Find least non-overlapping number from a given set of intervals

Given an array interval of pairs of integers representing the starting and ending points of the interval of size N. The task is to find the smallest non-negative integer which is a non-overlapping number from the given set of intervals.
Input constraints:

$1 \le N \le 10^{5}0 \le interval[i] \le 10^{9}$

Examples:

Input: interval = {{0, 4}, {6, 8}, {2, 3}, {9, 18}}
Output: 5
Explanation:
The smallest non-negative integer which is non-overlapping to all set of the intervals is 5.
Input: interval = {{0, 14}, {86, 108}, {22, 30}, {5, 17}}
Output: 18

Naive Approach:

Create a visited array of size MAX, and for every interval mark all value true from start to end.
Finally, iterate from 1 to MAX and find the smallest value which is not visited.

However, this approach will not work if the interval co-ordinates are up to 10 ⁹.
Time Complexity: O (N²)
Auxiliary Space: O (MAX)
Efficient Approach:

Instead of iterating from start to end just create a visited array and for each range, mark vis[start] = 1 and vis[end+1] = -1.
Take the prefix sum of the array.
Then iterate over the array to find the first integer with value 0.

Here is the implementation of the above approach:

C++

// C++ program to find the 
// least non-overlapping number 
// from a given set intervals 
 
#include <bits/stdc++.h> 
using namespace std; 
const int MAX = 1e5 + 5; 
 
// function to find the smallest 
// non-overlapping number 
void find_missing( 
    vector<pair<int, int> > interval) 
{ 
    // create a visited array 
    vector<int> vis(MAX); 
 
    for (int i = 0; i < interval.size(); ++i) { 
        int start = interval[i].first; 
        int end = interval[i].second; 
        vis[start]++; 
        vis[end + 1]--; 
    } 
 
    // find the first missing value 
    for (int i = 1; i < MAX; i++) { 
        vis[i] += vis[i - 1]; 
        if (!vis[i]) { 
            cout << i << endl; 
            return; 
        } 
    } 
} 
// Driver function 
int main() 
{ 
 
    vector<pair<int, int> > interval 
        = { { 0, 14 }, { 86, 108 }, 
            { 22, 30 }, { 5, 17 } }; 
    find_missing(interval); 
    return 0; 
} 

Java

// Java program to find the 
// least non-overlapping number 
// from a given set intervals 
class GFG{ 
   
static int MAX = (int) (1e5 + 5); 
static class pair
{ 
    int first, second; 
    public pair(int first, int second)  
    { 
        this.first = first; 
        this.second = second; 
    }    
} 
// function to find the smallest 
// non-overlapping number 
static void find_missing( 
    pair[] interval) 
{ 
    // create a visited array 
    int [] vis = new int[MAX];
 
    for (int i = 0; i < interval.length; ++i) 
    { 
        int start = interval[i].first; 
        int end = interval[i].second; 
        vis[start]++; 
        vis[end + 1]--; 
    } 
 
    // find the first missing value 
    for (int i = 1; i < MAX; i++) { 
        vis[i] += vis[i - 1]; 
        if (vis[i]==0) { 
            System.out.print(i +"\n"); 
            return; 
        } 
    } 
} 
// Driver function 
public static void main(String[] args) 
{ 
 
    pair []interval = {new pair( 0, 14 ), 
                       new pair( 86, 108 ), 
                       new pair( 22, 30 ), 
                       new pair( 5, 17 )}; 
    find_missing(interval); 
} 
} 
 
// This code is contributed by Rohit_ranjan

C#

// C# program to find the 
// least non-overlapping number 
// from a given set intervals 
using System;
 
class GFG{ 
 
static int MAX = (int)(1e5 + 5); 
class pair
{ 
    public int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
 
// Function to find the smallest 
// non-overlapping number 
static void find_missing(pair[] interval) 
{ 
     
    // Create a visited array 
    int [] vis = new int[MAX];
 
    for(int i = 0; i < interval.Length; ++i) 
    { 
        int start = interval[i].first; 
        int end = interval[i].second; 
         
        vis[start]++; 
        vis[end + 1]--; 
    } 
 
    // Find the first missing value 
    for(int i = 1; i < MAX; i++)
    { 
        vis[i] += vis[i - 1]; 
        if (vis[i] == 0)
        { 
            Console.Write(i + "\n"); 
            return; 
        } 
    } 
} 
 
// Driver code
public static void Main(String[] args) 
{ 
    pair []interval = { new pair(0, 14), 
                        new pair(86, 108), 
                        new pair(22, 30), 
                        new pair(5, 17) }; 
                         
    find_missing(interval); 
} 
} 
 
// This code is contributed by Amit Katiyar

Output:

Time Complexity: O (N)
Auxiliary Space: O (MAX)
However, this approach will also not work if the interval co-ordinates are up to 10 ⁹.
Efficient Approach:

Sort the range by their start-coordinate and for each next range.
Check if the starting point is greater than the maximum end-coordinate encountered so far, then a missing number can be found, and it will be previous_max + 1.

