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feat: non-overlapping-intervals solution
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/**
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* 알고리즘 복잡도
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* - 시간 복잡도: O(nlogn)
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* - 공간 복잡도: O(1)
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*/
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function eraseOverlapIntervals(intervals: number[][]): number {
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intervals.sort((a, b) => a[1] - b[1])
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let end = intervals[0][1]
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let count = 0
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for(let i = 1; i < intervals.length; i++) {
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if(intervals[i][0] < end) {
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count++
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} else {
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end = intervals[i][1]
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}
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}
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return count
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}

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