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mergeMeetings.js
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/*
Write a function mergeRanges() that takes an array of multiple meeting time ranges and returns an array of condensed ranges.
For example, given:
[
{startTime: 0, endTime: 1},
{startTime: 3, endTime: 5},
{startTime: 4, endTime: 8},
{startTime: 10, endTime: 12},
{startTime: 9, endTime: 10},
]
your function would return:
[
{startTime: 0, endTime: 1},
{startTime: 3, endTime: 8},
{startTime: 9, endTime: 12},
]
*/
function mergeRanges(meetings) {
// sort meetings by startTime: O(nlogn)
meetings.sort((firstRange, secondRange) => firstRange.startTime > secondRange.startTime);
let result = [];
// go through sorted meeting ranges: O(n)
for (let i = 1; i < meetings.length; i++) {
// check if current meeting overlaps with previous meeting
if (meetings[i].startTime <= meetings[i - 1].endTime) {
// merge current meeting with previous meeting
meetings[i].startTime = Math.min(meetings[i].startTime, meetings[i - 1].startTime);
meetings[i].endTime = Math.max(meetings[i].endTime, meetings[i - 1].endTime);
} else {
result.push(meetings[i - 1]);
}
// push last meeting into result
if (i === meetings.length - 1) {
result.push(meetings[meetings.length - 1]);
}
}
return result;
}
function mergeRangesAlt(meetings) {
meetings.sort((firstRange, secondRange) => firstRange.startTime > secondRange.startTime);
// initialize mergedMeetings with the earliest meeting
let mergedMeetings = [meetings[0]];
for (let i = 1; i < meetings.length; i++) {
let currentMeeting = meetings[i];
let lastMergedMeeting = mergedMeetings[mergedMeetings.length - 1];
// if the current meeting overlaps with the last merged meeting, use the later end time of the two
if (currentMeeting.startTime <= lastMergedMeeting.endTime) {
lastMergedMeeting.endTime = Math.max(lastMergedMeeting.endTime, currentMeeting.endTime);
// add the current meeting since it doesn't overlap
} else {
mergedMeetings.push(currentMeeting);
}
}
return mergedMeetings;
}