Leetcode with dani

🚀 Challenge for Aspiring Weightlifters! 🏋️‍♂️ Yaman, an aspiring weightlifter, is gearing up for his next tournament and needs your help! He has n barbell plates, each with a specific weight aᵢ and width bᵢ . Yaman's goal is to distribute these plates into two subsets: one for the left side of the barbell and one for the right. His coach has defined a balancing function f(l, r) as follows: f(l, r) = (| ∑(i ∈ l) bᵢ - ∑(i ∈ r) bᵢ )| Yaman has to ensure that the absolute difference in widths between the two sides falls within a specified range during his training. ▎Your Task: For each test case, determine the maximum total weight Yaman can lift while keeping the balancing function's value within the range [L, R]. If it's impossible to achieve this balance, you should return 0 . ▎Input Format: • The first line contains the number of test cases t ( 1 ≤ t ≤ 10⁵ ). • Each test case starts with two integers n (number of plates) and q (number of tests). • The next line lists n integers representing the weights of the plates. • The following line lists n integers representing the widths of the plates. • The next q lines each contain two integers Lᵢ and Rᵢ , specifying the desired range for the balancing function. ▎Output Format: For each test, output the maximum total weight Yaman can lift while maintaining the balancing function's value within the specified range. If it's not possible, output 0 . ▎Example: Input:

1
5 4
5 2 3 2 5
1 4 9 12 20
1 2
7 7
99 100
3 3

Output:

17
12
0
14

Can you help Yaman achieve his training goals? Share your solutions and strategies! 💪 🔗 GotoCodeForce

Not an Ad!

For those who didn't see the video, the speaker challenges the old idea that just knowing how to code will guarantee you a good job. With AI becoming more popular, the tech world is changing fast. Big companies like Google, Salesforce, Meta, and Amazon are now using advanced AI tools to do tasks that used to be done by human coders, and they can do them just as well as mid-level engineers. This shift is not only changing how companies hire but also leading to job cuts in the industry. But the speaker argues that coding isn't going away. Instead, we need to think differently: we should combine our coding skills with new AI tools. He shares a simple five-step plan for developers who want to succeed in this new environment. This plan focuses on having a strong coding foundation, using AI to work more efficiently, understanding how AI works, applying these skills to real projects, and always learning as the field changes. GPT's opinion : I believe that while AI can handle many routine tasks, the creative and problem-solving parts of coding are still really important. In my opinion, coding makes up about 80% of the development process, with AI helping out with the other 20%—making things easier and taking care of repetitive work. Embracing the combination of human creativity and AI technology is essential for success in the tech world moving forward.

post questions for all of you, so please choose a question based on your experience. Make sure to understand the sliding window and prefix sum technique before attempting to solve this question,

nums = [1, 1, 2, 1, 1]
k = 3
solution = Solution()
print(solution.numberOfSubarrays(nums, k))  # Output: Number of nice subarrays

💡 Test Yourself! Feel free to ask for any further modifications or explanations!

🌟 Day 9 Challenge: Count Number of Nice Subarrays 🌟 🔗 Problem Link: Count Number of Nice Subarrays 📝 Problem Statement: Given an array of integers nums and an integer k, a continuous subarray is called nice if it contains exactly k odd numbers. Return the number of nice subarrays. ▎Two Approaches to Solve the Problem ▎Approach 1: Sliding Window Technique This approach uses a sliding window to count the number of subarrays with exactly k odd numbers.

from typing import List

class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        def atMostK(k):
            count = 0
            left = 0
            for right in range(len(nums)):
                if nums[right] % 2 == 1:
                    k -= 1
                while k < 0:
                    if nums[left] % 2 == 1:
                        k += 1
                    left += 1
                count += right - left + 1
            return count
        
        return atMostK(k) - atMostK(k - 1)

Explanation: 1. Function atMostK(k): This helper function counts the number of subarrays that contain at most k odd numbers. It uses a sliding window where left and right pointers define the current window. 2. Counting Subarrays: For each position of right, if the number is odd, we decrement k. If k becomes negative, we increment left until we have at most k odd numbers again. The number of valid subarrays ending at right is added to the count. 3. Final Calculation: The main function returns the difference between the counts of subarrays with at most k and k-1 odd numbers, which gives us exactly k. ▎Approach 2: Using Indices of Odd Numbers This approach tracks the indices of odd numbers directly and counts the valid subarrays based on these indices.

class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        total = 0
        count_odd = 0
        l = 0
        myset = []
        
        for i in range(len(nums)):
            if nums[i] % 2:  # Check if the number is odd
                myset.append(i)  # Store the index of the odd number
                count_odd += 1
            
            if count_odd > k:
                # If we have more than k odd numbers, we need to move the left pointer
                if l:
                    l1 = (myset[l] - (myset[l - 1]))  # Distance between two odd indices
                    r = i - myset[-2]  # Distance from the second last odd index to current
                    total += l1 * r  # Count subarrays formed
                else:
                    r = i - myset[-2]
                    total += (myset[l] + 1) * r  # Count subarrays formed when l is 0
                
                l += 1  # Move the left pointer
                count_odd -= 1  # Decrease the count of odd numbers
            
        # Check for the last segment if count_odd >= k
        if count_odd >= k:
            if l:
                l1 = (myset[l] - (myset[l - 1]))
                r = (len(nums) - myset[-1])
                total += l1 * r
            else:
                r = (len(nums) - myset[-1])
                total += (myset[l] + 1) * r
        
        return total

Explanation: 1. Initialization: Similar to the first approach, we initialize variables to track counts and indices. 2. Iterate through the array: For each element, check if it’s odd and update our list of indices (myset) and counts. 3. Adjusting for excess odds: When the count exceeds k, adjust the left pointer to maintain exactly k odds and calculate valid subarrays based on distances between stored indices. 4. Final Count: After processing, check for any remaining valid segments to add to the total. ChatGPT 4 | Midjourney | Claude | Suno, [13/02/2025 17:30] ▎Complexity Analysis for Both Approaches: • Time Complexity: O(n), where n is the length of the input array. Both approaches iterate through the array a single time. • Space Complexity: O(n) in the worst case for storing indices in Approach 2. ▎Example Usage:

