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NOTE :- Don’t copy the full code — use the logic.
We're not responsible for plagiarism issues.
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Q1 - Problem Statement 1st

In the kingdom of Numerica, the wise sage Archimedes has discovered an integer N that holds the key to unlock powers.

To unlock these powers, Archimedes needs to find a special divisor of N. This divisor & must be greater than 1, and it will help in determining another number M which is defined as M-N/K.

Archimedes believes that the number M should have the maximum possible number of divisors, To assist Archimedes in his quest, you need to find the smallest A such that M has the maximum number of divisors.

Input Format

The first and only line of input consists of a single integer N representing the key to unlock powers.

Output Format

Print the smallest value of K that results in M having the maximum possible

Q1 Solutions -

import math

def count_divisors(n):

count = 0

i = 1

while i*i <= n:

if n % i == 0:

count += 2 if i != n // i else 1

i += 1

return count

def solve(n):

max_divisors = 0

best_k = 1

for k in range(1, int(math.sqrt(n)) + 1):

if n % k == 0:

for d in [k, n//k]:

m = n * d

divs = count_divisors(m)

if divs > max_divisors or (divs == max_divisors and d < best_k):

max_divisors = divs

best_k = d

return best_k

n = int(input())

print(solve(n))

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Q2 Solutions - Problem Statement 2st -

Number of jumps Bob can make between boxes.
Problem Understanding
Bob can jump between boxes with the following rules:
He can start at any box
From any box, he can jump left or right
He can only jump to adjacent boxes with equal or lower height
Consecutive boxes always have different heights
We need to find the maximum number of jumps possible from any starting position
Approach
For each box, calculate the maximum consecutive non-increasing sequence to the left
For each box, calculate the maximum consecutive non-increasing sequence to the right
The maximum jumps for each box is the sum of left and right sequences minus 1 (since the starting box is counted twice)
The answer is the maximum value found across all boxes

Q2 Solutions -

def max_jumps(heights):

n = len(heights)

left = [1] * n

right = [1] * n

for i in range(1, n):

if heights[i] <= heights[i - 1]:

left[i] = left[i - 1] + 1

for i in range(n - 2, -1, -1):

if heights[i] <= heights[i + 1]:

right[i] = right[i + 1] + 1

max_jumps = 0

for i in range(n):

max_jumps = max(max_jumps, left[i] + right[i] - 1)

return max_jumps

heights = list(map(int, input().split()))

print(max_jumps(heights))

Q3 Solutions - Problem Statement 3st

You have N tables numbered from 1 to N, each with a stack of dishes. The goal is to maximize the total number of dishes picked while moving sequentially from table 1 to N. You can pick the stack from table i (with Di dishes) under the following conditions:

You can pick Di if it is less than or equal to the smallest stack you've picked so far.

Alternatively, you can pick. Di if it is greater than or equal to the largest stack you've picked so far.

You can start picking dishes from any table, but once you move to the next table, you cannot go back.

Input Format

The first line of input contains an integer N representing the count of Table from 1 to N.

The second line Contains N space separated by integers represents dishes on ith table which is 1 << N.

Output Format

1234

Print the maximum number of dishes you can carry with given conditions.

Solution in py -

a = int(input())

b = list(map(int, input().split()))

c = 0

for d in range(a):

e = b[d]

f = b[d]

g = 1

for h in range(d + 1, a):

if b[h] <= e:

e = b[h]

g += 1

elif b[h] >= f:

f = b[h]

g += 1

else:

break

if g > c:

c = g

print(c)