Pattern 1: 7. Merge Intervals – Arrays

Description: Merge overlapping intervals.

Java

public int[][] merge(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
List<int[]> result = new ArrayList<>();
for (int[] interval : intervals) {
if (result.isEmpty() || result.get(result.size() - 1)[1] <
interval[0]) {
result.add(interval);
} else {

result.get(result.size() - 1)[1] = Math.max(result.get(result.size() - 1)[1], interval[1]); } } return result.toArray(new int[result.size()][]); }

C++

vector<vector<int>> merge(vector<vector<int>>& intervals)
{ sort(intervals.begin(), intervals.end());
vector<vector<int>> result;
for (auto& interval : intervals) {
if (result.empty() || result.back()[1] < interval[0])
{ result.push_back(interval);
} else {
result.back()[1] = max(result.back()[1], interval[1]);
}
}

return result; }

Python

def merge(intervals):

intervals.sort(key=lambda x: x[0]) result = [] for interval in intervals: if not result or result[-1][1] < interval[0]: result.append(interval) else: result[-1][1] = max(result[-1][1], interval[1]) return result