Illustration:
Consider the following example:
interval[][] = { { 0, 14 }, { 86, 108 }, { 22, 30 }, { 5, 17 } };
After sorting, interval[][] = { { 0, 14 }, { 5, 17 }, { 22, 30 }, { 86, 108 }};
Initial mx = 0 and after considering first interval mx = max(0, 15) = 15
Since mx = 15 and 15 > 5 so after considering second interval mx = max(15, 18) = 18
now 18 < 22 so 18 is least non-overlapping number.

Here is the implementation of the above approach:

C++

// C++ program to find the 
// least non-overlapping number 
// from a given set intervals 
 
#include <bits/stdc++.h> 
using namespace std; 
 
// function to find the smallest 
// non-overlapping number 
void find_missing( 
    vector<pair<int, int> > interval) 
{ 
    // Sort the intervals based on their 
    // starting value 
    sort(interval.begin(), interval.end()); 
 
    int mx = 0; 
 
    for (int i = 0; i < (int)interval.size(); ++i) { 
 
        // check if any missing vaue exist 
        if (interval[i].first > mx) { 
            cout << mx; 
            return; 
        } 
 
        else
            mx = max(mx, interval[i].second + 1); 
    } 
    // finally print the missing value 
    cout << mx; 
} 
// Driver function 
int main() 
{ 
 
    vector<pair<int, int> > interval 
        = { { 0, 14 }, { 86, 108 }, 
            { 22, 30 }, { 5, 17 } }; 
    find_missing(interval); 
    return 0; 
} 

Java

// Java program to find the 
// least non-overlapping number 
// from a given set intervals 
import java.util.*;
import java.io.*;
 
class GFG{
     
static class Pair implements Comparable<Pair>
{
    int start,end;
    Pair(int s, int e)
    {
        start = s;
        end = e;
    }
     
    public int compareTo(Pair p)
    {
        return this.start - p.start;
    }
}
 
// Function to find the smallest 
// non-overlapping number 
static void findMissing(ArrayList<Pair> interval) 
{ 
     
    // Sort the intervals based on their 
    // starting value 
    Collections.sort(interval); 
 
    int mx = 0; 
 
    for(int i = 0; i < interval.size(); ++i)
    { 
         
        // Check if any missing vaue exist 
        if (interval.get(i).start > mx)
        { 
            System.out.println(mx); 
            return; 
        } 
        else
            mx = Math.max(mx, interval.get(i).end + 1); 
    } 
     
    // Finally print the missing value 
    System.out.println(mx); 
} 
 
// Driver code
public static void main(String []args) 
{ 
    ArrayList<Pair> interval = new ArrayList<>();
    interval.add(new Pair(0, 14));
    interval.add(new Pair(86, 108));
    interval.add(new Pair(22, 30));
    interval.add(new Pair(5, 17));
     
    findMissing(interval); 
} 
}
 
// This code is contributed by Ganeshchowdharysadanala

Python3

# Python3 program to find the 
# least non-overlapping number 
# from a given set intervals 
 
# function to find the smallest 
# non-overlapping number 
def find_missing(interval):
 
    # Sort the intervals based  
    # on their starting value 
    interval.sort()
 
    mx = 0
 
    for i in range (len(interval)):
 
        # Check if any missing 
        # vaue exist 
        if (interval[i][0] > mx):
            print (mx)
            return
        
        else:
            mx = max(mx, 
                     interval[i][1] + 1)
 
    # Finally print the missing value 
    print (mx)
 
# Driver code
if __name__ == "__main__":
 
    interval = [[0, 14], [86, 108], 
                [22, 30], [5, 17]] 
    find_missing(interval);
    
# This code is contributed by Chitranayal

Output:

Time Complexity: O (N * logN)
Auxiliary Space: O (1)

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