Solving Next Greater Element I with Monotonic Stack

🚀 Day 29 of #100DaysOfCode – LeetCode Problem #643: Maximum Average Subarray I 🧩 Problem: Given an integer array nums and an integer k, find the contiguous subarray of length k that has the maximum average value. 💡 Approach: This problem is a great example of the Sliding Window Technique 🪟 Calculate the sum of the first k elements. Then slide the window by one element at a time — subtract the element that goes out, add the new one coming in. Keep track of the maximum sum seen. Return the maximum average (maxSum / k). 💻 Java Code: class Solution { public double findMaxAverage(int[] nums, int k) { double sum = 0; for (int i = 0; i < k; i++) { sum += nums[i]; } double max = sum; for (int i = k; i < nums.length; i++) { sum += nums[i] - nums[i - k]; max = Math.max(max, sum); } return max / k; } } ⏱️ Complexity: Time: O(n) Space: O(1) ✅ Result: Accepted (Runtime: 0 ms) This one reinforces how sliding window optimizations can drastically simplify what looks like a brute-force problem!

🚀 Day 30 of #100DaysOfCode – LeetCode Problem #88: Merge Sorted Array 🧩 Problem: You’re given two sorted arrays, nums1 and nums2, and need to merge them into a single sorted array — in-place inside nums1. The catch? You can’t return a new array, and nums1 already contains extra space for nums2. 💡 Approach: To avoid overwriting elements in nums1, we use a three-pointer approach: Start from the end of both arrays. Compare the elements and place the larger one at the back (nums1[pMerge]). Move pointers accordingly until all elements are merged. 💻 Java Code: class Solution { public void merge(int[] nums1, int m, int[] nums2, int n) { int p1 = m - 1, p2 = n - 1, pMerge = m + n - 1; while (p2 >= 0) { if (p1 >= 0 && nums1[p1] > nums2[p2]) { nums1[pMerge--] = nums1[p1--]; } else { nums1[pMerge--] = nums2[p2--]; } } } } ⏱️ Complexity: Time: O(m + n) Space: O(1) ✅ Result: Accepted (Runtime: 0 ms) A simple yet elegant example of solving array problems in-place using smart pointer manipulation! 

🚀 Day 34 of #100DaysOfCode – LeetCode Problem #228: Summary Ranges 🧩 Problem: Given a sorted, unique integer array nums, return the smallest list of ranges that covers all numbers in the array exactly. Each range is represented as: "a->b" if the range covers more than one number. "a" if it covers a single number. 📘 Example: Input: nums = [0,1,2,4,5,7] Output: ["0->2", "4->5", "7"] 💡 Approach: We traverse the array while tracking the start of each range. Whenever we detect a break (i.e., the next number isn’t consecutive), we close the range and store it. 🔍 Steps: Initialize start = nums[0]. Traverse the array. If the next element isn’t consecutive, push "start->current" (or "start" if equal). Move to the next starting number and continue. 💻 Java Code: class Solution { public List<String> summaryRanges(int[] nums) { List<String> result = new ArrayList<>(); if (nums.length == 0) return result; int i = 0; while (i < nums.length) { int start = nums[i]; int j = i; while (j + 1 < nums.length && nums[j + 1] == nums[j] + 1) { j++; } if (start == nums[j]) { result.add(String.valueOf(start)); } else { result.add(start + "->" + nums[j]); } i = j + 1; } return result; } } ⏱️ Complexity: Time: O(n) Space: O(1) ✅ Result: Accepted (Runtime: 3 ms) It’s a simple yet elegant problem — a great reminder that clean logic often beats complex tricks

