF:  
 
All candidates All Section Non-SDE Tech & Psychometric Assessment (1A/1B/1C) and Coding solution Complete in One PDF download here:

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