56. Merge Intervals LeetCode Solution

In this guide, you will get 56. Merge Intervals LeetCode Solution with the best time and space complexity. The solution to Merge Intervals problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Merge Intervals solution in C++
4. Merge Intervals solution in Java
5. Merge Intervals solution in Python
6. Additional Resources

Problem Statement of 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] 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

Complexity Analysis

• Time Complexity: O(\texttt{sort})
• Space Complexity: O(n)

56. Merge Intervals LeetCode Solution in C++

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

    ranges::sort(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;
  }
};
/* code provided by PROGIEZ */

56. Merge Intervals LeetCode Solution in Java

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

    Arrays.sort(intervals, (a, b) -> Integer.compare(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(int[][] ::new);
  }
}
// code provided by PROGIEZ

56. Merge Intervals LeetCode Solution 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
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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