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Leetcode with dani
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Leetcode with dani
#Q20 #leet_codeQ15 Medium noscript. 3Sum Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.…
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums = sorted(nums)
list1 = []
for i in range(len(nums)-1,-1,-1):
j = i - 1
left =0
if (i>1 and i!=(len(nums)-1) and nums[i]==nums[i+1]):
continue
while(j>-1 and left<=i-2 and j!=left):
total = nums[left] + nums[i] + nums[j]
if ( total)== 0:
list1.append([nums[left],nums[i],nums[j]])
left+=1
while(left<j and nums[left]==nums[left-1]):
left+=1
elif total > 0:
j-=1
else:
left+=1
return list1
after solving 3 sum try to solve the next question
#Q21 #leet_codeQ16 #3Sum_Closest
Medium

Given an integer array nums of length n and an integer target, find three integers in nums such that the sum is closest to target.

Return the sum of the three integers.

You may assume that each input would have exactly one solution.



Example 1:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
Example 2:

Input: nums = [0,0,0], target = 1
Output: 0
Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).


Constraints:

3 <= nums.length <= 500
-1000 <= nums[i] <= 1000
-104 <= target <= 104
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bubble sort
def bubbleSort(arr):
n = len(arr)

# Traverse through all array elements
for i in range(n):
swapped = False

# Last i elements are already in place
for j in range(0, n-i-1):

# Traverse the array from 0 to n-i-1
# Swap if the element found is greater
# than the next element
if arr[j] > arr[j+1]:
arr[j], arr[j+1] = arr[j+1], arr[j]
swapped = True
if (swapped == False):
break
insertion sort
def insertionSort(arr):
for i in range(1, len(arr)):
key = arr[i]
j = i - 1

# Move elements of arr[0..i-1], that are
# greater than key, to one position ahead
# of their current position
while j >= 0 and key < arr[j]:
arr[j + 1] = arr[j]
j -= 1
arr[j + 1] = key
Selection sort
def selection_sort(arr):
n = len(arr)
for i in range(n - 1):

# Assume the current position holds
# the minimum element
min_idx = i

# Iterate through the unsorted portion
# to find the actual minimum
for j in range(i + 1, n):
if arr[j] < arr[min_idx]:

# Update min_idx if a smaller element is found
min_idx = j

# Move minimum element to its
# correct position
arr[i], arr[min_idx] = arr[min_idx], arr[i]
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#Q21 #leet_codeQ1249 Medium noscript. Minimum Remove to Make Valid Parentheses
Given a string s of '(' , ')' and lowercase English characters.

Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

It is the empty string, contains only lowercase characters, or
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string.


Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"
Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.


Constraints:

1 <= s.length <= 105
s[i] is either '(' , ')', or lowercase English letter.
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Forwarded from A2SV - Community
Hello fellow coders,

After a brief break, we've returned! Get ready for our fun weekly community coding contest happening every Saturday 🗓. It's your chance to compete with other awesome programmers, learn more about coding, and, most importantly, have a great time! 🥳

🏆 Contest Details:
📅 Date: November 2, 2024
Time: 6:00 PM
🔗 Link for Contest: Community Weekly Contest Return - Contest No 1

P.S. Remember to fill out this form to secure your spot for our upcoming community classes and lectures. It’s your golden ticket to enriching experiences you won’t want to miss. 📚
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Forwarded from LeetCode VIP
🏆 LeetCode 490. The Maze (Premium) 🏆

There's a ball in a maze with:
- Empty spaces (0) and walls (1).

The ball can roll in any direction:
- Up, down, left, or right 🌐

But here's the catch:
- It won't stop until it hits a wall! 🚧
- Once it stops, it can choose another direction to roll.

### Task 🎯
Given:
- An m x n maze
- The ball's starting position and a destination

Determine:
- Can the ball stop at the destination?
- If yes, return true. Otherwise, return false.

📝 Assumption: All borders of the maze are walls.

---

### Examples 🔍

Example 1:

Input: 
maze = [
[0,0,1,0,0],
[0,0,0,0,0],
[0,0,0,1,0],
[1,1,0,1,1],
[0,0,0,0,0]
],
start = [0,4],
destination = [4,4]

Output: true

Explanation: One possible path: left ➡️ down ➡️ left ➡️ down ➡️ right ➡️ down ➡️ right 🎯

---

Example 2:

Input:
maze = [
[0,0,1,0,0],
[0,0,0,0,0],
[0,0,0,1,0],
[1,1,0,1,1],
[0,0,0,0,0]
],
start = [0,4],
destination = [3,2]

Output: false

Explanation: The ball can pass through the destination, but it cannot stop there.

---

Example 3:

Input:
maze = [
[0,0,0,0,0],
[1,1,0,0,1],
[0,0,0,0,0],
[0,1,0,0,1],
[0,1,0,0,0]
],
start = [4,3],
destination = [0,1]

Output: false


---

### Constraints 📏
- Maze Dimensions: 1 ≤ m, n ≤ 100
- Cells contain only 0 (empty) or 1 (wall).
- Start and destination are in empty spaces and won’t initially overlap.

Can you solve it? 🤔
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Forwarded from LeetCode VIP
Forwarded from LeetCode VIP
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Leetcode with dani
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share ur answer in the comment section , i will check it
Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.

Example:

Input : n =10
Output : 2 3 5 7


Input : n = 20
Output: 2 3 5 7 11 13 17 19
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Sieve of Eratosthenes

Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.

Example:

Input : n =10
Output : 2 3 5 7


Input : n = 20
Output: 2 3 5 7 11 13 17 19

The sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than n when n is smaller than 10 million or so.

Following is the algorithm to find all the prime numbers less than or equal to a given integer n by the Eratosthene’s method:
When the algorithm terminates, all the numbers in the list that are not marked are prime.

Explanation with Example:
Let us take an example when n = 100. So, we need to print all prime numbers smaller than or equal to 100.


We create a list of all numbers from 2 to 100.
According to the algorithm we will mark all the numbers which are divisible by 2 and are greater than or equal to the square of it.
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Now we move to our next unmarked number 3 and mark all the numbers which are multiples of 3 and are greater than or equal to the square of it.
We move to our next unmarked number 5 and mark all multiples of 5 and are greater than or equal to the square of it.
We continue this process, and our final table will look like below: