LeetCode Entry
3440. Reschedule Meetings for Maximum Free Time II
Max free time after moving 1 event
3440. Reschedule Meetings for Maximum Free Time II medium
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https://t.me/leetcode_daily_unstoppable/1045
Problem TLDR
Max free time after moving 1 event #medium #sorting
Intuition
// 0 17..19 24..25 41
// 17 5 16
// 2 1
* look for all free windows
* look around each event
* look if each event can fit into another window
* sort free windows
* track windows indices
Space optimization: do forward and backward pass to track the largest seen gap.
Approach
- we can do forward & backward pass in a single loop
Complexity
-
Time complexity: \(O(n)\)
-
Space complexity: \(O(n)\) or O(1)
Code
// 91ms
fun maxFreeTime(evt: Int, st: IntArray, et: IntArray): Int {
val free = IntArray(st.size)
val gaps = Array(3) { 0 to 0 }; gaps[0] = st[0] to 0
for (i in 0..<st.size) {
val s = st[i]; val e = et[i]
val prev = if (i == 0) 0 else et[i - 1]
val before = s - prev
val next = if (i < st.size - 1) st[i + 1] else evt
val after = next - e
free[i] = before + after
val g = (0..2).minBy { gaps[it].first }
if (gaps[g].first < after) gaps[g] = after to (i + 1)
}
var res = 0
for (i in free.indices) {
var fit = 0
val curr = et[i] - st[i]
for (g in 0..2) if (gaps[g].first >= curr && gaps[g].second !in i..i+1)
fit = curr
res = max(res, free[i] + fit)
}
return res
}
// 7ms
fun maxFreeTime(evt: Int, st: IntArray, et: IntArray): Int {
var left = 0; var right = 0; var res = 0; var j = st.size - 1
for (i in st.indices) {
var before = st[i] - if (i == 0) 0 else et[i - 1]
var after = (if (i < st.size - 1) st[i + 1] else evt) - et[i]
res = max(res, before + after + if (et[i] - st[i] <= left) et[i] - st[i] else 0)
left = max(left, before)
before = (if (j == st.size - 1) evt else st[j + 1]) - et[j]
after = st[j] - if (j > 0) et[j - 1] else 0
res = max(res, before + after + if (et[j] - st[j] <= right) et[j] - st[j] else 0)
right = max(right, before); j--
}
return res
}
// 4ms
pub fn max_free_time(evt: i32, st: Vec<i32>, et: Vec<i32>) -> i32 {
let (mut left, mut right, mut res, mut j) = (0, 0, 0, st.len() - 1);
for i in 0..st.len() {
let before = st[i] - if i == 0 { 0 } else { et[i - 1] };
let after = (if i < st.len() - 1 { st[i + 1] } else { evt }) - et[i];
res = res.max(before + after + if et[i] - st[i] <= left { et[i] - st[i] } else { 0 });
left = left.max(before);
let before = (if j == st.len() - 1 { evt } else { st[j + 1] }) - et[j];
let after = st[j] - if j > 0 { et[j - 1] } else { 0 };
res = res.max(before + after + if et[j] - st[j] <= right { et[j] - st[j] } else { 0 });
right = right.max(before); j -= 1
} res
}
// 0ms
int maxFreeTime(int evt, vector<int>& st, vector<int>& et) {
int left = 0, right = 0, res = 0, j = st.size() - 1;
for (int i = 0; i < st.size(); ++i) {
int before = st[i] - (i == 0 ? 0 : et[i - 1]);
int after = (i < st.size() - 1 ? st[i + 1] : evt) - et[i];
int dur = et[i] - st[i];
res = max(res, before + after + (dur <= left ? dur : 0));
left = max(left, before);
before = (j == st.size() - 1 ? evt : st[j + 1]) - et[j];
after = st[j] - (j > 0 ? et[j - 1] : 0);
dur = et[j] - st[j];
res = max(res, before + after + (dur <= right ? dur : 0));
right = max(right, before);
j--;
}
return res;
}