Sure! Here’s the formatted response for the "1588. Sum of All Odd Length Subarrays" problem: --- 🌟 Day 6 Challenge: Sum of All Odd Length Subarrays 🌟 🔗 Problem Link: Sum of All Odd Length Subarrays 📝 Problem Statement: Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr. A subarray is a contiguous subsequence of the array. ✨ Examples: • Input: arr = [1, 4, 2, 5, 3] → Output: 58 • Input: arr = [1, 2] → Output: 3 🔍 Approach: Counting Contributions Using Prefix Sums 1. Count Subarrays: For each element in the array, determine how many odd-length subarrays include that element. 2. Calculate Ranges: For an element at index i, calculate how many ways it can be included in odd-length subarrays by considering both left and right ranges. 3. Use Remainders: Count the occurrences of each remainder when dividing the prefix sums by 2 to identify odd-length contributions. 💻 Solution Code 1 :

class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        left = 0
        p = [0 for i in range(len(arr)+1)]
        p[1] = arr[0]
        for i in range(1,len(arr)):
            p[i+1] = p[i] +arr[i]
        total = 0
        for i in range(1,len(p)):
            l = 0
            while l<i:
                if (i-l)%2:
                    total +=  p[i]-p[l]
                l+=1
        return total

💻 Solution Code 2 optimal:

class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        total = 0
        n = len(arr)
        for i, val in enumerate(arr):
            left = i + 1          # ways to choose start position
            right = n - i         # ways to choose end position
            # Count of odd-length subarrays that include arr[i]
            odd_count = ((left + 1) // 2) * ((right + 1) // 2) + (left // 2) * (right // 2)
            total += val * odd_count
        return total

▎Example usage:

solution = Solution()
print(solution.sumOddLengthSubarrays([1, 4, 2, 5, 3]))  # Output: 58
print(solution.sumOddLengthSubarrays([1, 2]))            # Output: 3

📖 Explanation with Example: 1. For each index i, calculate how many odd-length subarrays can be formed that include arr[i]. 2. Use the counts of ways to select starting and ending positions to determine how many of those subarrays are odd-length. 3. Multiply the value of each element by its count of appearances in odd-length subarrays and sum these contributions for the final result. ⏳ Complexity Analysis: • Time Complexity: O(n) (single pass through the array). • Space Complexity: O(1) (no extra space used beyond a few variables). 💡 Test Yourself! Can you implement this solution efficiently? Dive in and share your approach with us! Let's see your coding skills shine! 🌟

🌟 Day 8 Challenge: Sum of All Odd Length Subarrays 🌟 🔗 Problem Link: Sum of All Odd Length Subarrays 📝 Problem Statement: Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr. A subarray is a contiguous subsequence of the array. ✨ Examples: • Input: arr = [1, 4, 2, 5, 3] → Output: 58 • Input: arr = [1, 2] → Output: 3 🔍 Approach: Counting Contributions Using Prefix Sums 1. Count Subarrays: For each element in the array, determine how many odd-length subarrays include that element. 2. Calculate Ranges: For an element at index i, calculate how many ways it can be included in odd-length subarrays by considering both left and right ranges. 3. Use Remainders: Count the occurrences of each remainder when dividing the prefix sums by 2 to identify odd-length contributions. code: 💻 Solution Code:

from typing import List

class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        total = 0
        n = len(arr)
        for i, val in enumerate(arr):
            left = i + 1          # ways to choose start position
            right = n - i         # ways to choose end position
            # Count of odd-length subarrays that include arr[i]
            odd_count = ((left + 1) // 2) * ((right + 1) // 2) + (left // 2) * (right // 2)
            total += val * odd_count
        return total

▎Example usage:

solution = Solution()
print(solution.sumOddLengthSubarrays([1, 4, 2, 5, 3]))  # Output: 58
print(solution.sumOddLengthSubarrays([1, 2]))            # Output: 3

📖 Explanation with Example: 1. For each index i, calculate how many odd-length subarrays can be formed that include arr[i]. 2. Use the counts of ways to select starting and ending positions to determine how many of those subarrays are odd-length. 3. Multiply the value of each element by its count of appearances in odd-length subarrays and sum these contributions for the final result. ⏳ Complexity Analysis: • Time Complexity: O(n) (single pass through the array). • Space Complexity: O(1) (no extra space used beyond a few variables).

Enjoy our content? Advertise on this channel and reach a highly engaged audience! 👉🏻 It's easy with Telega.io. As the leading platform for native ads and integrations on Telegram, it provides user-friendly and efficient tools for quick and automated ad launches. ⚡️ Place your ad here in three simple steps: 1 Sign up 2 Top up the balance in a convenient way 3 Create your advertising post If your ad aligns with our content, we’ll gladly publish it. Start your promotion journey now!

Enjoy our content? Advertise on this channel and reach a highly engaged audience! 👉🏻 It's easy with Telega.io. As the leading platform for native ads and integrations on Telegram, it provides user-friendly and efficient tools for quick and automated ad launches. ⚡️ Place your ad here in three simple steps: 1 Sign up 2 Top up the balance in a convenient way 3 Create your advertising post If your ad aligns with our content, we’ll gladly publish it. Start your promotion journey now!

--- 🌟 Day 7 Challenge: 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:

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).

🔍 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: 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 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

🌟 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

🌟 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