Find the missing elements from 1 to M in given N ranges

Given N segments as ranges [L, R] where ranges are non-intersecting and non-overlapping. The task is to find all number between 1 to M that doesn’t belong to any of the given ranges.

Examples:  

Input : N = 2, M = 6
        Ranges:
        [1, 2]
        [4, 5]
Output : 3, 6
Explanation: Only 3 and 6 are missing from
the above ranges.

Input : N = 1, M = 5
        Ranges:
        [2, 4]
Output : 1, 5

Approach: Given that we have N ranges, which are non-overlapping and non-intersecting. First of all, sort all segments based on starting value. After sorting, iterate from each segment and find the numbers which are missing.

Algorithm:

Step 1: Start
Step 2: Create a class called Pair and make a constructor which takes two integer values as parameters.
Step 3: Create a static function that takes two parameters as input one in ArrayList of type Pair and the second an integer value.
Step 4: Now sort the given ArrayList using the collections sort method and use a comparator if two values are equal then will sort               on the basis of the ending point.
Step 5: Now create an ArrayList called ans to store our answer
Step 6: Initialize a prev variable’s value to 0, Using a for-loop iterate across each range, saying for each range: first Create start and              end variables from the place where it begins and ends.
             , second Using a for-loop to iterate over all integers between prev and start, determine whether any elements are missing,              and add those that are to the ans ArrayList, third Change prev to end.
Step 7: Start a for-loop, and iterate through each number between prev and m to see if there are any missing elements. If there                 are, add each one to the ans ArrayList.
            Lastly, output any values that are less than or equal to m after iterating through the ans ArrayList.
Step 8: End

Below is the implementation of the above approach: 

C++

// C++ program to find missing elements
// from given Ranges

#include <bits/stdc++.h>
using namespace std;

// Function to find missing elements
// from given Ranges
void findMissingNumber(vector<pair<int, int> > ranges, int m)
{
    // First of all sort all the given ranges
    sort(ranges.begin(), ranges.end());

    // store ans in a different vector
    vector<int> ans;

    // prev is use to store end of
    // last range
    int prev = 0;

    // j is used as a counter for ranges
    for (int j = 0; j < ranges.size(); j++) {
        int start = ranges[j].first;
        int end = ranges[j].second;

        for (int i = prev + 1; i < start; i++)
            ans.push_back(i);

        prev = end;
    }

    // for last segment
    for (int i = prev + 1; i <= m; i++)
        ans.push_back(i);

    // finally print all answer
    for (int i = 0; i < ans.size(); i++) {
        if (ans[i] <= m)
            cout << ans[i] << " ";
    }
}

// Driver code
int main()
{
    int N = 2, M = 6;

    // Store ranges in vector of pair
    vector<pair<int, int> > ranges;
    ranges.push_back({ 1, 2 });
    ranges.push_back({ 4, 5 });

    findMissingNumber(ranges, M);

    return 0;
}

Java

// Java program to find missing elements
// from given Ranges
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

class GFG{

static class Pair
{
    int first, second;

    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find missing elements
// from given Ranges
static void findMissingNumber(ArrayList<Pair> ranges,
                              int m)
{

    // First of all sort all the given ranges
    Collections.sort(ranges, new Comparator<Pair>()
    {
        public int compare(Pair first, Pair second)
        {
            if (first.first == second.first)
            {
                return first.second - second.second;
            }
            return first.first - second.first;
        }
    });

    // Store ans in a different vector
    ArrayList<Integer> ans = new ArrayList<>();

    // prev is use to store end of
    // last range
    int prev = 0;

    // j is used as a counter for ranges
    for(int j = 0; j < ranges.size(); j++)
    {
        int start = ranges.get(j).first;
        int end = ranges.get(j).second;

        for(int i = prev + 1; i < start; i++)
            ans.add(i);

        prev = end;
    }

    // For last segment
    for(int i = prev + 1; i <= m; i++)
        ans.add(i);

    // Finally print all answer
    for(int i = 0; i < ans.size(); i++)
    {
        if (ans.get(i) <= m)
            System.out.print(ans.get(i) + " ");
    }
}

// Driver code
public static void main(String[] args)
{
    int N = 2, M = 6;

