Java LeetCode Challenge Day 89: Minimum Flips to Alternating Binary String

🚀 Day 97 of My 100 Days LeetCode Challenge | Java Today’s problem was a solid exercise in matrix processing and prefix sum optimization. The goal was to count the number of submatrices whose sum is less than or equal to a given value (k). A brute-force approach would be too slow, so the key was to use 2D prefix sums to efficiently compute submatrix sums. By converting the matrix into a prefix sum matrix, we can calculate the sum of any submatrix in constant time, making the overall solution much more efficient. ✅ Problem Solved: Count Submatrices With Sum ≤ K ✔️ All test cases passed (859/859) ⏱️ Runtime: 7 ms 🧠 Approach: 2D Prefix Sum 🧩 Key Learnings: ● Prefix sums are powerful for optimizing repeated range sum queries. ● 2D prefix sums extend the same idea from arrays to matrices. ● Preprocessing can drastically reduce computation time. ● Avoiding brute force is key in large input problems. ● Matrix problems often become easier with the right transformation. This problem reinforced how preprocessing techniques like prefix sums can turn complex problems into efficient solutions. 🔥 Day 97 complete — sharpening matrix optimization and prefix sum skills. #LeetCode #100DaysOfCode #Java #PrefixSum #Matrix #Algorithms #ProblemSolving #DSA #CodingJourney #Consistency

🚀 Day 536 of #750DaysOfCode 🚀 Today I solved Count Submatrices With Equal Frequency of X and Y (LeetCode 3212) using Java. 🔹 Problem Summary: Given a grid containing 'X', 'Y', and '.', we need to count the number of submatrices starting from the top-left corner (0,0) such that: • The number of 'X' and 'Y' is equal • The submatrix contains at least one 'X' 🔹 Approach: Instead of checking every possible submatrix (which would be too slow), I used the Prefix Sum technique. I maintained two prefix matrices to store the count of 'X' and 'Y' up to each cell. For every position (i, j), I checked: countX == countY and countX > 0 If true, that submatrix is valid. This reduces the complexity to O(n × m), which works efficiently for large grids. 🔹 Key Concepts Learned: ✅ Prefix Sum in 2D ✅ Matrix traversal optimization ✅ Handling constraints up to 1000 × 1000 ✅ Clean implementation in Java Consistent practice is making problem-solving faster and more structured every day. #750DaysOfCode #Day536 #LeetCode #Java #DataStructures #Algorithms #PrefixSum #CodingChallenge #ProblemSolving

🚀 Day 91 of My 100 Days LeetCode Challenge | Java Today’s challenge was a clever string + frequency analysis problem. The task was to reconstruct the original digits (0–9) from a scrambled string containing the letters of their English names (like zero, one, two…). At first glance it looks messy because letters overlap across multiple numbers. The key insight was noticing that certain letters are unique to specific digits: z → zero (0) w → two (2) u → four (4) x → six (6) g → eight (8) By identifying these unique markers first, we can determine those digits and then subtract their letter counts to uncover the remaining ones. ✅ Problem Solved: Reconstruct Original Digits from English ✔️ All test cases passed (24/24) ⏱️ Runtime: 4 ms 🧠 Approach: Frequency Counting + Greedy Deduction 🧩 Key Learnings: ● Character frequency counting can simplify complex string problems. ● Unique identifiers often unlock greedy solutions. ● Solving certain elements first can reveal hidden dependencies. ● Careful ordering of deductions avoids incorrect counts. ● Many string puzzles become manageable with pattern observation. This problem was a great reminder that recognizing unique patterns can turn a confusing problem into a structured solution. 🔥 Day 91 complete — improving pattern recognition and string processing skills. #LeetCode #100DaysOfCode #Java #Strings #Algorithms #ProblemSolving #DSA #CodingJourney #Consistency

🚀 Day 7 / 100 – LeetCode Challenge Today I solved Length of Last Word (LeetCode #58) using Java. 🔹 Problem: Given a string "s" containing words and spaces, return the length of the last word in the string. 📌 Approach: - Traverse the string from the end. - Ignore trailing spaces. - Start counting characters of the last word. - Stop when a space appears after counting begins. 💡 Key Concepts Practiced: - String traversal - "charAt()" usage - Reverse iteration - Conditional logic ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) ✅ Key Takeaway: Sometimes solving a problem becomes easier when we traverse the data from the end instead of the beginning. #100DaysOfCode #LeetCode #Java #DSA #ProblemSolving #CodingJourney

100 Days of Coding Challenge – Day 30 📌 Problem: Find the Index of the First Occurrence in a String 💻 Language: Java 🧠 Concept Used: String Matching (Brute Force) 🔍 Platform: LeetCode Today’s challenge was to find the first occurrence of a substring (needle) in a given string (haystack). If not found, return -1. Example: Input: "sadbutsad", "sad" Output: 0 Approach: ✔ Traverse the string from index 0 to n - m ✔ Extract substring of length m at each position ✔ Compare it with the target string ✔ Return index immediately when match is found ✔ If no match → return -1 Time Complexity: O(n × m) Space Complexity: O(1) 🔗 Problem Link: https://lnkd.in/g2ktYFFS 🔗 Code: https://lnkd.in/gynFixSQ #100DaysOfCode #Day30 #Java #DSA #LeetCode #Strings #ProblemSolving #CodingJourney

