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Add and organize new interval problems and notes
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‎README.md

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@@ -19,6 +19,7 @@ This project offers a curated collection of notes and resources covering fundame
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- [Dynamic Programming](./topics/dynamic-programming.md)
2020
- [Greedy](./topics/greedy.md)
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- [Shortest Path](./topics/shortest-path.md)
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- [Interval](./topics/interval.md)
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- [Backtracking](./topics/backtracking.md)
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## Problems & Solutions

‎leetcode/56.merge-intervals.md

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@@ -18,33 +18,34 @@ The idea is that we sort the array first by the `start`, so that we can iterate
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[1, 10], [15, 18]
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```
2020

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For two intervals `merged` and `toMerge`, there are four cases:
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* **Fully Overlapped**: To merge.
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* **Partially Overlapped**: To merge.
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* **Adjacent**: It's still overlapped, to merge.
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* **No Overlap**: To keep both, `[x, y]` and `[z, w]` => `[x, y]` and `[z, w]`, and move to the next interval.
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To merge the two intervals, we just need to keep the smaller `start` and the larger `end`, that is `min(merged[0], toMerge[0]), max(merged[1], toMerge[1])`
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```js
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// Fully Overlapped, just keep the longer interval [1, 4]
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[1 4]
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[2 3]
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// Or
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[2 3]
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[1 4]
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// Partially Overlapped, merged, [1, 5]
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[1 4]
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[3 5]
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// Or
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[3 5]
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[1 4]
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// Adjacent, extended, [1, 5]
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[1 4][4 5]
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// No Overlap, keep both, [1, 4][6, 8]
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// 1. Fully Overlapped: just keep the longer interval [1, 4]
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1 2 3 4 5 6 7 8
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|--------|
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|--|
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// Merged:
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1--------4
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// 2. Partially Overlapped: merged, [1, 5]
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1 2 3 4 5 6 7 8
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|--------|
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|-----|
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// Merged:
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1-----------5
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// 3. Adjacent: extended, [1, 5]
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1 2 3 4 5 6 7 8
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|--------|
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|--|
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// Merged:
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1-----------5
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// 4. No Overlap: keep both, [1, 4][6, 8]
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[1 4] [6 8]
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```
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results.add(merged)
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for (i in 1 until intervals.size) {
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val toMerge = intervals[i]
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// Overlapped
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// |---| merged
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// |-------| toMerge
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// |-------| toMerge
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//
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// We don't have to consider this case, because we have sorted the intervals by `start`.
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// |-----| merged
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// |------| toMerge
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// Or we can use the following function to check if overlapped.
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// if (overlapped(merged, toMerge)) {
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if (toMerge[0] <= merged[1]) {
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if (isOverlapped(merged, toMerge)) {
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merged[0] = minOf(merged[0], toMerge[0])
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merged[1] = maxOf(merged[1], toMerge[1])
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} else {
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for (i in 1 until intervals.size) {
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val toMerge = intervals[i]
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if (overlapped(merged, toMerge)) {
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if (isOverlapped(merged, toMerge)) {
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merged[0] = minOf(merged[0], toMerge[0])
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merged[1] = maxOf(merged[1], toMerge[1])
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} else {
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results.add(merged)
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return results.toTypedArray()
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}
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private fun overlapped(merged: IntArray, toMerge: IntArray): Boolean {
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// Not (Not overlapped) = Overlapped
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return !(
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merged[1] < toMerge[0] ||
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toMerge[1] < merged[0]
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)
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}
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private fun overlapped(merged: IntArray, toMerge: IntArray): Boolean {
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// Case 1:
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// |----| merged
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// |------| toMerge
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// Case 2:
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// |------|
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// |----|
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// Case 3:
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// |---|
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// |---------|
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// Case 4:
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// |---------|
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// |---|
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return (
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toMerge[0] in merged[0]..merged[1] || // Case 1
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merged[1] in toMerge[0]..toMerge[1] || // Case 2
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(toMerge[0] <= merged[0] && merged[1] <= toMerge[1]) || // Case 3
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(merged[0] <= toMerge[0] && toMerge[1] <= merged[1]) // Case 4
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)
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}
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```
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* **Time Complexity**: `O(n lg n)` for sorting, `O(n)` to merge, total time is `O(n lg n)`.
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* **Space Complexity**: `O(lg n)` for sorting.
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## Pitfalls
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There is a pitfall to check if two intervals are overlapped, we can't just check if `merged[0] < toMerge[1] || toMerged[0] < merged[1]`.
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```js
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|-------| merged
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|-------| toMerge
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|-------| merged
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|-------| toMerge
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```
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It failed in the following cases:
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```js
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// toMerge[1] < merged[0], but they are overlapped.
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|-------| merged
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|-------| toMerge
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// toMerge[0] < merged[1], but they are not overlapped.
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|-------| merged
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|-------| toMerge
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```
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> See [729. My Calendar I](../leetcode/729.my-calendar-i.md)
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* **Space Complexity**: `O(lg n)` for sorting.