🚀 Day 32 of #100DaysOfCode – LeetCode Problem #697: Degree of an Array 🧩 Problem: Given a non-empty array of non-negative integers, the degree of the array is the maximum frequency of any element. The goal is to find the smallest possible length of a contiguous subarray that has the same degree as the entire array. 💡 Approach: To solve this efficiently: Track the first and last occurrence of each element. Track their frequency count. Identify elements that contribute to the array’s maximum degree. For each of those elements, compute the subarray length = last[i] - first[i] + 1. Return the minimum of those lengths. 💻 Java Code: class Solution { public int findShortestSubArray(int[] nums) { Map<Integer, Integer> count = new HashMap<>(); Map<Integer, Integer> first = new HashMap<>(); int degree = 0, bestLen = nums.length; for (int i = 0; i < nums.length; i++) { int num = nums[i]; first.putIfAbsent(num, i); count.put(num, count.getOrDefault(num, 0) + 1); degree = Math.max(degree, count.get(num)); } for (int i = 0; i < nums.length; i++) { int num = nums[i]; if (count.get(num) == degree) { int len = i - first.get(num) + 1; bestLen = Math.min(bestLen, len); } } return bestLen; } } ⏱️ Complexity: Time: O(n) Space: O(n) ✅ Result: Accepted (Runtime: 0 ms) This problem beautifully blends hashmaps and frequency analysis to derive insights efficiently — a perfect example of thinking beyond brute force!

🚀 Day 31 of #100DaysOfCode – LeetCode Problem #704: Binary Search 🧩 Problem: Given a sorted array and a target value, find the index of the target using an algorithm with O(log n) time complexity. If the target doesn’t exist, return -1. 💡 Approach: This is a classic Binary Search problem. We use two pointers (i and j) to represent the current search range. Calculate the middle index mid. If nums[mid] equals the target → return mid. If target > nums[mid] → search the right half. Else → search the left half. Repeat until the range is empty. 💻 Java Code: class Solution { public int search(int[] nums, int target) { int i = 0, j = nums.length - 1; while (i <= j) { int mid_i = i + (j - i) / 2; int mid_val = nums[mid_i]; if (target < mid_val) { j = mid_i - 1; } else if (target > mid_val) { i = mid_i + 1; } else { return mid_i; } } return -1; } } ⏱️ Complexity: Time: O(log n) Space: O(1) ✅ Result: Accepted (Runtime: 0 ms) Binary Search — simple, efficient, and one of the most fundamental algorithms every developer must master.

Day 2 of my #50DaysOfCode challenge Today, I tackled a problem that focuses on careful string parsing and numeric conversions String to Integer (atoi).It’s a great exercise that deepens understanding of data types, character operations, and overflow handling in Java. Problem: String to Integer (atoi) Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer. The algorithm for myAtoi(string s) is as follows: Ignore any leading whitespace (" "),Determine the sign by checking if the next character is '-' or '+', assuming positivity if neither present. Read the integer by skipping leading zeros until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0. Solution: O(N) The solution iterates through each character of the string once. It first skips leading spaces, then checks for a sign, and builds the number digit by digit. At each step, it performs an overflow check to ensure the value stays within the 32-bit signed integer range. Finally, it multiplies the result by the sign and returns it as an integer. This efficient linear-time approach ensures both accuracy and robustness for different input formats.

Problem 27 : LeetCode 🎯 LeetCode Problem #645 Solved — Set Mismatch (Cyclic Sort Approach) 🧩 Today, I solved another interesting LeetCode Easy problem — “Set Mismatch”, which deepened my understanding of in-place sorting and index mapping using the Cyclic Sort technique. 🔍 Problem Overview Given an integer array nums where numbers are supposed to be from 1 to n, one number is duplicated, and one number is missing. The task: Find both numbers using O(n) time and O(1) space. 💡 My Approach — Cyclic Sort Technique I treated the array like a self-mapping structure where each number x should ideally be placed at index x - 1. If the current element isn’t in its correct place and its target isn’t already filled, I swap it. After the array is sorted in place, any index i where arr[i] != i + 1 indicates: arr[i] → the duplicate number i + 1 → the missing number ⚙️ Results ✅ Runtime: 3 ms — Beats 79.76% of Java submissions ✅ Memory: 47.45 MB — Beats 6.28% of submissions ✅ Complexity: O(n) time | O(1) space ✅ Status: Accepted ✅ (All 49 test cases passed) 🧩 Tech Stack Java | Cyclic Sort | Array Manipulation | In-Place Algorithm | DSA Another problem that reinforces my confidence in index-based logic and memory-efficient coding patterns — vital skills for backend and system optimization. #LeetCode #Java #ProblemSolving #CyclicSort #DataStructures #Algorithms #InPlaceAlgorithm #BackendDevelopment #SpringBoot #DSA #LearningInPublic #CodingJourney Question Url : https://lnkd.in/dixwjFE3

