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Leetcode with dani

Leetcode with dani

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Join us and let's tackle leet code questions together: improve your problem-solving skills Preparing for coding interviews learning new algorithms and data structures connect with other coding enthusiasts

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🌟 Day 7 Challenge 2: Minimum Size Subarray Sum 🌟 🔗 Problem Link: Minimum Size Subarray Sum 📝 Problem Statement: Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead. ✨ Examples: • Input: target = 7, nums = [2,3,1,2,4,3] → Output: 2 Explanation: The subarray [4,3] has the minimal length. • Input: target = 4, nums = [1,4,4] → Output: 1 • Input: target = 11, nums = [1,1,1,1,1,1,1,1] → Output: 0 🔍 Approach 1: O(n) Sliding Window This method uses two pointers to form a window and expands or shrinks it based on the current sum: 1. Initialize: Set two pointers (left, right), a running sum (curr_sum), and a variable (min_len) initialized to infinity. 2. Expand: Move the right pointer to add elements to curr_sum. 3. Shrink: Once curr_sum meets or exceeds the target, update min_len and shrink the window from the left until the sum is less than the target. 4. Return: If a valid subarray is found, return its length; otherwise, return 0. 💻 Solution Code:
def minSubArrayLen(target, nums):
    n = len(nums)
    left = 0
    curr_sum = 0
    min_len = float('inf')
    
    for right in range(n):
        curr_sum += nums[right]
        while curr_sum >= target:
            min_len = min(min_len, right - left + 1)
            curr_sum -= nums[left]
            left += 1
    
    return min_len if min_len != float('inf') else 0

# Testing the sliding window approach:
print(minSubArrayLen(7, [2, 3, 1, 2, 4, 3]))  # Expected Output: 2
print(minSubArrayLen(4, [1, 4, 4]))           # Expected Output: 1
print(minSubArrayLen(11, [1, 1, 1, 1, 1, 1, 1, 1]))  # Expected Output: 0
🔍 Approach 2: O(n log n) Prefix Sum with Binary Search This method involves creating a prefix sum array and using binary search to find the smallest index where the difference in prefix sums is at least the target. 1. Prefix Sum Array: Create an array such that prefix[i+1] = prefix[i] + nums[i] with prefix[0] = 0. 2. Binary Search: For each index i, compute the required bound (prefix[i] + target) and use binary search to find the smallest index j where prefix[j] is at least that bound. 3. Update Answer: Calculate the subarray length as j - i and update the minimal length. 4. Return: Return the minimum length found, or 0 if no valid subarray exists. 💻 Solution Code:
import bisect

def minSubArrayLenBinarySearch(target, nums):
    n = len(nums)
    # Build prefix sum array
    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] + nums[i]
    
    min_len = float('inf')
    # Iterate over prefix sum array
    for i in range(n):
        bound = prefix[i] + target
        # Find the minimal j such that prefix[j] >= bound
        j = bisect.bisect_left(prefix, bound)
        if j <= n:
            min_len = min(min_len, j - i)
    
    return min_len if min_len != float('inf') else 0

# Testing the O(n log n) approach:
print(minSubArrayLenBinarySearch(7, [2, 3, 1, 2, 4, 3]))  # Expected Output: 2
print(minSubArrayLenBinarySearch(4, [1, 4, 4]))           # Expected Output: 1
print(minSubArrayLenBinarySearch(11, [1, 1, 1, 1, 1, 1, 1, 1]))  # Expected Output: 0
📖 Explanation with Example: • In the Sliding Window approach, we continuously expand our window by moving the right pointer and check if we can shrink it from the left side once we reach or exceed the target sum. • In the Prefix Sum with Binary Search approach, we precompute cumulative sums and use binary search to efficiently find valid subarrays. ⏳ Complexity Analysis: • Sliding Window: • Time Complexity: O(n) (single pass through the array). • Space Complexity: O(1) (only a few variables used). • Prefix Sum with Binary Search: • Time Complexity: O(n log n) (due to binary search).

