153. Find Minimum in Rotated Sorted Array
#leet_code_Q10 #Medium #binary_search #Q_153
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.



Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Don't try to answer at first glance. it's more challenging than it seems

based on this poll we will start doing two pointer and sliding window respectively

344. Reverse String
#Topics #twopointers #strings

Write a function that reverses a string. The input string is given as an array of characters s
You must do this by modifying the input array in-place with O(1) extra memory
Example 1:

Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]
Example 2:

Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

link

#leet_code_Q_12 #easy #two_pointer #Q_202 202.Write an algorithm to determine if a number n is happy.

A happy number is a number defined by the following process:

Starting with any positive integer, replace the number by the sum of the squares of its digits.
Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.
Those numbers for which this process ends in 1 are happy.
Return true if n is a happy number, and false if not.



Example 1:

Input: n = 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
Example 2:

Input: n = 2
Output: false


Constraints:

1 <= n <= 231 - 1

አዳያመልጣችሁ አሪፍ ዕድል ነው

Answer:

How do you convert a tuple into a list in Python?
a) tuple.toList()
b) list(tuple)
c) tuple.list()

i recommend this video for those who want to learn sliding window algorithm , we will be working on this algorithm in the coming days

#leet_code_Q_13 #easy #Sliding_window #Q_219 219. Contains Duplicate II
Given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.



Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true
Example 2:

Input: nums = [1,0,1,1], k = 1
Output: true
Example 3:

Input: nums = [1,2,3,1,2,3], k = 2
Output: false

out of topic I just watched an amazing video that dives into the theory of relativity and how time works. ⏳

It’s fascinating how scientists always talk about light, and I used to think they had a beef with speed of light 😂 But now I understand why speed of light is so special and its profound connection to time. https://youtu.be/5BBHEZFearI?si=dGWI303LG9XXgQYh

share your ideas

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## Intuition
To solve this problem, we can use a sliding window approach along with a hash map (dictionary) to efficiently check for duplicate elements within a given range. The idea is to maintain a window of size k and check if any element within this window is duplicated.

## Algorithm
1. Initialize an empty hash map (dictionary) to store the most recent index of each element.
2. Iterate over the array nums:
- For each element nums[i]:
- If nums[i] is already in the hash map and the difference between the current index i and the stored index j is less than or equal to k, return true.
- Update the hash map to store the current index i of nums[i].
3. If no duplicates within the range k are found, return false.

## Code

def containsNearbyDuplicate(nums, k):
    num_indices = {}
    
    for i, num in enumerate(nums):
        if num in num_indices and i - num_indices[num] <= k:
            return True
        num_indices[num] = i
    
    return False

# Example 1
print(containsNearbyDuplicate([1, 2, 3, 1], 3))  # Output: True

# Example 2
print(containsNearbyDuplicate([1, 0, 1, 1], 1))  # Output: True

# Example 3
print(containsNearbyDuplicate([1, 2, 3, 1, 2, 3], 2))  # Output: False

## Complexity Analysis
- Time Complexity: O(n), where n is the length of the array nums. We only need to iterate through the array once.
- Space Complexity: O(min(n, k)), where n is the length of the array. The space used by the hash map is proportional to the number of unique elements within any window of size k. In the worst case, this is O(n), but usually, it's limited by k.

This algorithm ensures that we efficiently find any duplicates within a specified range, making use of the hash map for quick lookups and updates.

#leet_code_Q_14 #Q_2269 #easy #Sliding_window
2269. Find the K-Beauty of a Number
The k-beauty of an integer num is defined as the number of substrings of num when it is read as a string that meet the following conditions:

It has a length of k.
It is a divisor of num.
Given integers num and k, return the k-beauty of num.

Note:

Leading zeros are allowed.
0 is not a divisor of any value.
A substring is a contiguous sequence of characters in a string.



Example 1:

Input: num = 240, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
- "24" from "240": 24 is a divisor of 240.
- "40" from "240": 40 is a divisor of 240.
Therefore, the k-beauty is 2.
Constraints:

1 <= num <= 109
1 <= k <= num.length (taking num as a string)