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

Last Updated : 29 Mar, 2023

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 coordinates 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 
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

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:

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 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

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

Another Approach:

Approach:

Sort the intervals in ascending order based on their end points. This can be done using any efficient sorting algorithm such as merge sort or quicksort.

Initialize a variable “last” to the smallest possible integer value. This variable will keep track of the end point of the last interval that was added to the output.

Iterate through each interval in the sorted list. If the start point of the current interval is greater than “last”, add the end point of the current interval to the output and update “last” to be the end point of the current interval.

Return the value of “last” as the least non-overlapping number.

C++

#include <bits/stdc++.h>
using namespace std;
 
// Interval structure
struct Interval {
    int start;
    int end;
};
 
// Comparison function to sort intervals based on their end points
bool compare(const Interval& a, const Interval& b) {
    return a.end < b.end;
}
 
// Function to find the least non-overlapping number from a given set of intervals
int findLeastNonOverlappingNumber(Interval intervals[], int n) {
    // Sort intervals in ascending order based on their end points
    sort(intervals, intervals + n, compare);
 
    // Initialize last to the smallest possible integer value
    int last = -2147483648;
 
    // Iterate through each interval
    for (int i = 0; i < n; i++) {
        // If the start point of the current interval is greater than last,
        // add the end point of the current interval to the output and update last
        if (intervals[i].start > last) {
            last = intervals[i].end;
        }
    }
 
    // Return the value of last as the least non-overlapping number
    return last;
}
 
// Driver code
int main() {
    // Sample input
    Interval intervals[] = {{1, 3}, {2, 4}, {3, 6}, {5, 7}, {7, 8}};
    int n = sizeof(intervals) / sizeof(intervals[0]);
 
    // Find the least non-overlapping number from the input set of intervals
    int leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals, n);
 
    // Print the result
    cout << "The least non-overlapping number is " << leastNonOverlappingNumber << endl;
 
    return 0;
}

C

#include <stdio.h>
#include <stdlib.h>
 
// Interval structure
struct Interval {
    int start;
    int end;
};
 
// Comparison function to sort intervals based on their end points
int compare(const void* a, const void* b) {
    return ((struct Interval*)a)->end - ((struct Interval*)b)->end;
}
 
// Function to find the least non-overlapping number from a given set of intervals
int findLeastNonOverlappingNumber(struct Interval intervals[], int n) {
    // Sort intervals in ascending order based on their end points
    qsort(intervals, n, sizeof(struct Interval), compare);
 
    // Initialize last to the smallest possible integer value
    int last = -2147483648;
 
    // Iterate through each interval
    for (int i = 0; i < n; i++) {
        // If the start point of the current interval is greater than last,
        // add the end point of the current interval to the output and update last
        if (intervals[i].start > last) {
            last = intervals[i].end;
        }
    }
 
    // Return the value of last as the least non-overlapping number
    return last;
}
 
// Driver code
int main() {
    // Sample input
    struct Interval intervals[] = {{1, 3}, {2, 4}, {3, 6}, {5, 7}, {7, 8}};
    int n = sizeof(intervals) / sizeof(intervals[0]);
 
    // Find the least non-overlapping number from the input set of intervals
    int leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals, n);
 
    // Print the result
    printf("The least non-overlapping number is %d\n", leastNonOverlappingNumber);
 
    return 0;
}

Java

import java.util.Arrays;
import java.util.Comparator;
 
// Interval class
class Interval {
    int start;
    int end;
 
    public Interval(int start, int end) {
        this.start = start;
        this.end = end;
    }
}
 
public class Main {
    // Function to find the least non-overlapping number from a given set of intervals
    public static int findLeastNonOverlappingNumber(Interval[] intervals) {
        // Sort intervals in ascending order based on their end points
        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval a, Interval b) {
                return a.end - b.end;
            }
        });
 
        // Initialize last to the smallest possible integer value
        int last = Integer.MIN_VALUE;
 
        // Iterate through each interval
        for (int i = 0; i < intervals.length; i++) {
            // If the start point of the current interval is greater than last,
            // add the end point of the current interval to the output and update last
            if (intervals[i].start > last) {
                last = intervals[i].end;
            }
        }
 
        // Return the value of last as the least non-overlapping number
        return last;
    }
 
