| # | Topic | Problem | File Path | LeetCode Link | Solved | Crushed | Date Solved | Tags |
|---|---|---|---|---|---|---|---|---|
| 001 | Arrays | Longest Subarray with Sum K | arrays/_001_Longest_Subarray_With_Sum_K.java |
🔗 | ✅ | ❌ | 2025-06-23 | |
| 002 | Arrays | Max Subarray Sum | arrays/_002_Max_Subarray_Sum.java |
🔗 | ✅ | ❌ | 2025-06-24 | |
| 003 | Arrays | Best Time to Buy and Sell Stock | arrays/_003_Best_Time_to_Buy_and_Sell_Stock.java |
🔗 | ✅ | ✅ | 2025-06-30 | |
| 004 | Arrays | Container With Most Water | arrays/_004_Container_With_Most_Water.java |
🔗 | ✅ | ✅ | 2025-07-01 | Two Pointer |
| 005 | Arrays | Rearrange Array Elements by Sign | arrays/_005_Rearrange_Array_Elements_by_Sign.java |
🔗 | ✅ | ✅ | 2025-07-01 | |
| 006 | Arrays | Majority Element | arrays/_006_Majority_Element.java |
🔗 | ✅ | ❌ | 2025-07-02 |
🧠 Intuition for #001: Longest Subarray with Sum K
Brute Force:
Generate all possible subarrays and check if their sum equals k.
Better (Prefix Sum + HashMap):
Maintain a running prefix sum and store the earliest index where each sum occurs.
At each index, calculate rem = currentSum - k.
If this rem was seen before, it means a subarray summing to k exists — update max length.
Optimal (Sliding Window, Non-negative only):
Use a sliding window with two pointers.
Expand the window from the right and shrink from the left if the sum exceeds k.
When sum equals k, update the maximum subarray length.
🧠 Intuition for #002: Max Subarray Sum
Optimal: Optimal: Iterate through the array while maintaining a running sum. Reset the sum to zero when it becomes negative, and update the maximum sum and its indices whenever a higher sum is found.
🧠 Intuition for #003: Best Time to Buy and Sell Stock
Optimal: Maintain the minimum price seen so far while iterating through the array, and at each step, calculate the current profit. Update the maximum profit whenever the current profit exceeds it.
🧠 Intuition for #004: Container With Most Water
Optimal: Use two pointers left and right, move the pointer whose height is less. For each step find the water that can be stored.
🧠 Intuition for #005: Rearrange Array Elements by Sign
Optimal: Divide the array into two parts- one comprising of only positive integers and the other of negative integers. Merge the two parts to get the resultant array. It is not required to do the modifications in-place.
🧠 Intuition for #006: Majority Element
Brute Force: Nested for loops to check the frequency of each element.
Better: Use hashmap to store frequency of each element.
Optimal (Boyer-Moore Voting Algorithm): 🚀 Real-World Analogy: Voting System Imagine a voting scenario where each element in the array is a candidate.
- Each occurrence of a number is a vote.
- If a number gets canceled out by a different number (i.e., opposition), it loses a vote.
- The true majority candidate (if one exists) will outlast all others because it has more votes than any other.