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:

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 coordinates are up to 10 ⁹. 

Time Complexity: O (N ²)
Auxiliary Space: O (MAX)

Efficient Approach:

1. 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.
2. Take the prefix sum of the array.
3. 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
import java.io.*;
public 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

Python3

# Python3 program to find the
# least non-overlapping number
# from a given set intervals
MAX = int(1e5 + 5)

# Function to find the smallest
# non-overlapping number
def find_missing(interval):

    # Create a visited array
    vis = [0] * (MAX)

    for i in range(len(interval)):
        start = interval[i][0]
        end = interval[i][1]
        vis[start] += 1
        vis[end + 1] -= 1

    # Find the first missing value
    for i in range(1, MAX):
        vis[i] += vis[i - 1]

        if (vis[i] == 0):
            print(i)
            return

# Driver code
interval = [ [ 0, 14 ], [ 86, 108 ],
             [ 22, 30 ], [ 5, 17 ] ]

find_missing(interval)

# This code is contributed by divyeshrabadiya07

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

Javascript

<script>

// Javascript program to find the
// least non-overlapping number
// from a given set intervals
var MAX = 100005;

// function to find the smallest
// non-overlapping number
function find_missing(  interval)
{
    // create a visited array
    var vis = Array(MAX).fill(0);

    for (var i = 0; i < interval.length; ++i) {
        var start = interval[i][0];
        var end = interval[i][1];
        vis[start]++;
        vis[end + 1]--;
    }

    // find the first missing value
    for (var i = 1; i < MAX; i++) {
        vis[i] += vis[i - 1];
        if (!vis[i]) {
            document.write( i + "<br>");
            return;
        }
    }
}
// Driver function
var interval
    = [ [ 0, 14 ], [ 86, 108 ],
        [ 22, 30 ], [ 5, 17 ] ];
find_missing(interval);

</script>

Output

18

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

Efficient Approach:

1. Sort the range by their start-coordinate and for each next range.
2. 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 value 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 value 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
        # value 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

C#

// C# program to find the
// least non-overlapping number
// from a given set intervals
using System;
using System.Collections.Generic;
class GFG{

class Pair : IComparable<Pair>
{
    public int start,end;
    public 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(List<Pair> interval)
{

    // Sort the intervals based on their
    // starting value
    interval.Sort();
    int mx = 0;
    for(int i = 0; i < interval.Count; ++i)
    {

        // Check if any missing value exist
        if (interval[i].start > mx)
        {
            Console.WriteLine(mx);
            return;
        }
        else
            mx = Math.Max(mx, interval[i].end + 1);
    }

    // Finally print the missing value
    Console.WriteLine(mx);
}

// Driver code
public static void Main(String []args)
{
    List<Pair> interval = new List<Pair>();
    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 shikhasingrajput

Javascript

<script>
// Javascript program to find the
// least non-overlapping number
// from a given set intervals

// Function to find the smallest
// non-overlapping number
function findMissing(interval)
{

    // Sort the intervals based on their
    // starting value
    interval.sort(function(a,b){return a[0]-b[0];});
    let mx = 0;
    for(let i = 0; i < interval.length; ++i)
    {

        // Check if any missing value exist
        if (interval[i][0] > mx)
        {
            document.write(mx+"<br>");
            return;
        }
        else
            mx = Math.max(mx, interval[i][1] + 1);
    }

    // Finally print the missing value
    document.write(mx);
}

// Driver code
let interval = [[0, 14], [86, 108],
                [22, 30], [5, 17]];

findMissing(interval);

// This code is contributed by avanitrachhadiya2155.
</script>

Output

18

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