‎leetcode/57.insert-interval.md

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return results.toTypedArray()
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}
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/**
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a: |----|
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b: |----|
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a: |-----|
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b: |----|
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*/
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private fun isOverlapped(a: IntArray, b: IntArray): Boolean {
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return !(a[1] < b[0] || b[1] < a[0])
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}
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```
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Or equivalently, we can use for loop to iterate through the intervals. (but more complex, not recommended)
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if (!isAdded) results.add(merged)
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return results.toTypedArray()
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}
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/**
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a = [1,3]
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b = [2,5]
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*/
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private fun isOverlapped(a: IntArray, b: IntArray): Boolean {
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// Above: a
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// Below: b
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// |---|
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// |------|
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return b[0] in a[0]..a[1] ||
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// |----|
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// |----|
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a[0] in b[0]..b[1] ||
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// |----------|
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// |--|
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a[0] <= b[0] && b[1] <= a[1] ||
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// |--|
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// |----------|
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b[0] <= a[0] && a[1] <= b[1]
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}
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```
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* **Time Complexity**: `O(n)`.

‎leetcode/729.my-calendar-i.md

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## [729. My Calendar I](https://leetcode.com/problems/my-calendar-i/description/)
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## Pitfalls
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```js
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// Pitfall:
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if (interval[0] < end || start < interval[1]) return overlapped
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interval[0]
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^
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|-----| interval
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|-----| [start, end)
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*
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end
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|-----| // Also means `interval[0] < end`, but it's not overlapping.
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*
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end
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interval[1]
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^
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|-----| interval
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|-----| [start, end)
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*
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start
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|-| // Also means `start < interval[1]`, but it's not overlapping.
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*
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start
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```
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> See [56. Merge Intervals](../leetcode/56.merge-intervals.md)
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### Brute Force
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```kotlin
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class MyCalendar() {
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for (interval in intervals) {
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val noConflict = interval[1] <= start || end <= interval[0]
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if (!noConflict) return false
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// Or equivalent to
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// |-----| interval
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// |------------| start <= interval[0] && interval[1] <= end
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// Partical overlapped at `start`
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// |-----| interval
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// |--------- interval[0] <= start
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// |-----| start < interval[1]
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// Partial overlapped at `end`
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// |-----| interval
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// |----| interval[0] < end
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// |---| end <= interval[1]
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if (
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(start <= interval[0] && interval[1] <= end) ||
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start in interval[0] until interval[1] ||
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end in interval[0] + 1..interval[1]
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) return false
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}
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intervals.add(intArrayOf(start, end))
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return true

