Mitsogo Coding Questions & Solutions 2024 (NEW), Get Mitsogo Coding Questions

Mitsogo Coding Questions & Solutions 2024 | Mitsogo Coding Questions and Answers | Mitsogo Coding Questions | Mitsogo Round 2 Coding Questions 

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Mitsogo Coding Questions with Answers 

Q) Problem Description :

Stepping Numbers are numbers in which the adjacent digits differ by 1. For example, 123, 545, and 98 are stepping numbers, while 321, 444, and 75 are not. The task is to find all stepping numbers in a given range [n, m].

• For example
  • Range: [100, 500]
  • Stepping Numbers: 101, 121, 123, 210, 212, 232, 234, 321, 323, 343, 345
  • Explanation: The stepping numbers between 100 and 500 are 101, 121, 123, 210, 212, 232, 234, 321, 323, 343, and 345. These numbers have adjacent digits that differ by 1.

Write code to find out all the stepping numbers in the given range.

Input Format: First line contains two numbers N,M

Output Format: Print all the stepping numbers present in the range.

Constraints: 0 <= N < M <= 1,000,000,000

C++ Program 

#include<bits/stdc++.h>
using namespace std;

void displaySteppingNumbers(int n, int m) {
    queue q; 
    for (int i = 1; i <= 9; i++) { q.push(i); } while (!q.empty()) { int stepNum = q.front(); q.pop(); if (stepNum >= n && stepNum <= m) {
            cout << stepNum << " "; } if (stepNum == 0 || stepNum > m) {
            continue;
        }
        int lastDigit = stepNum % 10;
        int stepNumA = stepNum * 10 + (lastDigit - 1);
        int stepNumB = stepNum * 10 + (lastDigit + 1);
        if (lastDigit == 0) {
            q.push(stepNumB);
        } else if (lastDigit == 9) {
            q.push(stepNumA);
        } else {
            q.push(stepNumA);
            q.push(stepNumB);
        }
    }
}

int main() {
    int n, m;
    cout << "Enter the range [n, m]: "; cin >> n >> m;
    cout << "Stepping Numbers between " << n << " and " << m << ": ";
    displaySteppingNumbers(n, m);
    return 0;
}

Java Program 

import java.util.LinkedList;
import java.util.Queue;

public class Main {
    public static void displaySteppingNumbers(int n, int m) {
        Queue queue = new LinkedList<>();
        for (int i = 1; i <= 9; i++) { queue.add(i); } while (!queue.isEmpty()) { int stepNum = queue.poll(); if (stepNum >= n && stepNum <= m) { System.out.print(stepNum + " "); } if (stepNum == 0 || stepNum > m) {
                continue;
            }
            int lastDigit = stepNum % 10;
            int stepNumA = stepNum * 10 + (lastDigit - 1);
            int stepNumB = stepNum * 10 + (lastDigit + 1);
            if (lastDigit == 0) {
                queue.add(stepNumB);
            } else if (lastDigit == 9) {
                queue.add(stepNumA);
            } else {
                queue.add(stepNumA);
                queue.add(stepNumB);
            }
        }
    }

    public static void main(String[] args) {
        int n = 0, m = 100;
        System.out.print("Stepping Numbers between " + n + " and " + m + ": ");
        displaySteppingNumbers(n, m);
    }
}

Python Program 

def displaySteppingNumbers(n, m):
    queue = []
    for i in range(1, 10):
        queue.append(i)
    while queue:
        stepNum = queue.pop(0)
        if stepNum >= n and stepNum <= m: print(stepNum, end=' ') if stepNum == 0 or stepNum > m:
            continue
        lastDigit = stepNum % 10
        stepNumA = stepNum * 10 + (lastDigit - 1)
        stepNumB = stepNum * 10 + (lastDigit + 1)
        if lastDigit == 0:
            queue.append(stepNumB)
        elif lastDigit == 9:
            queue.append(stepNumA)
        else:
            queue.append(stepNumA)
            queue.append(stepNumB)
n = 0
m = 100
print("Stepping Numbers between", n, "and", m, ": ", end='')
displaySteppingNumbers(n, m)

Q) Problem Statement -:

In an airport, the Airport authority decides to charge a minimum amount to the passengers who are carrying luggage with them. They set a threshold weight value, say, T, if the luggage exceeds the weight threshold you should pay double the base amount. If it is less than or equal to threshold then you have to pay $1.

Function Description:

Complete the weightMachine function in the editor below. It has the following parameter(s):

Parameters:

Returns: The function must return an INTEGER denoting the required amount to be paid.

