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leetcode_java/src/main/java/LeetCodeJava/Greedy/NonOverlappingIntervals.java
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| package LeetCodeJava.Greedy; | ||
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| // https://leetcode.com/problems/non-overlapping-intervals/description/ | ||
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| import java.util.Arrays; | ||
| import java.util.Stack; | ||
| import java.util.Comparator; | ||
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| public class NonOverlappingIntervals { | ||
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| // V0 | ||
| // IDEA : 2 POINTERS + sorting + intervals (modified by GPT) | ||
| // https://github.com/yennanliu/CS_basics/blob/master/leetcode_python/Greedy/non-overlapping-intervals.py | ||
| public int eraseOverlapIntervals(int[][] intervals) { | ||
| Arrays.sort(intervals, Comparator.comparingInt(a -> a[1])); | ||
| int cnt = 0; | ||
| int[] last = intervals[0]; | ||
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| for (int i = 1; i < intervals.length; i++) { | ||
| if (intervals[i][0] < last[1]) { | ||
| cnt++; | ||
| } else { | ||
| last[1] = Math.max(intervals[i][1], last[1]); | ||
| } | ||
| } | ||
| return cnt; | ||
| } | ||
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| // V0' | ||
| // TODO : implement it | ||
| // https://github.com/yennanliu/CS_basics/blob/master/leetcode_python/Greedy/non-overlapping-intervals.py | ||
| // public int eraseOverlapIntervals(int[][] intervals) { | ||
| // return 0; | ||
| // } | ||
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| // V1 | ||
| // IDEA : STACK | ||
| // https://leetcode.com/problems/non-overlapping-intervals/solutions/4683650/4-line-solution-easy-to-understand-java-stack/ | ||
| public int eraseOverlapIntervals_1(int[][] intervals) { | ||
| Arrays.sort(intervals,(a, b)->a[1]-b[1]); | ||
| Stack<int[]> stk = new Stack<>(); | ||
| int res=0; | ||
| for(int[] i : intervals){ | ||
| if(!stk.isEmpty() && i[0]< stk.peek()[1]){ | ||
| res++; | ||
| }else{ | ||
| stk.push(i); | ||
| } | ||
| } | ||
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| return res; | ||
| } | ||
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| // V2 | ||
| // IDEA : SORT | ||
| // https://leetcode.com/problems/non-overlapping-intervals/solutions/3786438/java-easy-solution-using-sorting-explained/ | ||
| public int eraseOverlapIntervals_2(int[][] intervals) { | ||
| // Sort by ending time | ||
| Arrays.sort(intervals, (a, b) -> Integer.compare(a[1], b[1])); | ||
| int prev = 0, count = 1; | ||
| // if end is same, sort by start, bigger start in front | ||
| for(int i = 0; i < intervals.length; i ++) { | ||
| if(intervals[i][0] >= intervals[prev][1]) { | ||
| prev = i; | ||
| count ++; | ||
| } | ||
| } | ||
| return intervals.length - count; | ||
| } | ||
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| // V3 | ||
| // https://leetcode.com/problems/non-overlapping-intervals/submissions/1202459919/ | ||
| public int eraseOverlapIntervals_3(int[][] in) { | ||
| Arrays.sort(in, (a,b)->a[1]-b[1]); | ||
| int res=-1, p[]=in[0]; | ||
| for(int[] i: in) | ||
| if(i[0]<p[1]) res++; | ||
| else p=i; | ||
| return res; | ||
| } | ||
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| } |