🌟 Day 6 Challenge: Subarray Sums Divisible by K 🌟 🔗 Problem Link: Subarray Sums Divisible by K 📝 Problem Statement: Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k. A subarray is a contiguous part of an array. ✨ Examples: • Input: nums = [4, 5, 0, -2, -3, 1], k = 5 → Output: 7 • Input: nums = [5], k = 9 → Output: 0 🔍 Approach: Prefix Sum with Remainder Counting 1. Use a prefix sum to keep track of the cumulative sum of elements as you iterate through the array. 2. Calculate the remainder of the prefix sum when divided by k. If two prefix sums have the same remainder, the subarray between those indices has a sum divisible by k. 3. Utilize a hash map (or dictionary) to count occurrences of each remainder, allowing you to efficiently determine how many valid subarrays end at each position. 💻 Solution Code:
from collections import defaultdict

class Solution:
    def subarraysDivByK(self, nums: List[int], k: int) -> int:
        prefix_counts = defaultdict(int)
        prefix_counts[0] = 1  # Initialize with prefix sum 0 having remainder 0
        current_sum = 0
        result = 0
        
        for num in nums:
            current_sum += num
            mod = current_sum % k
            result += prefix_counts[mod]
            prefix_counts[mod] += 1
        
        return result
Example usage: solution = Solution() print(solution.subarraysDivByK([4, 5, 0, -2, -3, 1], 5)) # Output: 7 print(solution.subarraysDivByK([5], 9)) # Output: 0 📖 Explanation with Example: 1. As you iterate through the array, maintain a running total (current_sum) and calculate its remainder when divided by k. 2. For each element, check how many times this remainder has been seen before using the hash map. Each occurrence corresponds to a valid subarray. 3. Update the hash map with the current remainder count for future subarray calculations. ⏳ Complexity Analysis: • Time Complexity: O(n) (single pass through the array). • Space Complexity: O(k) (to store counts of remainders). 💡 Test Yourself! Can you solve this problem efficiently? Dive in and share your solution with us! Let's see your coding skills shine! 🌟

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Sure! Here’s the formatted response for the Minimum Average Difference problem, similar to your example: --- 🌟 Day 6 Challenge: Q 3 Minimum Average Difference 🌟 Medium 🔗 Problem Link: Minimum Average Difference 📝 Problem Statement: You are given a non-negative integer array nums. Your task is to find the index i such that the absolute difference between the average of the first i + 1 elements and the average of the last n - i - 1 elements is minimized. If there are multiple indices with the same minimum difference, return the smallest index. ✨ Examples: • Input: nums = [2, 5, 3, 9, 5, 3] → Output: 3 • Input: nums = [0] → Output: 0 • Input: nums = [1, 2, 3, 4, 5, 6] → Output: 2 🔍 Approach: Prefix Sum Calculation 1. Calculate the total sum of the array to help compute right averages. 2. Iterate through each index, maintaining a running sum (left sum) of elements from the start. 3. For each index, compute the left average and right average based on the sums. 4. Calculate the absolute difference between these two averages and track the minimum difference and corresponding index. 💻 Solution Code:
def minimumAverageDifference(nums):
    n = len(nums)
    total_sum = sum(nums)  # Calculate the total sum of the array
    left_sum = 0           # Initialize left sum
    min_diff = float('inf')  # Initialize minimum difference to infinity
    min_index = -1         # Initialize minimum index

    for i in range(n):
        left_sum += nums[i]  # Update left sum with the current element
        
        # Calculate left average
        left_avg = left_sum // (i + 1)
        
        # Calculate right average
        if i < n - 1:
            right_avg = (total_sum - left_sum) // (n - i - 1)
        else:
            right_avg = 0  # If it's the last element, right average is 0
        
        # Calculate absolute difference
        diff = abs(left_avg - right_avg)

        # Update minimum difference and index if a new minimum is found
        if diff < min_diff:
            min_diff = diff
            min_index = i

    return min_index

# Example usage:
print(minimumAverageDifference([2, 5, 3, 9, 5, 3]))  # Output: 3
print(minimumAverageDifference([0]))                  # Output: 0
📖 Explanation with Example (1994): 1. For each index, keep track of the left sum and calculate the left average. 2. Compute the right average using the total sum minus the left sum. 3. Calculate the absolute difference between left and right averages. 4. Track the smallest difference and its corresponding index. ⏳ Complexity Analysis: • Time Complexity: O(n) (single pass through the array). • Space Complexity: O(1) (only a few variables used for calculation). 💡 Test Yourself! Can you solve this problem efficiently? Dive in and share your solution with us! Let's see your coding skills shine! 🌟