    // Driver code
    public static void main(String[] args) {
        // Sample input
        Interval[] intervals = {new Interval(1, 3), new Interval(2, 4), new Interval(3, 6), new Interval(5, 7), new Interval(7, 8)};
 
        // Find the least non-overlapping number from the input set of intervals
        int leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals);
 
        // Print the result
        System.out.println("The least non-overlapping number is " + leastNonOverlappingNumber);
    }
}

Python3

from typing import List, Tuple
 
# Interval structure
class Interval:
    def __init__(self, start: int, end: int):
        self.start = start
        self.end = end
 
# Function to find the least non-overlapping number from a given set of intervals
def findLeastNonOverlappingNumber(intervals: List[Interval]) -> int:
    # Sort intervals in ascending order based on their end points
    intervals.sort(key=lambda x: x.end)
 
    # Initialize last to the smallest possible integer value
    last = float('-inf')
 
    # Iterate through each interval
    for interval in intervals:
        # If the start point of the current interval is greater than last,
        # add the end point of the current interval to the output and update last
        if interval.start > last:
            last = interval.end
 
    # Return the value of last as the least non-overlapping number
    return last
 
# Driver code
if __name__ == '__main__':
    # Sample input
    intervals = [Interval(1, 3), Interval(2, 4), Interval(3, 6), Interval(5, 7), Interval(7, 8)]
 
    # Find the least non-overlapping number from the input set of intervals
    leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals)
 
    # Print the result
    print(f'The least non-overlapping number is {leastNonOverlappingNumber}')

C#

using System;
using System.Collections.Generic;
 
// Make the Interval class public
public class Interval {
    public int start;
    public int end;
 
    public Interval(int start, int end)
    {
        this.start = start;
        this.end = end;
    }
}
 
public class Program {
    // Make the FindLeastNonOverlappingNumber method public
    // and static
    public static int
        findLeastNonOverlappingNumber(Interval[] intervals)
    {
        Array.Sort(intervals, (a, b) = > a.end - b.end);
 
        int last = int.MinValue;
 
        foreach(Interval interval in intervals)
        {
            if (interval.start > last) {
                last = interval.end;
            }
        }
 
        return last;
    }
 
    static void Main(string[] args)
    {
        Interval[] intervals
            = { new Interval(1, 3), new Interval(2, 4),
                new Interval(3, 6), new Interval(5, 7),
                new Interval(7, 8) };
        int leastNonOverlappingNumber
            = findLeastNonOverlappingNumber(intervals);
        Console.WriteLine(
            "The least non-overlapping number is "
            + leastNonOverlappingNumber);
    }
}

Javascript

// Interval structure
class Interval {
    constructor(start, end) {
        this.start = start;
        this.end = end;
    }
}
 
// Comparison function to sort intervals based on their end points
function compare(a, b) {
    return a.end - b.end;
}
 
// Function to find the least non-overlapping number from a given set of intervals
function findLeastNonOverlappingNumber(intervals) {
    const n = intervals.length;
    // Sort intervals in ascending order based on their end points
    intervals.sort(compare);
 
    // Initialize last to the smallest possible integer value
    let last = -2147483648;
 
    // Iterate through each interval
    for (let i = 0; i < n; i++) {
        // If the start point of the current interval is greater than last,
        // add the end point of the current interval to the output and update last
        if (intervals[i].start > last) {
            last = intervals[i].end;
        }
    }
 
    // Return the value of last as the least non-overlapping number
    return last;
}
 
// Driver code
(function() 
{
 
    // Sample input
    const intervals = [new Interval(1, 3), new Interval(2, 4), new Interval(3, 6), new Interval(5, 7), new Interval(7, 8)];
 
    // Find the least non-overlapping number from the input set of intervals
    const leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals);
 
    // Print the result
    console.log(`The least non-overlapping number is ${leastNonOverlappingNumber}`);
})();

Output

The least non-overlapping number is 7

time complexity of O(nlogn), Where n is the size of the intervals

space complexity of O(n), Where n is the size of the intervals

Suggest improvement

Find Non-overlapping intervals among a given set of intervals

Share your thoughts in the comments

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

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

Another Approach:

C++

C

Java

Python3

C#

Javascript

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