🚀 Day 15 of #50DaysLeetCode Challenge Today I solved the “Longest Common Prefix” problem on LeetCode using Java. 🔹 Problem: Find the longest common prefix among an array of strings. If no common prefix exists, return an empty string "". Example: Input: ["flower","flow","flight"] Output: "fl" Input: ["dog","racecar","car"] Output: "" 🔹 Approach I Used: ✔ Took the first string as the initial prefix ✔ Compared it with each string in the array ✔ If mismatch occurs, reduced the prefix step by step ✔ Continued until all strings share the same prefix 🔹 Key Insight: You don’t need complex logic — just keep shrinking the prefix until it matches all strings. 🔹 Concepts Practiced: • String manipulation • Iterative comparison • Edge case handling #LeetCode #DSA #Java #Algorithms #ProblemSolving #CodingChallenge #50DaysOfCode

100 Days of Coding Challenge – Day 24 📌 Problem: Two Sum II – Input Array Is Sorted 💻 Language: Java 🧠 Concept Used: Two Pointer Technique 🔍 Platform: LeetCode Today’s challenge was to find two numbers in a sorted array that add up to a given target and return their 1-indexed positions. The problem guarantees exactly one valid solution and requires constant extra space. Example: Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Approach: ✔ Use two pointers — one at the start (left) and one at the end (right) ✔ Calculate the sum of both elements ✔ If the sum equals the target → return their indices ✔ If the sum is smaller → move left forward ✔ If the sum is larger → move right backward Time Complexity: O(n) Space Complexity: O(1) 🔗 Problem Link: https://lnkd.in/gcM_dBA7 🔗 Code: https://lnkd.in/gaUrS-Ne #100DaysOfCode #Day24 #Java #DSA #LeetCode #TwoPointers #Arrays #ProblemSolving #CodingJourney

🚀 Day 93 of My 100 Days LeetCode Challenge | Java Today’s problem explored recursion and backtracking, focusing on generating valid strings under specific constraints. The challenge was to generate happy strings of length n, where: The string consists only of 'a', 'b', and 'c' No two adjacent characters are the same From all possible happy strings, the goal was to find the k-th lexicographical string. To solve this, I used a backtracking approach that builds the string step by step. At each step, we try adding characters while ensuring they don't match the previous character. This naturally generates all valid happy strings in lexicographical order. Once all valid strings are generated, we simply return the k-th one if it exists. ✅ Problem Solved: The k-th Lexicographical Happy String of Length n ✔️ All test cases passed (345/345) ⏱️ Runtime: 22 ms 🧠 Approach: Backtracking + Recursion 🧩 Key Learnings: ● Backtracking is ideal for generating constrained combinations. ● Pruning invalid states early improves efficiency. ● Recursion makes exploring possible paths intuitive. ● Lexicographical generation often comes naturally with ordered iteration. ● Constraint-based string generation is a common interview pattern. This problem was a great exercise in systematically exploring possibilities while enforcing constraints. 🔥 Day 93 complete — strengthening recursion and backtracking intuition. #LeetCode #100DaysOfCode #Java #Backtracking #Recursion #ProblemSolving #DSA #CodingJourney #Consistency

🚀 LeetCode Day 18 – Complement of Base 10 Integer Today I solved LeetCode Problem 1009: Complement of Base 10 Integer using Java. 🧠 Problem: Given a non-negative integer n, return the complement of its binary representation. The complement means flipping all bits in the binary form of the number. 📌 Example: Input: n = 5 Binary of 5 → 101 Complement → 010 Output → 2 💡 Approach: • Create a bitmask with all bits set to 1 up to the highest bit of n. • Use the XOR (^) operator with the mask to flip the bits. 💻 Java Code: class Solution { public int bitwiseComplement(int n) { if (n == 0) return 1; int mask = 0, temp = n; while (temp > 0) { mask = (mask << 1) | 1; temp >>= 1; } return n ^ mask; } } ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) 📚 Practicing Data Structures & Algorithms daily to improve problem-solving skills. #LeetCode #Java #DSA #ProblemSolving #CodingJourney #100DaysOfCode

🚀Day 24 of #75DaysofLeetCode LeetCode Problem Solved: Removing Stars From a String (2390) Today I solved the problem “Removing Stars From a String.” 🔹 Problem Statement: We are given a string containing characters and *. Each * removes the closest non-star character to its left, along with the * itself. The task is to return the final string after all stars are processed. 💡 Approach: The most efficient way to solve this is using a Stack-like approach. 1️⃣ Traverse the string from left to right. 2️⃣ If the character is not *, add it to the result (stack). 3️⃣ If the character is *, remove the last added character. 4️⃣ Continue until the string is processed. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 💻 Java Implementation: class Solution { public String removeStars(String s) { StringBuilder sb = new StringBuilder(); for(char c : s.toCharArray()) { if(c == '*') { sb.deleteCharAt(sb.length() - 1); } else { sb.append(c); } } return sb.toString(); } } ✨ Key Learning: Using a stack or StringBuilder helps efficiently simulate the removal of the closest character on the left. #LeetCode #Java #DataStructures #ProblemSolving #CodingPractice #100DaysOfCode