I post questions for all of you, so please choose a question based on your experience. Make sure to understand the sliding window technique before attempting to solve this question, as it is a difficult one. This question is intended for those who understand the sliding window concept and have practiced previously

Day 6 question 2 ▎LeetCode Question: Minimum Window SubstringProblem Statement: Given two strings s and t, find the minimum window in s which will contain all the characters in t. If there is no such window in s that covers all characters in t, return the empty string "". ▎Example:Input: s = "ADOBECODEBANC", t = "ABC"Output: "BANC"Constraints:1 <= s.length, t.length <= 10^5s and t consist of uppercase and lowercase English letters. ▎Solution Explanation: 1. Initialization: Create dictionaries to store the frequency of characters in string t and the current window. Use variables to track the number of unique characters required and formed in the window. 2. Sliding Window: Expand the window by moving the right pointer. Update the window_counts and formed variables. 3. Contraction: When the window contains all required characters, contract the window by moving the left pointer. Update the window_counts and formed variables. While contracting, check if the current window is smaller than the minimum window found so far. 4. Result: Return the minimum window substring. If no valid window is found, return an empty string. ▎Python Code: Here’s the solution using the sliding window technique:
def minWindow(s: str, t: str) -> str:
    if not t or not s:
        return ""

    dict_t = {}
    for char in t:
        dict_t[char] = dict_t.get(char, 0) + 1

    required = len(dict_t)
    formed = 0
    window_counts = {}

    left, right = 0, 0
    ans = float('inf'), None, None

    while right < len(s):
        character = s[right]
        window_counts[character] = window_counts.get(character, 0) + 1

        if character in dict_t and window_counts[character] == dict_t[character]:
            formed += 1

        while left <= right and formed == required:
            character = s[left]

            if right - left + 1 < ans[0]:
                ans = (right - left + 1, left, right)

            window_counts[character] -= 1
            if character in dict_t and window_counts[character] < dict_t[character]:
                formed -= 1

            left += 1

        right += 1

    return "" if ans[0] == float('inf') else s[ans[1]: ans[2] + 1]
Complexity Analysis:Time Complexity: O(N + M), where N is the length of string s and M is the length of string t. • Space Complexity: O(M), for storing the frequency of characters in t.

🌟 Day 6 Challenge: Integer to Roman 🌟 🔗 Problem Link: Integer to Roman 📝 Problem Statement: Roman numerals are fascinating representations of numbers, constructed using the following symbols: • I (1), V (5), X (10), L (50), C (100), D (500), M (1000). To form Roman numerals, we combine these symbols from largest to smallest, using subtraction for specific cases like 4 (IV) and 9 (IX). Your task is to convert any integer between 1 and 3999 into its Roman numeral equivalent! ✨ Examples: • Input: num = 3 → Output: "III" • Input: num = 58 → Output: "LVIII" (L = 50, V = 5, III = 3) • Input: num = 1994 → Output: "MCMXCIV" (M = 1000, CM = 900, XC = 90, IV = 4) 🔍 Approach: Greedy Algorithm with Predefined Values 1. Create a list of pairs (value, symbol) sorted in descending order, including subtractive combinations (e.g., 4, 9, 40). 2. Iterate through this list, subtracting the largest possible value from num, and appending the corresponding symbol to your result. 💻 Solution Code:
def intToRoman(num: int) -> str:
    val_sym = [
        (1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'),
        (100, 'C'), (90, 'XC'), (50, 'L'), (40, 'XL'),
        (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I')
    ]
    res = []
    for value, symbol in val_sym:
        while num >= value:
            res.append(symbol)
            num -= value
        if num == 0:
            break
    return ''.join(res)
📖 Explanation with Example (1994): 1. Subtract 1000 → 'M', remaining num = 994. 2. Subtract 900 → 'CM', remaining num = 94. 3. Subtract 90 → 'XC', remaining num = 4. 4. Subtract 4 → 'IV'. Result: "MCMXCIV". ⏳ Complexity Analysis:Time Complexity: O(1) (fixed number of iterations over 13 symbols). • Space Complexity: O(1) (output string has a maximum fixed length). 💡 Test Yourself! Can you solve this problem efficiently?

