Non-overlapping intervals in an array

Last Updated : 26 Mar, 2025

Given a 2d array arr[][] of time intervals, where each interval is of the form [start, end]. The task is to determine all intervals from the given array that do not overlap with any other interval in the set. If no such interval exists, return an empty list.

Examples:

Input: arr[] = [[1, 3], [2, 4], [3, 5], [7, 9]]
Output: [[5, 7]]
Explanation: The only interval which doesn’t overlaps with the other intervals is [5, 7].
Input: arr[][] = [[1, 3], [2, 6], [8, 10], [15, 18]]
Output: [[6, 8], [10, 15]]
Explanation: There are two intervals which don’t overlap with other intervals are [6, 8], [10, 15].
Input: arr[][] = [[1, 3], [9, 12], [2, 4], [6, 8]]
Output: [[4, 6], [8, 9]]
Explanation: There are two intervals which don’t overlap with other intervals are [4, 6], [8, 9].

Approach:

The idea is to sort the given time intervals according to starting time and if the consecutive intervals don't overlap then the difference between them is the free interval.

Steps to implement the above idea:

Sort the given set of intervals according to starting time.
Traverse all the set of intervals and check whether the consecutive intervals overlaps or not.
If the intervals(say interval a & interval b) do no't overlap then the set of pairs form by [a.end, b.start] is the non-overlapping interval.
If the intervals overlaps, then check for next consecutive intervals.

C++

#include <bits/stdc++.h> using namespace std;  vector<vector<int>> findGaps(vector<vector<int>> &arr) {     if (arr.empty()) return {};          // Sort intervals by start time     sort(arr.begin(), arr.end());          vector<vector<int>> res;     for (int i = 1; i < arr.size(); i++) {                  // End of previous and start of current are         // are checked and if non-overlapping, then         // added to the result         if (arr[i - 1][1] < arr[i][0]) {             res.push_back({arr[i - 1][1], arr[i][0]});         }     }     return res; }  int main() {     vector<vector<int>> arr = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};          for (auto &gap : findGaps(arr)) {         cout << "[" << gap[0] << ", " << gap[1] << "] ";     }     return 0; }

Java

import java.util.ArrayList; import java.util.Arrays; import java.util.List;  public class Main {     public static List<int[]> findGaps(int[][] arr) {         if (arr.length == 0) {             return new ArrayList<>();         }                  // Sort intervals by start time         Arrays.sort(arr, (a, b) -> Integer.compare(a[0], b[0]));                  List<int[]> res = new ArrayList<>();         for (int i = 1; i < arr.length; i++) {                          // End of previous and start of current are             // are checked and if non-overlapping, then             // added to the result             if (arr[i - 1][1] < arr[i][0]) {                 res.add(new int[]{arr[i - 1][1], arr[i][0]});             }         }         return res;     }          public static void main(String[] args) {         int[][] arr = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};                  for (int[] gap : findGaps(arr)) {             System.out.print("[" + gap[0] + ", " + gap[1] + "] ");         }     } }

Python

def find_gaps(arr):     if not arr:         return []          # Sort intervals by start time     arr.sort()          res = []     for i in range(1, len(arr)):                  # End of previous and start of current are         # are checked and if non-overlapping, then         # added to the result         if arr[i - 1][1] < arr[i][0]:             res.append([arr[i - 1][1], arr[i][0]])     return res  if __name__ == '__main__':     arr = [[1, 3], [2, 6], [8, 10], [15, 18]]          for gap in find_gaps(arr):         print(f'[{gap[0]}, {gap[1]}]', end=' ')

using System; using System.Collections.Generic; using System.Linq;  public class MainClass {     public static List<int[]> FindGaps(int[][] arr) {         if (arr.Length == 0) {             return new List<int[]>();         }                  // Sort intervals by start time         Array.Sort(arr, (a, b) => a[0].CompareTo(b[0]));                  List<int[]> res = new List<int[]>();         for (int i = 1; i < arr.Length; i++) {                          // End of previous and start of current are             // checked and if non-overlapping, then             // added to the result             if (arr[i - 1][1] < arr[i][0]) {                 res.Add(new int[] { arr[i - 1][1], arr[i][0] });             }         }         return res;     }          public static void Main(string[] args) {         int[][] arr = new int[][] { new int[] { 1, 3 }, new int[] { 2, 6 }, new int[] { 8, 10 }, new int[] { 15, 18 } };                  foreach (int[] gap in FindGaps(arr)) {             Console.Write("[" + gap[0] + ", " + gap[1] + "] ");         }     } }

JavaScript

function findGaps(arr) {     if (arr.length === 0) {         return [];     }          // Sort intervals by start time     arr.sort((a, b) => a[0] - b[0]);          const res = [];     for (let i = 1; i < arr.length; i++) {                  // End of previous and start of current are         // checked and if non-overlapping, then         // added to the result         if (arr[i - 1][1] < arr[i][0]) {             res.push([arr[i - 1][1], arr[i][0]]);         }     }     return res; }  const arr = [[1, 3], [2, 6], [8, 10], [15, 18]];  for (const gap of findGaps(arr)) {     console.log(`[${gap[0]}, ${gap[1]}]`); }

Output

[[6, 8], [10, 15]]

Time Complexity: O(n*log(n)), where n is the number of set of intervals.
Auxiliary Space: O(n)

Non-overlapping intervals in an array

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Non-overlapping intervals in an array

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