Merge Intervals Leetcode Solution

In this post, you will know how to solve the Merge Intervals Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

Problem

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] overlap, 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 Leetcode Solutions in 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

Merge Intervals Leetcode Solutions in CPP

class Solution {
 public:
  vector<vector<int>> merge(vector<vector<int>>& intervals) {
    vector<vector<int>> ans;
    sort(begin(intervals), end(intervals));
    for (const vector<int>& 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;
  }
};

Merge Intervals Leetcode Solutions in 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()][]);
  }
}

Note: This problem Merge Intervals is generated by Leetcode but the solution is provided by  BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

Next: Path Sum II Leetcode Solution