Minimum Operations to Reduce Array Sum to Zero with Sliding Window

Many subarray problems follow the same hidden pattern — once you recognize it, they become much easier. 🚀 Day 102/365 — DSA Challenge Solved: Count Number of Nice Subarrays Problem idea: We need to count the number of subarrays that contain exactly k odd numbers. Efficient approach: Use the same trick as previous problem: subarrays with exactly k odds = subarrays with ≤ k odds − subarrays with ≤ (k − 1) odds Steps: 1. Use a sliding window to count subarrays with at most k odd numbers 2. Expand window by moving right pointer 3. If odd count exceeds k, shrink window from left 4. Add valid subarrays ending at each index 5. Subtract results to get exact count This pattern works beautifully for binary-like conditions. ⏱ Time: O(n) 📦 Space: O(1) Day 102/365 complete. 💻 263 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #LeetCode #LearningInPublic

When a problem asks for maintaining a condition on both minimum and maximum, monotonic data structures become very powerful. 🚀 Day 110/365 — DSA Challenge Solved: Longest Continuous Subarray With Absolute Diff ≤ Limit Problem idea: We need to find the longest subarray such that the difference between max and min elements is ≤ limit. Efficient approach: Use Sliding Window + Two Monotonic Deques. Steps: 1. Use one deque to maintain maximum elements (decreasing order) 2. Use another deque to maintain minimum elements (increasing order) 3. Expand window by moving the right pointer 4. If (max − min) exceeds limit, shrink window from the left 5. Track the maximum valid window size This ensures we can get max and min in O(1) time. ⏱ Time: O(n) 📦 Space: O(n) Day 110/365 complete. 💻 255 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #Deque #LeetCode #LearningInPublic

Some problems break the standard sliding window pattern — especially when negative numbers are involved. 🚀 Day 104/365 — DSA Challenge Solved: Shortest Subarray with Sum at Least K Problem idea: We need to find the length of the shortest subarray whose sum is at least k. The challenge is that the array can contain negative numbers, so normal sliding window won't work. Efficient approach: Use Prefix Sum + Monotonic Deque. Steps: 1. Build a prefix sum array 2. Use a deque to store indices of useful prefix sums 3. While current sum − smallest prefix ≥ k → update answer 4. Maintain increasing order in deque by removing larger prefix sums from the back 5. This ensures we always get the shortest valid subarray This approach efficiently handles negative numbers while keeping optimal time complexity. ⏱ Time: O(n) 📦 Space: O(n) Day 104/365 complete. 💻 261 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #Deque #PrefixSum #LeetCode #LearningInPublic

Not all sliding window problems are about sums — sometimes the constraint is on the product, but the pattern still applies. 🚀 Day 103/365 — DSA Challenge Solved: Subarray Product Less Than K Problem idea: We need to count the number of subarrays where the product of all elements is strictly less than k. Efficient approach: Use a sliding window that maintains a valid product. Steps: 1. Expand the window by multiplying the current element 2. If product becomes ≥ k, shrink the window from the left 3. Keep dividing until product < k 4. Count all valid subarrays ending at each index This works because all numbers are positive, so the window can be adjusted greedily. ⏱ Time: O(n) 📦 Space: O(1) Day 103/365 complete. 💻 262 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #LeetCode #LearningInPublic

🖤 Day 44 of Solving DSA : Today, I implemented a solution to generate Pascal’s Triangle, a classic problem that helps strengthen problem-solving and array manipulation skills. 🔺 What is Pascal’s Triangle? It is a triangular array where: The first and last elements of each row are always 1 Every middle element is the sum of two elements from the previous row Example: [1] [1, 1] [1, 2, 1] [1, 3, 3, 1] 💡 Approach I Used: 1️⃣ Initialize Result List Create a list to store all rows of the triangle. 2️⃣ Iterate for Each Row Loop from 0 to numRows - 1 to build each row. 3️⃣ Set First and Last Elements Each row starts and ends with 1. 4️⃣ Calculate Middle Elements Use values from the previous row: 👉 current[j] = prev[j-1] + prev[j] 5️⃣ Store Each Row Add the constructed row to the result list. #DSA #java #100dayschallenge #leetcode #consistency