🚀 Day 38 of #100DaysOfCode – LeetCode Problem #747: Largest Number At Least Twice of Others 💡 Problem Summary: Given an integer array nums, find whether the largest number is at least twice as large as all other numbers. If it is, return the index of the largest element; otherwise, return -1. 📘 Example: Input: nums = [3,6,1,0] Output: 1 Input: nums = [1,2,3,4] Output: -1 🧠 Approach: Find the largest and second largest numbers in the array. If largest >= 2 * secondLargest, return the index of the largest. Otherwise, return -1. 💻 Java Solution: class Solution { public int dominantIndex(int[] nums) { int max = -1, second = -1, index = -1; for (int i = 0; i < nums.length; i++) { if (nums[i] > max) { second = max; max = nums[i]; index = i; } else if (nums[i] > second) { second = nums[i]; } } return second * 2 <= max ? index : -1; } } ⚙️ Complexity: Time: O(n) Space: O(1) ✅ Result: Accepted (Runtime: 0 ms) 🎯 Key Takeaway: Sometimes, solving problems isn’t about complex algorithms — it’s about thinking clearly and tracking key elements efficiently.

✅ Day 35 of #100DaysOfCode Challenge 📘 LeetCode Problem 112: Path Sum 🧩 Problem Statement:  Given the root of a binary tree and an integer targetSum, return true if there exists a root-to-leaf path whose sum of node values equals targetSum. Example: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22   Output: true Explanation:  Path 5 → 4 → 11 → 2 gives the sum = 22 ✅ 💡 Simple Approach: If the tree is empty → return false If it’s a leaf node → check if its value equals targetSum Otherwise → subtract node value and check left and right recursively 💻 Easiest Java Code: class Solution {   public boolean hasPathSum(TreeNode root, int sum) {     if (root == null) return false;     if (root.left == null && root.right == null && root.val == sum)       return true;     sum = sum - root.val;     return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);   } } ⚙ Complexity: ⏱ Time: O(n) → visit each node once 💾 Space: O(h) → recursion stack (h = height of tree) 🌿 Small code, big concept — recursion makes trees easy 🌱  #Day35 #100DaysOfCode #LeetCode #Java #DSA #BinaryTree #Recursion #CodingChallenge

✅ Day 35 of #100DaysOfCode Challenge 📘 LeetCode Problem 112: Path Sum 🧩 Problem Statement: Given the root of a binary tree and an integer targetSum, return true if there exists a root-to-leaf path whose sum of node values equals targetSum. Example: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: Path 5 → 4 → 11 → 2 gives the sum = 22 ✅ 💡 Simple Approach: If the tree is empty → return false If it’s a leaf node → check if its value equals targetSum Otherwise → subtract node value and check left and right recursively 💻 Easiest Java Code: class Solution { public boolean hasPathSum(TreeNode root, int sum) { if (root == null) return false; if (root.left == null && root.right == null && root.val == sum) return true; sum = sum - root.val; return hasPathSum(root.left, sum) || hasPathSum(root.right, sum); } } ⚙️ Complexity: ⏱️ Time: O(n) → visit each node once 💾 Space: O(h) → recursion stack (h = height of tree) 🌿 Small code, big concept — recursion makes trees easy 🌱 #Day35 #100DaysOfCode #LeetCode #Java #DSA #BinaryTree #Recursion #CodingChallenge