19 months ago

buzz • 10

Bengaluru

1.     Given an integer N, the task is to count the minimum steps required to reduce the value of N to 0 by performing the following two operations:

• Consider integers A and B where N = A * B (A != 1 and B != 1), reduce N to max(A, B)
• Decrease the value of N by 1

2.  Given a string S containing letter and digit and an integer K where, and . The task is to return the K-th letter of the new string S’.
The new string S’ is formed from old string S by following steps: 
1. If the character read is a letter, that letter is added at end of S’. 
2. If the character read is a digit, then entire string S’ repeatedly written d-1 more times in total.
Note: The new string is guaranteed to have less than 2^63 letters.

same gfg question

3. one more tough DP question

 Remaining MCQs based on linux commands mostly, SQL question,OS round robin question .

22 months ago

minion • 0

Varanasi

which college questions are these?

20 months ago

KUMARVRUSHNI VYAWAHARE • 0

Mumbai

#Problem 1: Minimum Steps to Reduce N to 0

from collections import deque

def min_steps_to_zero(N):
    queue = deque([(N, 0)])  # (current value, steps)
    visited = set()
    
    while queue:
        current, steps = queue.popleft()
        
        if current == 0:
            return steps
        
        # Decrement operation
        if current - 1 not in visited:
            visited.add(current - 1)
            queue.append((current - 1, steps + 1))
        
        # Factorization operation
        for i in range(2, int(current**0.5) + 1):
            if current % i == 0:
                A, B = i, current // i
                next_n = max(A, B)
                if next_n not in visited:
                    visited.add(next_n)
                    queue.append((next_n, steps + 1))
    
    return -1  # In case no solution found

#Problem 2: K-th Character in Expanded String

def find_kth_character(S, K):
    current_length = 0
    stack = []
    
    for char in S:
        if char.isalpha():
            current_length += 1
            if current_length == K:
                return char
        elif char.isdigit():
            digit = int(char)
            if current_length * digit >= K:
                # If we would surpass K, we need to find the K-th character in the expanded form
                stack.append((current_length, digit))
            current_length *= digit
            
    # If we exit the loop, we haven't found it in the letters directly
    return None  # if K is out of bounds