Counting Special Positions in Binary Matrix with LeetCode Solution

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g6uRPkE6 💡 My thought process: The first approach calculates the bitwise complement by examining each bit of the number. It starts by determining how many bits are needed to represent n using log2(n) + 1. For each bit position i, a mask (1 << i) is used to isolate that bit from n. If the extracted bit is 0, the result for that position is set by adding the same mask, effectively flipping the bit. This creates a new number where all bits within the effective bit length of n are inverted, resulting in the complement. The second approach uses the XOR operation to flip all relevant bits at once. It first calculates the number of bits in n and creates a mask filled with 1s in all those positions using (1 << digits) - 1. This mask represents the highest value that can be formed with the same number of bits. Applying n ^ mask flips every bit within that range because XOR with 1 changes the bit and XOR with 0 leaves it the same, giving the bitwise complement of n. 👉 My Solution: https://lnkd.in/gGCdUW3H If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gx6PQVMB 💡 My thought process: The brute force method checks all possible rotations of the binary string by moving the first character to the end repeatedly. For each rotated string, it counts the number of mismatches needed to create an alternating pattern that starts with 1. The opposite alternating pattern, which starts with 0, would require different changes. The method takes the minimum of the two counts for each rotation and returns the lowest value overall. The optimized method does not generate all rotations. Instead, it doubles the string and uses a sliding window of size n to represent every possible rotation. It keeps track of the mismatch count for forming an alternating pattern that starts with 1. As the window slides, it updates this count by subtracting the contribution of the character that leaves the window and adding the incoming character. At each step, it compares both alternating patterns to find the minimum number of flips needed. 👉 My Solution: https://lnkd.in/gP2FWVYP If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gSiw_GQq 💡 My thought process: It uses a two-pointer approach. One pointer starts at the top row of the submatrix, and the other starts at the bottom row. For each pair of rows, it goes through all columns within the submatrix range and swaps the elements column by column. This process continues until the top and bottom pointers meet or cross. This ensures the submatrix is reversed in place without using any extra space. The updated grid is then returned. 👉 My Solution: https://lnkd.in/g_cAfrtb If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g5ggMZxw 💡 My thought process: The approach checks how many mismatches occur if the string follows an alternating pattern starting with 0 (0,1,0,1...), counts the number of positions that violate this pattern, and since the opposite alternating pattern starting with 1 (1,0,1,0...) would require the remaining changes, the minimum of the two mismatch counts represents the minimum number of operations required. 👉 My Solution: https://lnkd.in/gVuWTeKh If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gx6PQVMB 💡 My thought process: It is mentioned that the size of the array also denotes the size of each string and the number of unique strings in the array. I took an unordered set, and I generated all the possible binary strings of length n. I only generated n+1 strings because we only need to return any one of them. When the size of the set goes above n, I stop finding anymore binary strings. Then, I iterate over the array and remove those binary strings present in the array from the set. Just simply return the remaining string in the set. 👉 My Solution: https://lnkd.in/gJKMj6vX If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gBvd5iWy 💡 My thought process: For this question, I am precalculating the total sum of the grid and row-wise and column-wise sums. Store the precalculated values in a vector and then simply do a cumulative sum over the vector. If the cumulative sum equals the total sum - cumulative sum, this means that a single cut is possible to split the array. The calculation part is done in another method, and I simply call the method twice for row-wise check and then column-wise. 👉 My Solution: https://lnkd.in/gNYsk6iz If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gbtaBXYW 💡 My thought process: The algorithm first calculates a normalized shift value using k % n. Shifting a row by its total length, n, returns the original configuration. Therefore, any shift k is the same as k modulo n. Instead of rotating memory physically or creating a temporary matrix, the code uses a virtual mapping strategy. It goes through each cell (i, j) and finds the index of the element that would be in that position after the shift. For even rows (left shift), to find the original value that moves into position j after a left shift, the code looks at the index (j - shift). It manages negative results by wrapping around with an addition of n. For odd rows (right shift), it checks the index (j + shift) % n. The function has a fail-fast mechanism; it returns false right away if it finds the first difference between a cell and its shifted counterpart. 👉 My Solution: https://lnkd.in/g5myRYRj If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gVK8xX3p 💡 My thought process: The solution uses a top-down dynamic programming approach with memoization. Each cell stores a pair of values: the maximum product and the minimum product achievable from that cell to the destination. The minimum value is necessary because multiplying two negative numbers can lead to a larger positive result later. The recursive function explores both possible moves—right and down—from the current cell and retrieves their stored results. For each direction, it considers both the maximum and minimum products returned and multiplies them by the current cell’s value. All combinations are evaluated, and the overall maximum and minimum are selected and stored in the DP table for that cell. Memoization ensures that each cell is computed only once, reducing the time complexity to O(m × n). The base case occurs at the bottom-right cell, where both maximum and minimum values equal the cell’s value. Invalid positions return sentinel values to indicate they should not be considered. Finally, the function checks the result from the starting cell. If the maximum product is negative, it returns -1. Otherwise, it returns the result modulo 1e9 + 7. 👉 My Solution: https://lnkd.in/gcKfTVNq If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

✅ Solved: Palindrome Number — LeetCode #9 Accepted with all 11,511 / 11,511 test cases passed! 🎯 The problem asks: Given an integer x, return true if x is a palindrome, and false otherwise. My approach: Immediately return false for negative numbers (can't be palindromes) Convert the integer to a String using String.valueOf() Reverse the string by iterating from end to start Compare original and reversed — if equal, it's a palindrome ✔️ Runtime: 16 ms | Memory: 46.48 MB A clean beginner-friendly solution. Next, I'd like to optimize it using a mathematical digit-reversal approach — no String conversion needed, better memory efficiency! 🚀 If you're grinding LeetCode, don't skip the easy problems. They build the intuition you need for the hard ones. 💪 #LeetCode #Java #DSA #DataStructures #Algorithms #CodingChallenge #Programming #SoftwareDevelopment #ProblemSolving #PalindromeNumber

🚀 Day 26 | 100 Days of Coding Challenge #DrGViswanathanChallenge 📘 Problem Solved: Single Number (LeetCode) 🔧 Approach Used (Brute Force): • Checked each element and counted its occurrences • Returned the element that appears only once ⏳ Complexity: Time: O(n²) Space: O(1) 🧠 Key Learning: Brute force works, but understanding constraints helps us optimize further. 💡 Optimal Insight: This problem can be solved in O(n) using XOR: 👉 a ^ a = 0 and a ^ 0 = a → all duplicates cancel out, leaving the single number. 📂 Topics Covered: Arrays, Brute Force, Bit Manipulation #100DaysOfCode #DSA #CPP #LeetCode #CodingJourney #CompetitiveProgramming