LeetCode 747: Dominant Index in Array with Python

Day 30/100 – #100DaysOfCode 🚀 Solved LeetCode #1299 – Replace Elements with Greatest Element on Right Side (Python). Today I practiced reverse traversal to efficiently replace each element with the greatest element on its right. Approach: 1) Initialize max_right = -1. 2) Traverse the array from right to left. 3) Store the current element in a temporary variable. 4) Replace the current element with max_right. 5) Update max_right as the maximum of max_right and the stored value. 6) Continue until the entire array is processed. Time Complexity: O(n) Space Complexity: O(1) Learning how reverse traversal can simplify problems efficiently 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode

Day 39/100 – #100DaysOfCode 🚀 Solved LeetCode #2215 – Find the Difference of Two Arrays (Python). Today I practiced set operations to efficiently find distinct elements between two arrays. Approach: 1) Convert both arrays into sets to remove duplicates. 2) Find elements present in nums1 but not in nums2 using set difference. 3) Find elements present in nums2 but not in nums1. 4) Convert both results back to lists. 5) Return the final list of differences. Time Complexity: O(n + m) Space Complexity: O(n + m) Understanding how set operations simplify comparison problems 💪 #LeetCode #Python #DSA #Sets #Arrays #ProblemSolving #100DaysOfCode

Day 42/100 – #100DaysOfCode 🚀 Solved LeetCode #2574 – Left and Right Sum Differences (Python). Today I practiced prefix sum logic to calculate the absolute difference between left and right sums for each index. Approach: 1) Calculate the total sum of the array. 2) Initialize leftSum = 0. 3) Traverse the array. 4) For each index, compute rightSum = total - leftSum - nums[i]. 5) Calculate the absolute difference and append it to the result. 6) Update leftSum by adding nums[i]. Time Complexity: O(n) Space Complexity: O(n) Understanding prefix sum helps solve problems efficiently 💪 #LeetCode #Python #DSA #Arrays #PrefixSum #ProblemSolving #100DaysOfCode

Day 46/100 – #100DaysOfCode 🚀 Solved LeetCode #2733 – Neither Minimum nor Maximum (Python). Today I practiced sorting and simple array logic to find any element that is neither the minimum nor the maximum in the array. Approach: 1) Sort the array. 2) Check the length of the array. 3) If the size is less than or equal to 2, return -1 (no valid element exists). 4) Otherwise, return the second element as it will be neither min nor max. Time Complexity: O(n log n) Space Complexity: O(1) Understanding how sorting simplifies boundary-based problems 💪 #LeetCode #Python #DSA #Arrays #Sorting #ProblemSolving #100DaysOfCode

Day 31/100 – #100DaysOfCode 🚀 Solved LeetCode #1346 – Check If N and Its Double Exist (Python). Today I practiced brute-force comparison to check whether there exist two indices i and j such that arr[i] = 2 * arr[j]. Approach: 1) Use two nested loops to check all possible pairs. 2) Ensure that i ≠ j. 3) For each pair, check if arr[i] == 2 * arr[j]. 4) If condition is satisfied, return True. 5) If no such pair is found, return False. Time Complexity: O(n²) Space Complexity: O(1) Starting with brute force helps build understanding before optimization 💪 #LeetCode #Python #DSA #Arrays #BruteForce #ProblemSolving #100DaysOfCode

Day 43/100 – #100DaysOfCode 🚀 Solved LeetCode #2610 – Convert an Array Into a 2D Array With Conditions (Python). Today I practiced hashmap (frequency counting) to construct a 2D array based on given conditions. Approach: 1) Create a frequency map to count occurrences of each element. 2) Initialize an empty result list. 3) While the frequency map is not empty: 4)  Create a new row. 5)  Iterate through keys and add each number once to the row. 6)  Decrease its frequency and remove it if it becomes zero. 7) Add the row to the result. 8) Return the final 2D array. Time Complexity: O(n) Space Complexity: O(n) Learning how frequency maps help in structuring data efficiently 💪 #LeetCode #Python #DSA #HashMap #Arrays #ProblemSolving #100DaysOfCode

Day 49/60 #geekstreak60 Solved Segregate 0s and 1s today. A simple yet useful problem that teaches the power of the two-pointer approach — solving efficiently in O(n) time and O(1) space without extra memory. Small problems like these build strong problem-solving habits step by step. 💻 #geekstreak60 #npci #DSA #Python #ProblemSolving #CodingJourney

Day 33/100 – #100DaysOfCode 🚀 Solved LeetCode #1480 – Running Sum of 1d Array (Python). Today I practiced prefix sum logic to compute the running sum of an array. Approach: 1) Initialize an empty list to store the running sum. 2) Maintain a variable sum = 0. 3) Traverse the array and keep adding each element to sum. 4) Append the updated sum to the result list. 5) Return the final running sum array. Time Complexity: O(n) Space Complexity: O(n) Understanding prefix sums helps solve many array problems efficiently 💪 #LeetCode #Python #DSA #Arrays #PrefixSum #ProblemSolving #100DaysOfCode

Day 50/100 – #100DaysOfCode 🚀 Solved LeetCode #28 – Find the Index of the First Occurrence in a String (Python). Today I practiced string matching using a brute-force approach to find the first occurrence of a substring. Approach: 1) Traverse the main string (haystack). 2) For each index, try to match the substring (needle). 3) Compare characters one by one. 4) If all characters match, return the starting index. 5) If mismatch occurs, break and move to the next index. 6) If no match is found, return -1. Time Complexity: O(n × m) Space Complexity: O(1) Understanding basic string matching techniques step by step 💪 #LeetCode #Python #DSA #Strings #ProblemSolving #100DaysOfCode