Python Tips: Two Confusing methods.
strip() and split() Methods
strip(): This method removes any leading or trailing spaces or specific characters from a string.
split(): This method splits a string into a list of substrings based on a specified separator.
To see
strip() and split() Methods
strip(): This method removes any leading or trailing spaces or specific characters from a string.
text = " Hello, World! "
print(text.strip()) # Output: "Hello, World!"
split(): This method splits a string into a list of substrings based on a specified separator.
data = "apple,banana,cherry"
print(data.split(',')) # Output: ['apple', 'banana', 'cherry']
To see
split() in action, check out the [Compare Version Numbers](https://leetcode.com/problems/compare-version-numbers/) problem on LeetCode, where you'll split version strings using a dot (.) as the separator.LeetCode
Compare Version Numbers - LeetCode
Can you solve this real interview question? Compare Version Numbers - Given two version strings, version1 and version2, compare them. A version string consists of revisions separated by dots '.'. The value of the revision is its integer conversion ignoring…
⚡3👍1
1. 3Sum (Hard)
Denoscription: Given an integer array
- Example: For input
Denoscription: Given an integer array
nums, return all unique triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. - Example: For input
nums = [-1,0,1,2,-1,-4], the output should be [[-1,-1,2],[-1,0,1]].2. Minimum Window Substring (Hard)
Denoscription: Given two strings
- Example: For
Denoscription: Given two strings
s and t, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return an empty string.- Example: For
s = "ADOBECODEBANC" and t = "ABC", the minimum window is "BANC".3. Fruit Into Baskets (Easy)
Denoscription: You are visiting a farm and have a basket that can hold at most two types of fruits. You want to maximize the number of fruits you collect. Given an array of integers representing the types of fruits in a row, find the maximum number of fruits you can collect in one basket.
- Example: For input
Denoscription: You are visiting a farm and have a basket that can hold at most two types of fruits. You want to maximize the number of fruits you collect. Given an array of integers representing the types of fruits in a row, find the maximum number of fruits you can collect in one basket.
- Example: For input
[1,2,1], the maximum number of fruits collected is 3.
Leetcode with dani
3. Fruit Into Baskets (Easy) Denoscription: You are visiting a farm and have a basket that can hold at most two types of fruits. You want to maximize the number of fruits you collect. Given an array of integers representing the types of fruits in a row, find…
1. 3Sum (Hard)
Answer: The solution involves sorting the array and using a two-pointer technique to find triplets that sum to zero. The time complexity is O(n^2).
2. Minimum Window Substring (Hard)
Answer: Use a sliding window approach to maintain a count of characters and expand/shrink the window until the minimum substring is found. The time complexity is O(n).
3. Fruit Into Baskets (Easy)
Answer: Utilize a sliding window to track the types of fruits and their counts, ensuring you only have two types at any time. The maximum length of the window gives the answer. The time complexity is O(n).
Answer: The solution involves sorting the array and using a two-pointer technique to find triplets that sum to zero. The time complexity is O(n^2).
2. Minimum Window Substring (Hard)
Answer: Use a sliding window approach to maintain a count of characters and expand/shrink the window until the minimum substring is found. The time complexity is O(n).
3. Fruit Into Baskets (Easy)
Answer: Utilize a sliding window to track the types of fruits and their counts, ensuring you only have two types at any time. The maximum length of the window gives the answer. The time complexity is O(n).