    // Store ranges in vector of pair
    ArrayList<Pair> ranges = new ArrayList<>();
    ranges.add(new Pair(1, 2));
    ranges.add(new Pair(4, 5));

    findMissingNumber(ranges, M);
}
}

// This code is contributed by sanjeev2552

Python3

# Python3 program to find missing
# elements from given Ranges

# Function to find missing elements
# from given Ranges
def findMissingNumber(ranges, m):

    # First of all sort all the
    # given ranges
    ranges.sort()

    # store ans in a different vector
    ans = []

    # prev is use to store end
    # of last range
    prev = 0

    # j is used as a counter for ranges
    for j in range(len(ranges)):
        start = ranges[j][0]
        end = ranges[j][1]

        for i in range(prev + 1, start):
            ans.append(i)

        prev = end

    # for last segment
    for i in range(prev + 1, m + 1):
        ans.append(i)

    # finally print all answer
    for i in range(len(ans)):
        if ans[i] <= m:
            print(ans[i], end = " ")

# Driver Code
if __name__ == "__main__":

    N, M = 2, 6

    # Store ranges in vector of pair
    ranges = []
    ranges.append([1, 2])
    ranges.append([4, 5])

    findMissingNumber(ranges, M)

# This code is contributed
# by Rituraj Jain

C#

// C# program to find missing elements
// from given Ranges
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

class sortHelper : IComparer
{
   int IComparer.Compare(object a, object b)
   {
      Pair first = (Pair)a;
      Pair second = (Pair)b;
      if (first.first == second.first)
      {
        return first.second - second.second;
      }

      return first.first - second.first;
   }
}

public class Pair
{
    public int first, second;

    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}


// Function to find missing elements
// from given Ranges
static void findMissingNumber(ArrayList ranges, int m)
{
    IComparer myComparer = new sortHelper();
    ranges.Sort(myComparer);

    // Store ans in a different vector
    ArrayList ans = new ArrayList();

    // prev is use to store end of
    // last range
    int prev = 0;

    // j is used as a counter for ranges
    for(int j = 0; j < ranges.Count; j++)
    {
        int start = ((Pair)ranges[j]).first;
        int end = ((Pair)ranges[j]).second;

        for(int i = prev + 1; i < start; i++)
            ans.Add(i);

        prev = end;
    }

    // For last segment
    for(int i = prev + 1; i <= m; i++)
        ans.Add(i);

    // Finally print all answer
    for(int i = 0; i < ans.Count; i++)
    {
        if ((int)ans[i] <= m)
            Console.Write(ans[i] + " ");
    }
}

// Driver code
public static void Main(string[] args)
{
    int  M = 6;

    // Store ranges in vector of pair
    ArrayList ranges = new ArrayList();
    ranges.Add(new Pair(1, 2));
    ranges.Add(new Pair(4, 5));

    findMissingNumber(ranges, M);
}
}

// This code is contributed by rutvik_56

PHP

<?php
// PHP program to find missing
// elements from given Ranges

// Function to find missing elements
// from given Ranges
function findMissingNumber($ranges, $m)
{
    // First of all sort all the
    // given ranges
    sort($ranges);

    // store ans in a different vector
    $ans = [];

    // prev is use to store end
    // of last range
    $prev = 0;

    // j is used as a counter for ranges
    for ($j = 0; $j < count($ranges); $j++)
    {
        $start = $ranges[$j][0];
        $end = $ranges[$j][1];

        for ($i = $prev + 1; $i < $start; $i++)
            array_push($ans, $i);

        $prev = $end;
    }

    // for last segment
    for ($i = $prev + 1; $i < $m + 1; $i++)
            array_push($ans, $i);

    // finally print all answer
    for ($i = 0; $i < count($ans); $i++)
    {
        if ($ans[$i] <= $m)
            echo "$ans[$i] ";
    }
}

// Driver Code
$N = 2;
$M = 6;

// Store ranges in vector of pair
$ranges = [];
array_push($ranges, [1, 2]);
array_push($ranges, [4, 5]);

findMissingNumber($ranges, $M)

// This code is contributed
// by Srathore
?>

Javascript

<script>
// Javascript program to find missing elements
// from given Ranges

// Function to find missing elements
// from given Ranges
function findMissingNumber(ranges, m)
{
    // First of all sort all the given ranges
    ranges.sort();

    // store ans in a different vector
    let ans=[];

    // prev is use to store end of
    // last range
    let prev = 0;

    // j is used as a counter for ranges
    for (let j = 0; j < ranges.length; j++) {
        let start = ranges[j][0];
        let end = ranges[j][1];

        for (let i = prev + 1; i < start; i++)
            ans.push(i);

        prev = end;
    }

    // for last segment
    for (let i = prev + 1; i <= m; i++)
        ans.push(i);

    // finally print all answer
    for (let i = 0; i < ans.length; i++) {
        if (ans[i] <= m)
            document.write(ans[i]," ");
    }
}

// Driver code

    let N = 2;
    let M = 6;

    // Store ranges in vector of pair
    let ranges=[];
    ranges.push([ 1, 2 ]);
    ranges.push([ 4, 5 ]);
    findMissingNumber(ranges, M);

// This code is contributed by Pushpesh Raj
</script>

Time Complexity: O(n * log(n)), where n is the length of the vector

Auxiliary Space: O(n)