Constraints:

• 1 <= N <= 10^5
• 1 <= weights[i] <= 10^5
• 1 <= T <= 10^5

Input Format for Custom Testing:

• The first line contains an integer, N, denoting the number of luggage.
• Each line i of the N subsequent lines (where 0 <= i <n) contains an integer describing the weight of ith luggage.
• The next line contains an integer, T, denoting the threshold weight of the boundary wall.

Sample Cases:

• Sample Input 1
  4
  1
  2
  3
  4
  3
• Sample Output 1
  5
• Explanation:
  Here all weights are less than threshold weight except the luggage with weight 4 (at index 3) so all pays base fare and it pays double fare.

C Program 

#include<stdio.h>
long int weightMachine(long int N,long int weights[],long int T)
{
    long int amount=0,i;
    for(i=0;i<N;i++) { amount++; if(weights[i]>T)
        {
            amount++;
        }
    }
    return amount;
}
int main()
{
    long int N,i,T;
    scanf("%ld",&N);
    long int weights[N];
    for(i=0;i<N;i++)
    {        scanf("%ld",&weights[i]);
    }
    scanf("%ld",&T);
    printf("%ld",weightMachine(N,weights,T));
    return 0;
}

C++ Program 

#include <bits/stdc++.h>
using namespace std;
long int weightMachine(long int N,long int weights[],long int T)
{
    long int amount=0,i;
    for(i=0;i<N;i++) { amount++; if(weights[i]>T)
        {
            amount++;
        }
    }
    return amount;
}
int main()
{
    long int N,i,T;
    cin>>N;
    long int weights[N];
    for(i=0;i<N;i++) { cin>>weights[i];
    }
    cin>>T;  cout<<weightMachine(N,weights,T);
    return 0;
}

Java Program 

import java.util.*;
class Main
{
  static int weightMachine (int N, int weights[],int T)
  {
    int amount = 0, i;
    for (i = 0; i < N; i++) { amount++; if (weights[i] > T)
	   {
	      amount++;
	    }
    }
    return amount;
  }

  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    int n = sc.nextInt ();
    int weights[]= new int[n];    
    for(int i=0; i<n; i++)
      weights[i] = sc.nextInt(); 
    int t = sc.nextInt ();
    System.out.println (weightMachine(n, weights, t));      
  }
}

Python Program 

def weightMachine(N,weights,T):
    amount=0
    for i in weights:
        amount+=1 
        if(i>T):
            amount+=1 
    return amount
N=int(input())
weights=[]
for i in range(N):
    weights.append(int(input()))
T=int(input())
print(weightMachine(N,weights,T))

Q) You are given a list of daily prices of a stock. You can buy a stock on one day and sell it later on another day after the day you bought the stock. You can perform the above operation only once. What is the maximum loss possible?

Example

Prices=[10,4,2,9]

The greatest loss is incurred when you buy at a price of 10 and sell at a price of 2. Return the difference:9.

Example

Price=[1,2,3,4]

The Price went up every day. Return 0.

Sample Input for Custom Testing

STDIN                   Function

———–               ————–

•    7   → Prices []  size n=7
•    1 →       prices =[1,8,4,2,10,3,2]
•    8
•    4
•    2
•   10
•    3
•    2

Sample Output

•   8

Explanation

Using zero-based index notation, the correct answer is a[4]-a[6]=10-2=8. There is a greater difference between 10 and 1 but that would imply selling before buying, and short selling is not allowed in this problem.

C++ Program 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int solve(vector v)
{
    int n = v.size();
    if (n == 0)
        return 0;
    int mx = v[0];
    for (int i = 1; i < n; i++)
        mx = max(mx, v[i]);
        if (mx <= 0)
            return 0;
        int mxSum = 0;
        int cSum = 0;
        for (int i = 0; i < n; i++)
        {
            cSum += v[i];
            if (cSum < 0) cSum = 0; mxSum = max(mxSum, cSum); } return mxSum; } int main() { int n; cin >> n;
    int price[n];
    for (int i = 0; i < n; i++) { cin >> price[i];
    }
    vector diff;
    for (int i = n-2; i >=0 ; i--)
        diff.push_back(price[i] - price[i+1]);
    int ans = solve(diff);
    if(ans<0)
    {
        cout<<0<< endl;
    }
    else
    {
        cout<< ans<< endl;
    }
    return 0;
}

n=int(input())
arr=[]

ans=[0]

for i in range(n):

    arr.append(int(input()))

for i in range(n-1):

    x=min(arr[i+1:])-arr[i]
    if x<0:
        ans.append(x)

print(-1*min(ans))

