Given an array of meeting time intervals where each element represents a meeting time as [start, end], determine whether a person can attend all the meetings without any overlap between meeting times.
intervals of size n x 2, where n is the number of meetings and each interval [start, end] represents a meeting starting at time start and ending at time endInput: intervals = [[83,99]]Output: trueExplanation: There is only one meeting, so there is no possibility of overlap.
Input: intervals = [[1,5],[5,10],[10,15]]Output: trueExplanation: The meetings are back-to-back but do not overlap.
Input: intervals = [[8,10],[1,3],[2,6],[15,18]]Output: falseExplanation: The meetings [1, 3] and [2, 6] overlap, so it is not possible to attend all meetings.
intervals.length <= 1000intervals[i].length == 2start < end <= 1,000,000
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Given an array of meeting time intervals where each element represents a meeting time as [start, end], determine whether a person can attend all the meetings without any overlap between meeting times.
intervals of size n x 2, where n is the number of meetings and each interval [start, end] represents a meeting starting at time start and ending at time endInput: intervals = [[83,99]]Output: trueExplanation: There is only one meeting, so there is no possibility of overlap.
Input: intervals = [[1,5],[5,10],[10,15]]Output: trueExplanation: The meetings are back-to-back but do not overlap.
Input: intervals = [[8,10],[1,3],[2,6],[15,18]]Output: falseExplanation: The meetings [1, 3] and [2, 6] overlap, so it is not possible to attend all meetings.
intervals.length <= 1000intervals[i].length == 2start < end <= 1,000,000Sponsor: SwagOverflowA true front end engineer never reuses, they only reinvent. This sleek black tee sums up the reality of constantly rebuilding component libraries instead of using the one from last month. Clean, sharp, and painfully accurate, it's the perfect fit for devs who can't resist a fresh start.
console.log() statements will appear here.