✅ Day 80 of 100 Days LeetCode Challenge Problem: 🔹 #3876 – Construct Uniform Parity Array II 🔗 https://lnkd.in/gENuCXJc Learning Journey: 🔹 Today’s problem focused on determining whether we can construct an array where all elements are either all even or all odd using allowed operations. 🔹 The key observation was around parity (odd/even nature) of numbers. 🔹 Since we can subtract elements, the parity depends on whether differences can produce consistent odd or even values. 🔹 I simplified the logic by checking: • If the minimum element is odd → we can make all elements odd • Otherwise, verify if all elements are already even Concepts Used: 🔹 Parity (Odd/Even Analysis) 🔹 Mathematical Observation 🔹 Greedy Logic Key Insight: 🔹 Subtracting numbers preserves or changes parity in predictable ways. 🔹 The smallest element plays a crucial role in determining whether all elements can be transformed into the same parity. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers
Construct Uniform Parity Array II LeetCode Challenge
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✅ Day 87 of 100 Days LeetCode Challenge Problem: 🔹 #2840 – Check if Strings Can be Made Equal With Operations II 🔗 https://lnkd.in/gY73RBb5 Learning Journey: 🔹 Today’s problem extended the previous one, allowing swaps between indices where the difference is even. 🔹 I observed that this again partitions the string into two independent groups: • Even indices (0, 2, 4, …) • Odd indices (1, 3, 5, …) 🔹 I extracted characters from even and odd positions separately for both strings. 🔹 Then, I sorted these groups and compared them between s1 and s2. 🔹 If both even-index groups and odd-index groups match, the strings can be made equal. Concepts Used: 🔹 String Manipulation 🔹 Index Grouping (Parity-based) 🔹 Sorting 🔹 Greedy Observation Key Insight: 🔹 Since swaps are allowed only between indices with even distance, characters can only move within their parity group. 🔹 Therefore, the problem reduces to checking if both parity groups have identical character distributions. Complexity: 🔹 Time: O(n log n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 88 of 100 Days LeetCode Challenge Problem: 🔹 #1252 – Cells with Odd Values in a Matrix 🔗 https://lnkd.in/g5eE9A-e Learning Journey: 🔹 Today’s problem focused on simulating row and column increment operations on a matrix. 🔹 I initialized an m × n matrix with all zeros. 🔹 For each index [ri, ci], I incremented: • All elements in row ri • All elements in column ci 🔹 After applying all operations, I traversed the matrix to count how many cells contain odd values. Concepts Used: 🔹 Matrix Simulation 🔹 Nested Loops 🔹 Conditional Counting 🔹 Basic Array Manipulation Key Insight: 🔹 Direct simulation works efficiently due to small constraints. 🔹 Each operation affects an entire row and column, so tracking increments carefully is key. Complexity: 🔹 Time: O(m × n + k × (m + n)) 🔹 Space: O(m × n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 89 of 100 Days LeetCode Challenge Problem: 🔹 #1480 – Running Sum of 1D Array 🔗 https://lnkd.in/gSTZrxF7 Learning Journey: 🔹 Today’s problem focused on computing the running (prefix) sum of an array. 🔹 Instead of using an extra array, I optimized the solution by modifying the input array in-place. 🔹 Starting from index 1, I updated each element as: • nums[i] += nums[i-1] 🔹 This way, each index stores the cumulative sum up to that point. 🔹 Finally, returned the modified array. Concepts Used: 🔹 Prefix Sum 🔹 In-place Computation 🔹 Array Traversal Key Insight: 🔹 The previous element already stores the prefix sum, so we can reuse it directly. 🔹 Eliminates the need for extra space while maintaining linear time. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 81 of 100 Days LeetCode Challenge Problem: 🔹 #1897 – Redistribute Characters to Make All Strings Equal 🔗 https://lnkd.in/gZ7wZ7JE Learning Journey: 🔹 Today’s problem required checking if we can redistribute characters so that all strings become identical. 🔹 I combined all strings and used a frequency counter to count occurrences of each character. 🔹 Since we can freely move characters between strings, the only requirement is that each character count must be divisible by the number of strings. 🔹 If any character fails this condition, it’s impossible to evenly distribute it. Concepts Used: 🔹 Hash Map / Frequency Counting (Counter) 🔹 String Concatenation 🔹 Divisibility Check Key Insight: 🔹 The exact arrangement doesn’t matter—only the total frequency distribution matters. 🔹 If every character can be equally divided among all strings, a valid configuration always exists. Complexity: 🔹 Time: O(n * k) (n = number of strings, k = average length) 🔹 Space: O(1) (bounded by alphabet size) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 83 of 100 Days LeetCode Challenge Problem: 🔹 #832 – Flipping an Image 🔗 https://lnkd.in/ghfEcHHT Learning Journey: 🔹 Today’s problem involved two operations on a binary matrix: horizontal flip and bit inversion. 🔹 First, I reversed each row to achieve the horizontal flip. 🔹 Then, I inverted every bit (0 → 1 and 1 → 0) using a helper function. 🔹 This approach ensured the transformation was done in-place without extra space. Concepts Used: 🔹 Array Manipulation 🔹 Matrix Traversal 🔹 Two-Pointer / In-place Operations 🔹 Bit Manipulation Key Insight: 🔹 Instead of treating flipping and inversion as separate heavy operations, they can be efficiently combined or done sequentially in-place. 