Some subarray counting problems become much easier when we transform them into “at most” problems. 🚀 Day 101/365 — DSA Challenge Solved: Binary Subarrays With Sum Problem idea: We need to count the number of subarrays whose sum equals a given goal in a binary array. Efficient approach: Instead of directly counting subarrays with exact sum, we use a trick: subarrays with sum = goal = subarrays with sum ≤ goal − subarrays with sum ≤ (goal − 1) Steps: 1. Create a helper function to count subarrays with sum at most k 2. Use a sliding window to maintain a valid window where sum ≤ k 3. Add the number of valid subarrays ending at each index 4. Subtract results to get the exact count for the goal This converts the problem into an efficient sliding window solution. ⏱ Time: O(n) 📦 Space: O(1) Day 101/365 complete. 💻 264 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #LeetCode #LearningInPublic

Many array problems that involve maximizing a window can be solved using the sliding window technique. 🚀 Day 98/365 — DSA Challenge Solved: Max Consecutive Ones III Problem idea: We need to find the maximum number of consecutive 1's in a binary array if we are allowed to flip at most k zeros. Efficient approach: Use a sliding window that allows at most k zeros inside the window. Steps: 1. Expand the window by moving the right pointer 2. Count how many zeros are inside the window 3. If zero count becomes greater than k, move the left pointer until zeros ≤ k 4. Track the maximum window length This keeps the window valid while maximizing the number of 1's. ⏱ Time: O(n) 📦 Space: O(1) Day 98/365 complete. 💻 267 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #LeetCode #LearningInPublic

Some problems look difficult at first, but once you know the right data structure, they become much easier. 🚀 Day 99/365 — DSA Challenge Solved: Sliding Window Maximum Problem idea: We are given an array and a window of size k that moves from left to right. For each window position, we need to find the maximum element inside that window. Efficient approach: Use a Deque (Double Ended Queue) to keep track of useful elements for the current window. Steps: 1. Store indices in the deque 2. Remove indices that are outside the current window 3. Remove elements smaller than the current element from the back 4. The front of the deque always stores the index of the maximum element 5. Record the maximum once the first window is formed This ensures we always know the maximum of the current window efficiently. ⏱ Time: O(n) 📦 Space: O(k) Day 99/365 complete. 💻 266 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #Deque #LeetCode #LearningInPublic

Linked list problems often test how well you can manipulate pointers without losing track. 🚀 Day 114/365 — DSA Challenge Solved: Swap Nodes in Pairs Problem idea: We need to swap every two adjacent nodes in a linked list without changing values, only pointers. Efficient approach: Use a dummy node and carefully adjust pointers in pairs. Steps: 1. Use a dummy node pointing to head 2. Maintain a pointer prev before the current pair 3. Identify two nodes: first and second 4. Swap them by updating pointers 5. Move prev forward to the next pair This ensures all pairs are swapped correctly. ⏱ Time: O(n) 📦 Space: O(1) Day 114/365 complete. 💻 251 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #LinkedList #LeetCode #LearningInPublic

Day 46/75 — Frequency of the Most Frequent Element Today’s problem was about maximizing the frequency of an element using at most k increments. Approach: • Sort the array • Use sliding window • Maintain sum of current window • Expand right pointer • Shrink window when operations exceed k Key logic: while ((long) nums[right] * (right - left + 1) - total > k) {  total -= nums[left];  left++; } maxFreq = Math.max(maxFreq, right - left + 1); Time Complexity: O(n log n) (sorting) Space Complexity: O(1) Key Insight: Make all elements in window equal to the largest element — check if cost ≤ k. 46/75 🚀 #Day46 #DSA #SlidingWindow #TwoPointers #Java #Algorithms #LeetCode