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| 1 | +/* |
| 2 | +
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| 3 | +
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| 4 | +-* 57. Insert Interval *- |
| 5 | +
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| 6 | +You are given an array of non-overlapping intervals intervals where intervals[i] = [start-i, end-i] represent the start and the end of the ith interval and intervals is sorted in ascending order by start-i. You are also given an interval newInterval = [start, end] that represents the start and end of another interval. |
| 7 | +
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| 8 | +Insert newInterval into intervals such that intervals is still sorted in ascending order by start-i and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary). |
| 9 | +
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| 10 | +Return intervals after the insertion. |
| 11 | +
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| 12 | + |
| 13 | +
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| 14 | +Example 1: |
| 15 | +
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| 16 | +Input: intervals = [[1,3],[6,9]], newInterval = [2,5] |
| 17 | +Output: [[1,5],[6,9]] |
| 18 | +Example 2: |
| 19 | +
|
| 20 | +Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8] |
| 21 | +Output: [[1,2],[3,10],[12,16]] |
| 22 | +Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10]. |
| 23 | + |
| 24 | +
|
| 25 | +Constraints: |
| 26 | +
|
| 27 | +0 <= intervals.length <= 104 |
| 28 | +intervals[i].length == 2 |
| 29 | +0 <= start-i <= end-i <= 105 |
| 30 | +intervals is sorted by start-i in ascending order. |
| 31 | +newInterval.length == 2 |
| 32 | +0 <= start <= end <= 105 |
| 33 | +
|
| 34 | +*/ |
| 35 | + |
| 36 | +import 'dart:math'; |
| 37 | + |
| 38 | +class A { |
| 39 | + bool doesIntersect(List<int> intervalA, List<int> intervalB) { |
| 40 | + return intervalA[0] <= intervalB[1] && intervalB[0] <= intervalA[1]; |
| 41 | + } |
| 42 | + |
| 43 | + List<List<int>> insert(List<List<int>> intervals, List<int> newInterval) { |
| 44 | + List<List<int>> mergedIntervals = []; |
| 45 | + List<int> merged = [newInterval[0], newInterval[1]]; |
| 46 | + bool isInserted = false; |
| 47 | + |
| 48 | + for (List<int> interval in intervals) { |
| 49 | + if (interval[0] > merged[1]) { |
| 50 | + if (!isInserted) { |
| 51 | + mergedIntervals.add(merged); |
| 52 | + isInserted = true; |
| 53 | + } |
| 54 | + mergedIntervals.add(interval); |
| 55 | + } else { |
| 56 | + // does intersect |
| 57 | + if (doesIntersect(merged, interval)) { |
| 58 | + merged[0] = min(merged[0], interval[0]); |
| 59 | + merged[1] = max(merged[1], interval[1]); |
| 60 | + } else { |
| 61 | + // not intersecting |
| 62 | + mergedIntervals.add(interval); |
| 63 | + } |
| 64 | + } |
| 65 | + } |
| 66 | + |
| 67 | + // new interval is at the end |
| 68 | + if (!isInserted) { |
| 69 | + mergedIntervals.add(merged); |
| 70 | + isInserted = true; |
| 71 | + } |
| 72 | + |
| 73 | + // Java specific to copy to arr[][]... |
| 74 | + List.filled(mergedIntervals.length, 0).map((e) => List.filled(2, 0)); |
| 75 | + List<List<int>> res = List.filled(mergedIntervals.length, 0) |
| 76 | + .map((e) => List.filled(2, 0)) |
| 77 | + .toList(); |
| 78 | + for (int i = 0; i < mergedIntervals.length; i++) { |
| 79 | + res[i] = mergedIntervals.elementAt(i); |
| 80 | + } |
| 81 | + return res; |
| 82 | + } |
| 83 | +} |
| 84 | + |
| 85 | +class B { |
| 86 | + List<List<int>> insert(List<List<int>> intervals, List<int> newInterval) { |
| 87 | + List<List<int>> ans = []; |
| 88 | + List<int> toAdd = newInterval; |
| 89 | + for (int i = 0; i < intervals.length; i++) { |
| 90 | + if (intervals[i][0] > toAdd[1]) { |
| 91 | + ans.add(toAdd); |
| 92 | + toAdd = intervals[i]; |
| 93 | + } else if (intervals[i][1] >= toAdd[0]) |
| 94 | + toAdd = [ |
| 95 | + min(intervals[i][0], toAdd[0]), |
| 96 | + max(intervals[i][1], toAdd[1]) |
| 97 | + ]; |
| 98 | + else |
| 99 | + ans.add(intervals[i]); |
| 100 | + } |
| 101 | + ans.add(toAdd); |
| 102 | + return List.filled(ans.length, 0).map((e) => List.filled(2, 0)).toList(); |
| 103 | + } |
| 104 | + // List.filled(res.length, 0).map((e) => List.filled(2, 0)).toList(); |
| 105 | +} |
| 106 | + |
| 107 | +class C { |
| 108 | + List<List<int>> insert(List<List<int>> intervals, List<int> newInterval) { |
| 109 | + List<List<int>> ans = []; |
| 110 | + int idx = 0; |
| 111 | + while (idx < intervals.length && intervals[idx][0] < newInterval[0]) { |
| 112 | + ans.add(intervals[idx++]); |
| 113 | + } |
| 114 | + if (ans.length == 0 || newInterval[0] > ans.elementAt(ans.length - 1)[1]) { |
| 115 | + ans.add(newInterval); |
| 116 | + } else { |
| 117 | + List<int> lastInterval = ans.elementAt(ans.length - 1); |
| 118 | + lastInterval[1] = max(lastInterval[1], newInterval[1]); |
| 119 | + } |
| 120 | + while (idx < intervals.length) { |
| 121 | + List<int> lastInterval = ans.elementAt(ans.length - 1); |
| 122 | + if (lastInterval[1] >= intervals[idx][0]) { |
| 123 | + lastInterval[1] = max(lastInterval[1], intervals[idx++][1]); |
| 124 | + } else |
| 125 | + ans.add(intervals[idx++]); |
| 126 | + } |
| 127 | + return List.filled(ans.length, 0).map((e) => <int>[]).toList(); |
| 128 | + // return ans.toArray(new int[ans.size()][]); |
| 129 | + } |
| 130 | +} |
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