Java Program 

import java.util.*;
class Solution {
  public static int solve(ArrayList < Integer > list) {
    int n = list.size();
    if (n == 0)
      return 0;
    int max = list.get(0);
    for (int i = 1; i < n; i++)
      max = Math.max(max, list.get(i));
    if (max <= 0)
      return 0;
    int maxSum = 0;
    int sum = 0;

    for (int i = 0; i < n; i++) {
      sum = sum + list.get(i);
      if (sum < 0)
        sum = 0;
      maxSum = Math.max(maxSum, sum);
    }
    return maxSum;
  }

  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();
    ArrayList < Integer > list = new ArrayList();

    for (int i = n - 2; i >= 0; i--)
      list.add(arr[i] - arr[i + 1]);
    int res = solve(list);

    if (res < 0)
      System.out.println(0);
    else
      System.out.println(res);
  }
}

C++ Program 

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n;cin>>n;
    string s;cin>>s;
    int prime[128];
    prime[0]=prime[1]=0;
    int p=2;
    for(int i=2;i<128;i++) prime[i]=1;
    while(p*p<=127)
    {
        if(prime[p])
        {
            for(int i=p*p;i<=127;i+=p)
            prime[i]=0;
        }
        p++;
    }
    string s1,s2;
    for(auto i:s)
    {
        if(prime[i]) s1+=i;
        else s2+=i;
    }
    sort(s1.begin(),s1.end());
    sort(s2.begin(),s2.end());
    reverse(s2.begin(),s2.end());
    cout<< s1<< s2;


}

Java Program 

import java.util.Collections;
import java.util.Scanner;
import java.util.Vector;
public class Applicartion22 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.next();
        boolean primes[] = new boolean[127];
        for (int i = 0; i < 127; i++) {
            primes[i] = true;
        }
        primes[0] = false;
        primes[1] = false;
        int p = 2;
        while (p * p <= 126) {
            if (primes[p]) {
                for (int i = p * p; i < 127; i += p) {
                    primes[i] = false;
                }
            }
            p += 1;
        }
        Vector < Character > s1 = new Vector < > ();
        Vector < Character > s2 = new Vector < > ();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (primes[(int) ch]) {
                s1.add(ch);
            } else {
                s2.add(ch);
            }
        }
        Collections.sort(s1);
        Collections.sort(s2);
        for (int i = 0; i < s1.size(); i++) { System.out.print(s1.get(i)); } for (int i = s2.size() - 1; i >= 0; i--) {
            System.out.print(s2.get(i));
        }
    }
}

Python Program 

n = int(input())
s = input()
prime = [True] * 127
prime[0] = prime[1] = False
p = 2
while p * p <= 126:
    if prime[p]:
        for i in range(p * p, 127, p):
            prime[i] = False
    p += 1


s1 = []
s2 = []
for i in s:
    if prime[ord(i)]:
        s1.append(i)
    else:
        s2.append(i)

s1.sort()
s2.sort(reverse=True)
s1.extend(s2)
print("".join(s1))

#include<bits/stdc++.h>

using namespace std;

bool isPrime(int n) {
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    for (int i = 5; i * i <= n; i += 6)
        if ((n % i == 0) || (n % (i + 2) == 0))
            return false;
    return true;
}

int solve(int n, int k) {
    int c = 0;
    vector list;

    // Find k numbers greater than n with the property that the sum of their digits is prime
    for (int i = n + 1; c < k; i++) { int temp = i; int sum = 0; while (temp != 0) { sum += temp % 10; temp /= 10; } if (isPrime(sum)) { list.push_back(i); c++; } } return list[k - 1]; } int main() { int n; cin >> n;

    int k;
    cin >> k;

    cout << solve(n, k);

    return 0;
}

import java.util.*;
class Main {
    public static boolean isPrime(int n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        for (int i = 5; i * i <= n; i = i + 6)
            if ((n % i == 0) || (n % (i + 2) == 0))
                return false;
        return true;
    }
    public static int solve(int n, int k) {
        int c = 0;
        ArrayList < Integer > list = new ArrayList < > ();
        for (int i = n + 1; c < k; i++) {
            int temp = i;
            int sum = 0;
            while (temp != 0) {
                sum = sum + temp % 10;
                temp = temp / 10;
            }
            if (isPrime(sum)) {
                list.add(i);
                c++;
            }
        }
        return list.get(k - 1);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        System.out.println(solve(n, k));
    }
}

Python Program 

import math

def isPrime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    for i in range(5, int(math.sqrt(n)) + 1, 6):
        if n % i == 0 or n % (i + 2) == 0:
            return False
    return True


def solve(n, k):
    c = 0
    list = []
    for i in range(n + 1, 1000000000):
        temp = i
        sum = 0
        while temp != 0:
            sum = sum + temp % 10
            temp = temp // 10
        if isPrime(sum):
            list.append(i)
            c += 1
        if c == k:
            break
    return list[k - 1]


n, k = map(int, input().split())
print(solve(n, k))

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