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{
    'David': 20,
    'Bob': 25,
    'Eve': 28,
    'Alice': 30,
    'Charlie': 35
}

people = {
    'Alice': 30,
    'Bob': 25,
    'Charlie': 35,
    'David': 20,
    'Eve': 28
}
Sorting by Age: You can sort the dictionary using the sorted() function combined with a lambda function. Here’s the code: # Sort the dictionary by age sorted_by_age = dict(sorted(people.items(), key=lambda item: item[1])) # Display the sorted dictionary print(sorted_by_age) { 'David': 20, 'Bob': 25, 'Eve': 28, 'Alice': 30, 'Charlie': 35 }

Hey Coders! 👋 Today, let's dive into a classic challenge: 3Sum! 🎯 ▎Problem Statement: Given an integer array nums, your task is to return all the unique triplets [nums[i], nums[j], nums[k]] such that: •  i ≠ j ≠ k •  nums[i] + nums[j] + nums[k] = 0 ▎Example: Input: nums = [-1, 0, 1, 2, -1, -4]  Output: [[-1, -1, 2], [-1, 0, 1]]  Explanation: • The triplet [-1, 0, 1] sums to zero. • The triplet [-1, -1, 2] also sums to zero. ▎Constraints: •  3 ≤ nums.length ≤ 3000 •  -10⁵ ≤ nums[i] ≤ 10⁵ ▎Approach: 1. Sorting: The array is sorted to make it easier to find the triplets. 2. Fixed Pointer: The outer loop fixes the first element of the triplet. 3. Two-pointer Technique: The inner loop uses two pointers to find the other two elements that sum up to zero with the fixed element. 4. Avoiding Duplicates: The code skips over duplicate elements to ensure that only unique triplets are added to the result. ▎Complexity Analysis:Time Complexity:  O(n²) • Space Complexity:  O(1)  (excluding the space required for the output) ▎Challenge: Can you solve it with a time complexity of  O(n²) ? 🤔 ▎Hint: Think about how you can reduce the problem to a two-sum problem after fixing one element. ▎Solution Code (Python):
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

🌟 Day 4 Coding Challenges: Solutions Explanations 🌟 Welcome to Day 4 of our coding journey! Today, we tackle some exciting problems that challenge our data structure skills. Let’s dive in! --- ▎🏗 Q1: Insert Delete GetRandom O(1)Problem Statement: Design a data structure that supports insert, remove, and getRandom operations in average O(1) time. ▎💡 Solution: To achieve O(1) time complexity for all operations, we can employ a clever combination of a dynamic array and a hash map: 1. Dynamic Array: Stores elements, allowing for O(1) random access. 2. Hash Map: Tracks the indices of elements for O(1) insertion and deletion. 3. Removal Strategy: Swap the target element with the last element to avoid shifting. ▎🔍 Python Implementation:
import random

class RandomizedSet:
    def __init__(self):
        self.array = []  # For O(1) random access
        self.indices = {}  # Maps elements to their indices

    def insert(self, val: int) -> bool:
        if val in self.indices:
            return False  # Element already exists
        self.indices[val] = len(self.array)
        self.array.append(val)
        return True

    def remove(self, val: int) -> bool:
        if val not in self.indices:
            return False  # Element does not exist
        index = self.indices[val]
        last_element = self.array[-1]
        
        # Swap and remove
        self.array[index] = last_element
        self.indices[last_element] = index
        self.array.pop()
        del self.indices[val]
        return True

    def getRandom(self) -> int:
        return random.choice(self.array)
🚀 Example Usage:
randomizedSet = RandomizedSet()
print(randomizedSet.insert(1))  # True
print(randomizedSet.remove(2))  # False
print(randomizedSet.insert(2))  # True
print(randomizedSet.getRandom())  # Either 1 or 2
print(randomizedSet.remove(1))  # True
print(randomizedSet.insert(2))  # False
print(randomizedSet.getRandom())  # 2
--- ▎🏆 Q2: Find Players With Zero or One LossesProblem Statement: Given match outcomes as [[winner, loser]], return two sorted lists: 1. Players with 0 losses. 2. Players with exactly 1 loss. ▎💡 Solution: We can efficiently solve this using: • A hash map to count losses. • A set to track all players. ▎🔍 Python Implementation:
def findWinners(matches):
    loss_counts = {}
    players = set()

    for winner, loser in matches:
        players.add(winner)
        players.add(loser)
        loss_counts[loser] = loss_counts.get(loser, 0) + 1

    zero_losses = sorted([player for player in players if loss_counts.get(player, 0) == 0])
    one_loss = sorted([player for player in players if loss_counts.get(player, 0) == 1])

