Special Binary String LeetCode Solution

Day 62 of 365 Days of code Qn1) insert into a binary search🌳 Approach: recursion def insert(root,val): if root==None: set root=Treenode(val) return root elif root.val<val: root.right=insert(root.right,val) else: root.left=insert(root.left,val) return root Sleep well during exams bud, a piece of advice :"). #365daysOfCode #NeetCode  #leetcode  #DSA  #python #LeetCode #ProblemSolving  #Algorithms #365dayschallenge

🔥 Day 30 of #100DaysOfCode 📌 LeetCode 26 – Remove Duplicates from Sorted Array Today’s problem focused on in-place array manipulation and strengthening the two-pointer technique. 🧠 Problem Summary Given a sorted array, remove duplicates such that each unique element appears only once. Return the count of unique elements (k). Important: Do it in-place with O(1) extra space. 💡 Key Insight Because the array is already sorted, duplicates are always adjacent. So instead of checking all elements repeatedly: 👉 Compare current element with previous one. 👉 If different → place it at the next unique index. This is where the Two Pointer Approach shines: Pointer i → traverses array Pointer k → tracks position of next unique element 📊 Complexity ⏱ Time: O(n) 📦 Space: O(1) Consistency > Motivation 30 Days Done. Still going strong 💪 #Day30 #LeetCode #DSA #100DaysOfCode #Python #CodingJourney

Merge Intervals (LeetCode 56) - Medium Key Learnings: Sorting: By sorting intervals based on their start times, we can process them in a single pass (O(N)). Overlap Logic: Comparing the current_start with the previous_end time is the secret sauce to identifying overlaps. Space-Time Tradeoff: Sorting takes O(N log N) time, and we use O(N) space to store the results. #CodingJourney #LeetCode #Blind75 #SDEPreparation #SoftwareEngineering #Python #DataStructures

LeetCode Problem 3070: "Count submatrices with top left element and sum less than k": You are given a 0-indexed integer matrix grid and an integer k. Return the number of submatrices that contain the top-left element of the grid, and have a sum less than or equal to k. Approach: use Prefix sum to calculate the total value of submatrix including the top left element upto a given cell. This approach is optimal and straightforward. #Python #DSA #LeetCode #ProblemSolving #CompetitiveProgramming #DailyCoding

🚀 #100DaysOfCode – Day 52 📌 Problem: Sort Colors (LeetCode 75) 💡 Problem Idea: You are given an array containing only 0s, 1s, and 2s, representing three colors. The task is to sort the array in-place so that all 0s come first, then 1s, and then 2s. ⚡ Key Insight: Instead of using a sorting algorithm, we can solve this in one pass using three pointers. We maintain three regions: Left pointer → position for next 0 Right pointer → position for next 2 Current pointer (i) → traverses the array 🎯 Approach (Dutch National Flag Algorithm): If element is 0 → swap with left, move both left and i If element is 1 → just move i If element is 2 → swap with right, decrease right 📈 Complexity: Time Complexity: O(n) Space Complexity: O(1) (in-place sorting) ✅ Efficient because the array is processed only once. 💭 Learning: This problem is a great example of how pointer techniques can optimize sorting problems without using built-in sort functions. #DSA #LeetCode #Python #CodingJourney #ProblemSolving #Algorithms #100DaysOfCode #LinkedInLearning

🔹 LinkedIn Post – Day 42 | LeetCode 162: Find Peak Element 🚀 Problem: Find Peak Element A peak element is an element in the array that is strictly greater than its neighbors. The task is to return the index of any peak element present in the array. 💡 Key Idea Instead of checking every element sequentially, we can use a Binary Search approach to reduce the search space efficiently. ⚙️ Logic Compare the middle element with its next element. If the middle element is greater, it means a peak exists on the left side (including mid). If the next element is greater, the peak must lie on the right side. Continue reducing the search range until both pointers meet. The final position will be the index of a peak element. ⏱️ Complexity Time Complexity: O(log n) Space Complexity: O(1) 🎯 Insight The array can have multiple peaks, but the problem only requires returning any one peak index. Binary search works because the array always moves towards a peak when following the increasing slope. #Day41 #LeetCode #DSA #BinarySearch #ProblemSolving #CodingJourney #Python

LeetCode Problem 1545: "Given two positive integers n and k, the binary string Sn is formed as follows: S1 = "0" Si = Si - 1 + "1" + reverse(invert(Si - 1)) for i > 1 Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0)." Approach: Initialize an array with a single element '0' which serves as the string1 then iteratively calculate the consecutive strings and store them in the array. Finally, form the final string using the join() function and return the kth character of the string. #Python #LeetCode #String #ProblemSolving #CompetitiveProgramming #DataStructures #Algorithm

Day 68 of 365 Days of code Qn 1) Min stack Approach: use an extra stack(to store minimum element up to the particular index) for push: push the value into the stack, check if the val is lesser than the element at the top of minstack, if yes, push the value, else push the copy of the top element in minstack for pop: pop from both of the stacks for returnmin: return minstack[top] :) good night #365daysOfCode #NeetCode  #leetcode  #DSA  #python #LeetCode #ProblemSolving  #Algorithms #365dayschallenge

Day 67 of 365 Days of code the infamous brainrot number Qn 1) Reorder list You are given the head of a singly linked-list. The positions of a linked list of length = 7 for example, can intially be represented as: [0, 1, 2, 3, 4, 5, 6] Reorder the nodes of the linked list to be in the following order: [0, 6, 1, 5, 2, 4, 3] Notice that in the general case for a list of length = n the nodes are reordered to be in the following order: [0, n-1, 1, n-2, 2, n-3, ...] You may not modify the values in the list's nodes, but instead you must reorder the nodes themselves. approach: use 5 pointers(scary :")) #365daysOfCode #NeetCode  #leetcode  #DSA  #python #LeetCode #ProblemSolving  #Algorithms #365dayschallenge

feb-21-26 LeetCode 762 – Prime Number of Set Bits in Binary Representation Solved today’s problem focused on bit manipulation and prime checking! 🔹 Problem: Given two integers left and right, count how many numbers in that range have a prime number of set bits (1s) in their binary representation. 💡 Approach: Iterate through each number in the range. Count set bits using bitwise operations (& and right shift). Check whether the count of set bits is a prime number. Keep track of valid cases. 🧠 Concepts Used: Bit Manipulation Prime Number Checking Mathematical Optimization Simple problem on the surface, but a great way to strengthen understanding of binary representation and efficient computation. Consistency + Daily Practice = Growth 📈🔥 #LeetCode #BitManipulation #PrimeNumbers #Python #CodingPractice #ProblemSolving