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In this video we solve LeetCode 435 β "Non-overlapping Intervals" using a classic greedy algorithm in C++. The goal: given a list of intervals, remove the minimum number so that the rest do not overlap. Key idea: β’ Sort intervals by their END time. β’ Always keep the interval that ends earliest. β’ Whenever the next interval overlaps with the one we kept, we "remove" it (increase the removal count) and keep the one with the smaller end. This is the same pattern as the activity selection / interval scheduling problem. Time complexity: β’ Sorting: O(n log n) β’ One pass over the intervals: O(n) β overall O(n log n) time, O(1) extra space (ignoring sorting). What youβll learn in this video: β’ How to detect overlapping vs non-overlapping intervals β’ Why sorting by end time gives the optimal greedy choice β’ Step-by-step C++ implementation β’ Common pitfalls and edge cases β’ How this pattern appears in many other interval problems If this video helps you, please LIKE, SUBSCRIBE, and check my other LeetCode videos in C++! #leetcode #leetcodecoding #leetcodesolution #leetcodechallenge #faang #fang #computerscience #cs #codingtiktok #coding #codinglife #boostofhope #fyp #codinginterview #softwareengineer #swe #tech #foryou #python #learntocode #google #techtok #programming #algorithms #learntocode
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