    return [zero_losses, one_loss]
🚀 Example Usage:
matches = [[1, 3], [2, 1], [4, 2], [5, 2]]
result = findWinners(matches)
print(result)  # Output: [[4, 5], [1, 3]]
--- ▎🔄 Q3: Shuffle StringProblem Statement: Reconstruct a string using an index array. For s = "abc" and indices = [0, 2, 1], return "acb". ▎💡 Solution: We create a result array of the same length as s and place each character at its corresponding index from the indices array. ▎🔍 Python Implementation:
def restoreString(s, indices):
    result = [''] * len(s)
    for i, idx in enumerate(indices):
        result[idx] = s[i]
    return ''.join(result)
🚀 Example Usage:
s = "codeleet"
indices = [4, 5, 6, 7, 0, 2, 1, 3]
result = restoreString(s, indices)
print(result)  # Output: "leetcode"

Certainly! Let's go through each problem and provide solutions with clear explanations. --- ### Day 4 Q1: Insert Delete GetRandom O(1) #### Problem: Design a data structure that supports insert, remove, and getRandom operations in average O(1) time. #### Solution: To achieve O(1) time complexity for all operations, we can use the following approach: 1. Use a dynamic array (list) to store elements for O(1) random access. 2. Use a hash map to track the indices of elements in the array for O(1) insertion and deletion. 3. For removal, swap the target element with the last element in the array to avoid shifting elements. Here is the Python implementation:
import random

class RandomizedSet:
    def __init__(self):
        self.array = []  # Store elements for O(1) random access
        self.indices = {}  # Map elements to their indices in the array

    def insert(self, val: int) -> bool:
        if val in self.indices:
            return False  # Element already exists
        self.indices[val] = len(self.array)
        self.array.append(val)
        return True

    def remove(self, val: int) -> bool:
        if val not in self.indices:
            return False  # Element does not exist
        index = self.indices[val]
        last_element = self.array[-1]
        
        # Swap the target element with the last element
        self.array[index] = last_element
        self.indices[last_element] = index
        
        # Remove the last element from the array and the hash map
        self.array.pop()
        del self.indices[val]
        return True

    def getRandom(self) -> int:
        return random.choice(self.array)
#### Example:
randomizedSet = RandomizedSet()
print(randomizedSet.insert(1))  # True
print(randomizedSet.remove(2))  # False
print(randomizedSet.insert(2))  # True
print(randomizedSet.getRandom())  # Either 1 or 2
print(randomizedSet.remove(1))  # True
print(randomizedSet.insert(2))  # False
print(randomizedSet.getRandom())  # 2
--- ### Day 4 Q2: Find Players With Zero or One Losses #### Problem: Given match outcomes as [winner, loser], return two lists: 1. Players with 0 losses. 2. Players with exactly 1 loss. Both lists should be sorted in ascending order. #### Solution: We can solve this using: 1. A hash map to count the number of losses for each player. 2. A set to track all players (including winners who have 0 losses). 3. Filter players based on their loss counts and sort the results. Here is the Python implementation:
def findWinners(matches):
    loss_counts = {}
    players = set()

    for winner, loser in matches:
        players.add(winner)
        players.add(loser)
        loss_counts[loser] = loss_counts.get(loser, 0) + 1

    zero_losses = sorted([player for player in players if loss_counts.get(player, 0) == 0])
    one_loss = sorted([player for player in players if loss_counts.get(player, 0) == 1])

    return [zero_losses, one_loss]
#### Example:
matches = [[1, 3], [2, 1], [4, 2], [5, 2]]
result = findWinners(matches)
print(result)  # Output: [[4, 5], [1, 3]]
--- ### Day 4 Q3: Shuffle String #### Problem: Reconstruct a string using an index array. For s = "abc" and indices = [0, 2, 1], return "acb". #### Solution: We can create a result array of the same length as s and place each character at its corresponding index from the indices array. Finally, convert the array back to a string. Here is the Python implementation:
def restoreString(s, indices):
    result = [''] * len(s)
    for i, idx in enumerate(indices):
        result[idx] = s[i]
    return ''.join(result)
#### Example:
s = "codeleet"
indices = [4, 5, 6, 7, 0, 2, 1, 3]
result = restoreString(s, indices)
print(result)  # Output: "leetcode"