Merge Intervals LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Merge Intervals | Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:

  • 1 <= intervals.length <= 104
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 104

Merge Intervals Solutions

Time: O(nlogn)

Space: O(n)O(n)
Space: O(n)

C++

class Solution {
 public:
  vector<vector<int>> merge(vector<vector<int>>& intervals) {
    vector<vector<int>> ans;

    sort(begin(intervals), end(intervals));

    for (const auto& interval : intervals)
      if (ans.empty() || ans.back()[1] < interval[0])
        ans.push_back(interval);
      else
        ans.back()[1] = max(ans.back()[1], interval[1]);

    return ans;
  }
};

Java

 class Solution {
  public int[][] merge(int[][] intervals) {
    List<int[]> ans = new ArrayList<>();

    Arrays.sort(intervals, (a, b) -> (a[0] - b[0]));

    for (int[] interval : intervals)
      if (ans.isEmpty() || ans.get(ans.size() - 1)[1] < interval[0])
        ans.add(interval);
      else
        ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], interval[1]);

    return ans.toArray(new int[ans.size()][]);
  }
}

Python

  class Solution:
  def merge(self, intervals: List[List[int]]) -> List[List[int]]:
    ans = []

    for interval in sorted(intervals):
      if not ans or ans[-1][1] < interval[0]:
        ans.append(interval)
      else:
        ans[-1][1] = max(ans[-1][1], interval[1])

    return ans

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