🔹 Using simple transformations keeps the solution clean and optimal. Complexity: 🔹 Time: O(n × m) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 95 of 100 Days LeetCode Challenge Problem: 🔹 #869 – Reordered Power of 2 🔗 https://lnkd.in/gkNXaSFM Learning Journey: 🔹 Today’s problem focused on checking whether the digits of a number can be rearranged to form a power of 2. 🔹 I created a helper function to extract and store the digits of a number. 🔹 Then I sorted the digits of the input number for comparison. 🔹 Next, I generated powers of 2 iteratively and compared their sorted digit lists with the input. 🔹 If any match was found, I returned True. Otherwise, continued until the digit length exceeded the input. Concepts Used: 🔹 Digit Extraction 🔹 Sorting 🔹 Simulation of Powers of 2 🔹 Brute Force Optimization Key Insight: 🔹 Instead of generating permutations (which is expensive), sorting digits allows quick comparison. 🔹 Any valid rearrangement must have the same digit frequency as some power of 2. Complexity: 🔹 Time: O(log n * d log d) 🔹 Space: O(d) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 97 of 100 Days LeetCode Challenge Problem: 🔹 #1281 – Subtract the Product and Sum of Digits of an Integer 🔗 https://lnkd.in/gxTAZc6U Learning Journey: 🔹 Today’s problem involved extracting digits of a number and performing two operations simultaneously. 🔹 I initialized two variables: one for product (pr) and one for sum (sm). 🔹 Using a while loop, I extracted each digit using n % 10. 🔹 Updated the product by multiplying the digit and updated the sum by adding it. 🔹 Reduced the number using integer division (n //= 10) after each step. 🔹 Finally returned the difference between product and sum. Concepts Used: 🔹 Digit Extraction 🔹 While Loop 🔹 Arithmetic Operations 🔹 Number Manipulation Key Insight: 🔹 Both product and sum can be computed in a single traversal of digits. 🔹 Efficient use of modulus and division avoids converting the number to a string. Complexity: 🔹 Time: O(d) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 82 of 100 Days LeetCode Challenge Problem: 🔹 #2906 – Construct Product Matrix 🔗 https://lnkd.in/gdb7GZNB Learning Journey: 🔹 Today’s problem required constructing a matrix where each cell contains the product of all other elements except itself, modulo 12345. 🔹 I flattened the 2D matrix into a 1D list to simplify processing. 🔹 Then I used the prefix and postfix product technique: • pre[i] → product of all elements before index i • post[i] → product of all elements after index i 🔹 Multiplying pre[i] * post[i] gives the required result for each position. 🔹 Finally, I mapped the computed values back into the original matrix shape. Concepts Used: 🔹 Prefix Product 🔹 Postfix Product 🔹 Array Flattening 🔹 Modular Arithmetic Key Insight: 🔹 Using prefix and postfix arrays avoids recomputing products for every cell, reducing time complexity. 🔹 This is an extension of the classic “product of array except self” problem. Complexity: 🔹 Time: O(n × m) 🔹 Space: O(n × m) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers
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✅ Day 99 of 100 Days LeetCode Challenge Problem: 🔹 #999 – Available Captures for Rook 🔗 https://lnkd.in/gAw_XpbJ Learning Journey: 🔹 Today’s problem focused on simulating rook movement on a chessboard. 🔹 First, I located the position of the rook 'R' on the board. 🔹 Then, I explored all four directions: up, down, left, and right. 🔹 In each direction, I moved step-by-step until: • I hit a bishop 'B' → stop (blocked) • I found a pawn 'p' → increment count and stop • Or reached the board boundary 🔹 Summed all valid captures and returned the result. Concepts Used: 🔹 Matrix Traversal 🔹 Simulation 🔹 Direction Vectors 🔹 Boundary Checking Key Insight: 🔹 The rook’s movement is linear in 4 directions, and each direction is independent. 🔹 Early stopping (on bishop or pawn) is critical for correctness and efficiency. Complexity: 🔹 Time: O(1) (fixed 8×8 board, constant work) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #100DaysOfCode #Python #CodingJourney #ProblemSolving #LearningInPublic
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Day 56 of #GeekStreak60: The Phantom Pointer Trick! 🕵️♂️📈 Tackled the "Sorted subsequence of size 3" problem on @GeeksforGeeks today. Key Learning: Finding an increasing triplet is easy if you use extra arrays to track minimums and maximums, but that violates the O(1) space constraint. The optimal solution is to use "Greedy State Tracking." By iterating through the array in a single pass, I maintained three variables: the absolute smallest number seen (num1), a valid middle number (num2), and a snapshot of num1 locked in at the exact moment num2 was discovered. If the loop encounters any number strictly greater than num2, the valid triplet is instantly formed! This perfectly eliminates the need for O(n) memory arrays while keeping the time complexity to a strict O(n). Just 4 days left! The logic is feeling sharper than ever. 🚀 #geekstreak60 #npci #coding #Algorithms #Python #DataStructures #Optimization #SoftwareEngineering
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