Leetcode with dani
Guys, have you heard of the BIT (Binary Indexed Tree)? I learned about it a few days ago — it transforms both prefix‐sum queries and point updates from O(n) down to O(log n)! In problems with frequent range‐sum queries and multiple updates, a simple prefix‐sum…
I know this one is a little bit harder question — I don’t expect you to get the solution right away.
But I’m sharing it because trying to solve this will help you deeply understand the Binary Indexed Tree (BIT) concept.
You might not get it in one go — and that’s okay.
You can refer to this video after you’ve tried solving it yourself:
🔗 Problem: Create Sorted Array through Instructions
🎥 Helpful video/Ans: YouTube Explanation
Once you solve this, you’ll be confident with how BIT works in real problems — from prefix sums to range frequency queries.
But I’m sharing it because trying to solve this will help you deeply understand the Binary Indexed Tree (BIT) concept.
You might not get it in one go — and that’s okay.
You can refer to this video after you’ve tried solving it yourself:
🔗 Problem: Create Sorted Array through Instructions
🎥 Helpful video/Ans: YouTube Explanation
Once you solve this, you’ll be confident with how BIT works in real problems — from prefix sums to range frequency queries.
LeetCode
Create Sorted Array through Instructions - LeetCode
Can you solve this real interview question? Create Sorted Array through Instructions - Given an integer array instructions, you are asked to create a sorted array from the elements in instructions. You start with an empty container nums. For each element…
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Check this website to practice for the INSA test. There are some rumors that INSA takes questions from this site.
Mensa IQ Test
Leetcode with dani
I'll share any new info as soon as I get it!
Mensa IQ Test
Leetcode with dani
I'll share any new info as soon as I get it!
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can u solve this question? the logic is simple but it may take while to figure it out
70. Climbing Stairs
Solved
Easy
You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Example 1:
Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
Constraints:
1 <= n <= 45
see the question in leetcode
70. Climbing Stairs
Solved
Easy
You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Example 1:
Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:
Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
Constraints:
1 <= n <= 45
see the question in leetcode
Leetcode with dani
can u solve this question? the logic is simple but it may take while to figure it out 70. Climbing Stairs Solved Easy You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways…
What if the question asks to use 1,2 and 3 steps?
Leetcode with dani
can u solve this question? the logic is simple but it may take while to figure it out 70. Climbing Stairs Solved Easy You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways…
a, b= 1, 2
for i in range(3,n+1):
a, b = b, a + b
return b
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Leetcode with dani
What if the question asks to use 1,2 and 3 steps?
a, b, c = 1, 2, 4
for i in range(4, n+1):
a, b, c = b, c, a + b + c
return c
Hey Developers! Let me introduce you to an amazing project: an unlimited free phone verification API created by my friend Yonatan. Check it out!
Forwarded from Yonathan
Built something cool! 🚀
Verify is my personal project—an API for free, unlimited phone verification using Telegram. No SMS costs, simple integration, and optional self-hosting.
🔗 Check it out: https://verify.yonathan.tech
Would love your feedback!
Verify is my personal project—an API for free, unlimited phone verification using Telegram. No SMS costs, simple integration, and optional self-hosting.
🔗 Check it out: https://verify.yonathan.tech
Would love your feedback!
i want to make this channel more usefull for u so .what type of post do u want to see more often?
Anonymous Poll
23%
tech news
54%
algorithm explanation with question
33%
more questions with similar pattern
25%
get matched with other memebers to do mock interviews
25%
just to see the result
if u have better ideas, i would love to hear and collaborate with u .please feel free to contact me with this bot @zprogramming_bot
here is the first question of A2SV weekly contest For G6 i will post each question with their answer
▎🖊 A. Pens and Pencils
▎Problem Statement
Tomorrow is a difficult day for Polycarp: he has to attend lectures and practical classes at the university! He writes lectures with pens and practicals with pencils.
• One pen lasts for
• One pencil lasts for
• His pencil case can hold at most
Can Polycarp pack enough pens and pencils to cover the day?
---
▎Input Format
• First line: An integer
• Each test case: Five integers
–
–
–
–
–
---
▎Output Format
For each test case, output:
• Two integers
–
–
• Or output
---
▎Example Input
▎Example Output
---
▎Explanation
• Test Case 1:
– Needs
– Total =
• Test Case 2:
– Needs
– Total =
• Test Case 3:
– Needs
– Total =
---
▎Sample Code (Python)
▎🖊 A. Pens and Pencils
▎Problem Statement
Tomorrow is a difficult day for Polycarp: he has to attend lectures and practical classes at the university! He writes lectures with pens and practicals with pencils.
• One pen lasts for
c lectures.• One pencil lasts for
d practicals.• His pencil case can hold at most
k writing tools in total.Can Polycarp pack enough pens and pencils to cover the day?