Leetcode with dani
1. 3Sum (Hard) Answer: The solution involves sorting the array and using a two-pointer technique to find triplets that sum to zero. The time complexity is O(n^2). 2. Minimum Window Substring (Hard) Answer: Use a sliding window approach to maintain…
▎1. 3Sum
def threeSum(nums):
nums.sort()
result = []
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left, right = i + 1, len(nums) - 1
while left < right:
total = nums[i] + nums[left] + nums[right]
if total < 0:
left += 1
elif total > 0:
right -= 1
else:
result.append([nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
return result
▎2. Minimum Window Substring
from collections import Counter
def minWindow(s, t):
if not t or not s:
return ""
dict_t = Counter(t)
required = len(dict_t)
l, r = 0, 0
formed = 0
window_counts = {}
ans = float("inf"), None, None
while r < len(s):
character = s[r]
window_counts[character] = window_counts.get(character, 0) + 1
if character in dict_t and window_counts[character] == dict_t[character]:
formed += 1
while l <= r and formed == required:
character = s[l]
if r - l + 1 < ans[0]:
ans = (r - l + 1, l, r)
window_counts[character] -= 1
if character in dict_t and window_counts[character] < dict_t[character]:
formed -= 1
l += 1
r += 1
return "" if ans[0] == float("inf") else s[ans[1]: ans[2] + 1]
▎3. Fruit Into Baskets
def totalFruits(fruits):
left, right = 0, 0
basket = {}
max_fruits = 0
while right < len(fruits):
basket[fruits[right]] = basket.get(fruits[right], 0) + 1
while len(basket) > 2:
basket[fruits[left]] -= 1
if basket[fruits[left]] == 0:
del basket[fruits[left]]
left += 1
max_fruits = max(max_fruits, right - left + 1)
right += 1
return max_fruits
nums.sort()
result = []
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left, right = i + 1, len(nums) - 1
while left < right:
total = nums[i] + nums[left] + nums[right]
if total < 0:
left += 1
elif total > 0:
right -= 1
else:
result.append([nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
return result
▎2. Minimum Window Substring
from collections import Counter
if not t or not s:
return ""
dict_t = Counter(t)
required = len(dict_t)
l, r = 0, 0
formed = 0
window_counts = {}
ans = float("inf"), None, None
while r < len(s):
character = s[r]
window_counts[character] = window_counts.get(character, 0) + 1
if character in dict_t and window_counts[character] == dict_t[character]:
formed += 1
while l <= r and formed == required:
character = s[l]
if r - l + 1 < ans[0]:
ans = (r - l + 1, l, r)
window_counts[character] -= 1
if character in dict_t and window_counts[character] < dict_t[character]:
formed -= 1
l += 1
r += 1
return "" if ans[0] == float("inf") else s[ans[1]: ans[2] + 1]
▎3. Fruit Into Baskets
left, right = 0, 0
basket = {}
max_fruits = 0
while right < len(fruits):
basket[fruits[right]] = basket.get(fruits[right], 0) + 1
while len(basket) > 2:
basket[fruits[left]] -= 1
if basket[fruits[left]] == 0:
del basket[fruits[left]]
left += 1
max_fruits = max(max_fruits, right - left + 1)
right += 1
return max_fruits
👍8
🚀 Explore Palindrome Challenges with the Two-Pointer Technique!🧑💻
Palindromes are a classic problem type that can be efficiently tackled using the two-pointer technique. Whether you're just starting or looking to refine your skills, these questions will help you master this approach. Check out these palindrome problems:
1.Valid Palindrome
🔗 [Link](https://leetcode.com/problems/valid-palindrome/)
Denoscription: Determine if a string is a palindrome, considering only alphanumeric characters and ignoring cases.
2. Valid Palindrome II
🔗 [Link](https://leetcode.com/problems/valid-palindrome-ii/)
Denoscription: Can the string become a palindrome by deleting just one character?
3. Palindrome Linked List
🔗 [Link](https://leetcode.com/problems/palindrome-linked-list/)
Denoscription: Check if a singly linked list is a palindrome using two pointers and a bit of list manipulation.
4. Longest Palindromic Substring
🔗 [Link](https://leetcode.com/problems/longest-palindromic-substring/)
Denoscription: Find the longest palindromic substring in a given string, with pointers expanding from each character.
5. Palindromic Substrings
🔗 [Link](https://leetcode.com/problems/palindromic-substrings/)
Denoscription: Count how many palindromic substrings are present in the string using two pointers.
6. Shortest Palindrome.
🔗 [Link](https://leetcode.com/problems/shortest-palindrome/)
Denoscription: Find the shortest palindrome by adding characters to the start of the string.
7. Palindrome Partitioning
🔗 [Link](https://leetcode.com/problems/palindrome-partitioning/)
Denoscription: Partition a string into all possible palindromic substrings.
8. Palindrome Pairs
🔗 [Link](https://leetcode.com/problems/palindrome-pairs/)
Denoscription: Given a list of words, find all pairs of distinct indices that form palindromes.
Ready to boost your problem-solving skills? Dive into these problems, practice your two-pointer technique, and ace those palindrome challenges! 💥
Happy Coding! 💻💡
Palindromes are a classic problem type that can be efficiently tackled using the two-pointer technique. Whether you're just starting or looking to refine your skills, these questions will help you master this approach. Check out these palindrome problems:
1.Valid Palindrome
🔗 [Link](https://leetcode.com/problems/valid-palindrome/)
Denoscription: Determine if a string is a palindrome, considering only alphanumeric characters and ignoring cases.
2. Valid Palindrome II
🔗 [Link](https://leetcode.com/problems/valid-palindrome-ii/)
Denoscription: Can the string become a palindrome by deleting just one character?