‎problems/interval-problems.md

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## Intervals
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### Basic
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| Problem | Difficulty |
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|------------------|------------|
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|[56. Merge Intervals](../leetcode/56.merge-intervals.md)|Medium|
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|[57. Insert Interval](../leetcode/57.insert-interval.md)|Medium|
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|**[56. Merge Intervals](../leetcode/56.merge-intervals.md)**|Medium|
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|**[57. Insert Interval](../leetcode/57.insert-interval.md)**|Medium|
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|**[986. Interval List Intersections](../leetcode/986.interval-list-intersections.md)**|Medium|
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|[729. My Calendar I](../leetcode/729.my-calendar-i.md)|Medium|
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|[452. Minimum Number of Arrows to Burst Balloons](../leetcode/452.minimum-number-of-arrows-to-burst-balloons.md)|Medium|
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|[1094. Car Pooling](../leetcode/1094.car-pooling.md)|Medium|
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|**[729. My Calendar I](../leetcode/729.my-calendar-i.md)**|Medium|
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|[55. Jump Game](../leetcode/55.jump-game.md)|Medium|
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|**[1094. Car Pooling](../leetcode/1094.car-pooling.md)**|Medium|
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|[2779. Maximum Beauty of an Array After Applying Operation](../leetcode/2779.maximum-beauty-of-an-array-after-applying-operation.md)|Medium (1638)|
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### Resources
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* Study guidelines: https://leetcode.com/discuss/study-guide/2166045/ (Problem listing: https://leetcode.com/problem-list/mzw3cyy6/)
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* [Greedy problem listing - 二、区间贪心](https://huxulm.github.io/lc-rating/list/greedy)
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* Video: https://www.bilibili.com/video/BV1qY411n7Qs/?vd_source=2f62e0e1762a6703c96771d3baa35968
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> * https://leetcode.com/problems/points-that-intersect-with-cars/description/ 1229
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> * https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered/description/ 1307
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> * https://leetcode.com/problems/maximum-population-year/description/ 1370
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> * https://leetcode.com/problems/count-days-without-meetings/description/ 1483
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> * **https://leetcode.com/problems/my-calendar-ii/description/** 2k m
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> * https://leetcode.com/problems/count-ways-to-group-overlapping-ranges/description/ 1631
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> * https://leetcode.com/problems/employee-free-time/description/ 1710 (Premium)
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> * **https://leetcode.com/problems/number-of-flowers-in-full-bloom/** 2022
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> * **https://leetcode.com/problems/my-calendar-iii/description/** h
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> * **https://leetcode.com/problems/maximum-profit-in-job-scheduling/**
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> * https://leetcode.com/problems/the-skyline-problem/description/ h
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> * https://leetcode.cn/problems/my-calendar-i/solutions/1646264/by-jiang-hui-4-pyfn/
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### 2.1 不相交区间
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| Problem | Difficulty |
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|------------------|------------|
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> * https://leetcode.com/problems/maximum-length-of-pair-chain/description/ m
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> * https://leetcode.com/problems/non-overlapping-intervals/description/ 8k m
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> * https://leetcode.com/problems/number-of-flowers-in-full-bloom 1k h
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> * https://leetcode.com/problems/points-that-intersect-with-cars/description/
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> * https://leetcode.com/problems/my-calendar-ii/description/ 2k m
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> * https://leetcode.com/problems/maximum-profit-in-job-scheduling/
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> * https://leetcode.com/problems/number-of-flowers-in-full-bloom/
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> * https://leetcode.com/problems/maximum-population-year/description/ 1k e
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### 2.2 区间分组
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| Problem | Difficulty |
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|------------------|------------|
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> * https://leetcode.com/problems/divide-intervals-into-minimum-number-of-groups 1713
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> * https://leetcode.com/problems/the-skyline-problem/description/ h
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> * https://leetcode.com/problems/meeting-rooms-ii/ (Premium)
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### 2.3 区间选点
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| Problem | Difficulty |
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|------------------|------------|
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|**[452. Minimum Number of Arrows to Burst Balloons](../leetcode/452.minimum-number-of-arrows-to-burst-balloons.md)**|Medium|
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### 2.4 区间覆盖
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| Problem | Difficulty |
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|------------------|------------|
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|[45. Jump Game II](../leetcode/45.jump-game-ii.md)|Medium|
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> * https://leetcode.com/problems/video-stitching/description/ 1746
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> * https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/description/ 1885
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### 2.6 其他区间贪心
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| Problem | Difficulty |
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|------------------|------------|
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> * https://leetcode.com/problems/remove-covered-intervals/description/ 1375
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> * https://leetcode.com/problems/two-best-non-overlapping-events/description/ 1883
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> * https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended/description/ 2015
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### Resources
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* ✅ [Study guidelines](https://leetcode.com/discuss/study-guide/2166045/)
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* ✅ [Greedy problem listing - 二、区间贪心](https://huxulm.github.io/lc-rating/list/greedy)
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* [Video](https://www.bilibili.com/video/BV1qY411n7Qs/?vd_source=2f62e0e1762a6703c96771d3baa35968)
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* [Segment Tree](https://leetcode.cn/problems/my-calendar-i/solutions/1646264/by-jiang-hui-4-pyfn/)

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