---
▎Input Format
• First line: An integer
t (1 ≤ t ≤ 100), the number of test cases.• Each test case: Five integers
a, b, c, d, k–
a: Number of lectures–
b: Number of practical classes–
c: Lectures per pen–
d: Practicals per pencil–
k: Maximum tools in the pencil case---
▎Output Format
For each test case, output:
• Two integers
x y, where:–
x: Number of pens–
y: Number of pencils• Or output
-1 if it’s not possible to pack enough tools.---
▎Example Input
3
7 5 4 5 8
7 5 4 5 2
20 53 45 26 4
▎Example Output
2 1
-1
1 3
---
▎Explanation
• Test Case 1:
– Needs
ceil(7/4) = 2 pens and ceil(5/5) = 1 pencil.– Total =
2 + 1 = 3 ≤ 8 (possible).• Test Case 2:
– Needs
2 pens and 1 pencil.– Total =
2 + 1 = 3 > 2 (not possible).• Test Case 3:
– Needs
1 pen (ceil(20/45)) and 3 pencils (ceil(53/26)).– Total =
1 + 3 = 4 = 4 (possible).---
▎Sample Code (Python)
t = int(input())
for _ in range(t):
a, b, c, d, k = map(int, input().split())
pens_needed = (a + c - 1) // c # Ceiling of a/c
pencils_needed = (b + d - 1) // d # Ceiling of b/d
if pens_needed + pencils_needed <= k:
print(pens_needed, pencils_needed)
else:
print(-1)
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▎B. Integer Points
▎Problem (brief)
DLS draws
▎Key Fact
The intersection of y = x + p and y = -x + q is given by:
For both x and y to be integers, q and p must have the same parity.
▎Solution
1. Count the number of even and odd integers in sets P and Q.
2. The total number of intersecting pairs is calculated as:
▎Implementation
▎Example
▎Input
▎Output
▎Problem (brief)
DLS draws
n lines of the form y = x + pᵢ and JLS draws m lines of the form y = -x + qⱼ . Count how many pairs (one from each set) intersect at integer coordinates (x, y).▎Key Fact
The intersection of y = x + p and y = -x + q is given by:
x = q - p / 2, y = q + p / 2
For both x and y to be integers, q and p must have the same parity.
▎Solution
1. Count the number of even and odd integers in sets P and Q.
2. The total number of intersecting pairs is calculated as:
Answer = (even_P × even_Q) + (odd_P × odd_Q)
▎Implementation
for _ in range(int(input())):
n, P = int(input()), list(map(int, input().split()))
m, Q = int(input()), list(map(int, input().split()))
eP = sum(p % 2 == 0 for p in P) # Count evens in P
oP = n - eP # Count odds in P
eQ = sum(q % 2 == 0 for q in Q) # Count evens in Q
oQ = m - eQ # Count odds in Q
print(eP * eQ + oP * oQ)
▎Example
▎Input
3
3
1 3 2
2
0 3
1
1
1
1
1
2
1
1
▎Output
3
1
0
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▎C. Labs (Simplified)
You have n² labs numbered from 1 (lowest) to n² (highest).
Any lab u can send 1 unit of water to any lower lab v if u > v .
▎Problem Statement
Split the labs into n groups of size n .
For each ordered pair of groups (A, B) , let f(A, B) be the total units sendable from all labs in A to all in B .
Objective: Maximize the minimum f(A, B) over all A ≠ B .
▎Output
Print any grouping that achieves this.
▎Example
Input:
Output (one possible grouping):
Here, n = 3 , meaning there are labs numbered from 1 to 9.
Every group has 3 labs.
The smallest f(A, B) among all 6 ordered pairs is 4, which is optimal.
Solution
You have n² labs numbered from 1 (lowest) to n² (highest).
Any lab u can send 1 unit of water to any lower lab v if u > v .
▎Problem Statement
Split the labs into n groups of size n .
For each ordered pair of groups (A, B) , let f(A, B) be the total units sendable from all labs in A to all in B .
Objective: Maximize the minimum f(A, B) over all A ≠ B .
▎Output
Print any grouping that achieves this.
▎Example
Input:
3
Output (one possible grouping):
2 8 5
9 3 4
7 6 1
Here, n = 3 , meaning there are labs numbered from 1 to 9.
Every group has 3 labs.
The smallest f(A, B) among all 6 ordered pairs is 4, which is optimal.
Solution
def main():
n = int(input())
groups = [[] for _ in range(n)]
num = 1
for row in range(n):
# decide direction: left‑to‑right on even rows, right‑to‑left on odd
cols = range(n) if row % 2 == 0 else range(n - 1, -1, -1)
for col in cols:
groups[col].append(num)
num += 1
# output
for g in groups:
print(*g)
if __name__ == "__main__":
main()
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