3. Palindrome Linked List
🔗 [Link](https://leetcode.com/problems/palindrome-linked-list/)
Denoscription: Check if a singly linked list is a palindrome using two pointers and a bit of list manipulation.
4. Longest Palindromic Substring
🔗 [Link](https://leetcode.com/problems/longest-palindromic-substring/)
Denoscription: Find the longest palindromic substring in a given string, with pointers expanding from each character.
5. Palindromic Substrings
🔗 [Link](https://leetcode.com/problems/palindromic-substrings/)
Denoscription: Count how many palindromic substrings are present in the string using two pointers.
6. Shortest Palindrome.
🔗 [Link](https://leetcode.com/problems/shortest-palindrome/)
Denoscription: Find the shortest palindrome by adding characters to the start of the string.
7. Palindrome Partitioning
🔗 [Link](https://leetcode.com/problems/palindrome-partitioning/)
Denoscription: Partition a string into all possible palindromic substrings.
8. Palindrome Pairs
🔗 [Link](https://leetcode.com/problems/palindrome-pairs/)
Denoscription: Given a list of words, find all pairs of distinct indices that form palindromes.
Ready to boost your problem-solving skills? Dive into these problems, practice your two-pointer technique, and ace those palindrome challenges! 💥
Happy Coding! 💻💡
LeetCode
Valid Palindrome - LeetCode
Can you solve this real interview question? Valid Palindrome - A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters…
👍4
#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.
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
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.
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
👍4
In my opinion, if you're preparing for an interview and have no enough time left, it’s best not to spend more than 45 minutes on a single problem. Instead, I believe you should focus on understanding different patterns and concepts. This approach can help you become more versatile and better equipped to tackle various questions during the interview. what do you think?
👍6❤1
✨ Master Python Division:
Ever wonder about the difference between
- Performs regular division first (
- Then, truncates the result to an integer using
- Example:
-
-
- Directly performs floor division, rounding down to the nearest integer.
- Example:
-
-
Key Differences:
- Rounding:
-
-
- Negative numbers:
- For positive results, both are the same.
- For negative results,
Example with Negative Numbers:
-
-
int(x / y) vs. x // y ✨Ever wonder about the difference between
int(x / y) and x // y in Python? Let’s break it down! 🧠👇int(x / y):- Performs regular division first (
x / y), giving a floating-point result.- Then, truncates the result to an integer using
int(), removing the decimal part.- Example:
-
int(7 / 3) ➡️ 2 (since 7 / 3 ≈ 2.3333, truncates to 2).-
int(-7 / 3) ➡️ -2 (since -7 / 3 ≈ -2.3333, truncates to -2).x // y (Floor Division):- Directly performs floor division, rounding down to the nearest integer.
- Example:
-
7 // 3 ➡️ 2 (largest integer ≤ 2.3333).-
-7 // 3 ➡️ -3 (largest integer ≤ -2.3333).Key Differences:
- Rounding:
-
int(x / y) truncates towards zero. -
x // y rounds down towards negative infinity.- Negative numbers:
- For positive results, both are the same.
- For negative results,
int(x / y) truncates, but x // y rounds down.Example with Negative Numbers:
-
int(-7 / 3) ➡️ -2 -
-7 // 3 ➡️ -3✍4
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.…
YouTube
3Sum - Leetcode 15 - Python
🚀 https://neetcode.io/ - A better way to prepare for Coding Interviews
🧑💼 LinkedIn: https://www.linkedin.com/in/navdeep-singh-3aaa14161/
🥷 Discord: https://discord.gg/ddjKRXPqtk
🐦 Twitter: https://twitter.com/neetcode1
🐮 Support the channel: https://w…
🧑💼 LinkedIn: https://www.linkedin.com/in/navdeep-singh-3aaa14161/
🥷 Discord: https://discord.gg/ddjKRXPqtk
🐦 Twitter: https://twitter.com/neetcode1
🐮 Support the channel: https://w…
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
#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
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
👍3
this question is asked by amazon in interviews frequently
https://leetcode.com/problems/the-kth-factor-of-n/?envType=study-plan-v2&envId=amazon-spring-23-high-frequency
https://leetcode.com/problems/the-kth-factor-of-n/?envType=study-plan-v2&envId=amazon-spring-23-high-frequency
LeetCode
The kth Factor of n - LeetCode
Can you solve this real interview question? The kth Factor of n - You are given two positive integers n and k. A factor of an integer n is defined as an integer i where n % i == 0.
Consider a list of all factors of n sorted in ascending order, return the…
Consider a list of all factors of n sorted in ascending order, return the…
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]
👍3