Maximum ranges that can be uniquely represented by any integer from the range

Given an array arr[] consisting of N ranges of the form {L, R}, the task is to find the maximize the number of ranges such that each range can be uniquely represented by any integer from that range.

Examples:

Input: arr[] = {{1, 2}, {2, 3}, {3, 4}}
Output: 3
Explanation:
Number of ranges can be maximized by following representations:

1. Range {1, 2} can be represented by 1.
2. Range {2, 3} can be represented by 2.
3. Range {3, 4} can be represented by 3.

Therefore, the maximized count of ranges is 3.

Input: arr[] = {{1, 4}, {4, 4}, {2, 2}, {3, 4}, {1, 1}}
Output: 4

Naive Approach: The simplest approach to solve the given problem is to sort the ranges and iterate from the minimum value of L to the maximum value of R to greedily assign an integer for a particular range and keep a count of the possible assigned integers. After completing the above steps, print the count obtained as the result. 
Time Complexity: O(M * N), where M is the maximum element among the given ranges.
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the Greedy Approach, the idea is to assign the minimum unique value from each range by selecting the given ranges in increasing order. Follow the steps to solve the problem:

• Sort the given array of ranges in ascending order.
• Initialize a vector, say prev as arr[] that stores the previously assigned values to the given ranges.
• Initialize a variable, say count as 1 to store the maximum number of ranges where each range can be uniquely represented by an integer from the range.
• Traverse the given array arr[] over the range [1, N – 1] and perform the following steps:
  • Initialize a vector, say current[] as arr[i] that stores the values assigned to the current range.
  • If the value of max(current[i][1], prev[i][1]) – max(current[i][0], prev[i][0] – 1) is positive, then update the value of prev as {1 + max(current[i][0], prev[i][0] – 1), max(current[i][1], prev[i][1])} and increment the value of count by 1.
  • Otherwise, check of the next range.
• After completing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to find the maximum number
// of ranges where each range can be
// uniquely represented by an integer
int maxRanges(vector<vector<int> > arr,
              int N)
{
    // Sort the ranges in ascending order
    sort(arr.begin(), arr.end());

    // Stores the count of ranges
    int count = 1;

    // Stores previously assigned range
    vector<int> prev = arr[0];

    // Traverse the vector arr[]
    for (int i = 1; i < N; i++) {

        vector<int> last = arr[i - 1];
        vector<int> current = arr[i];

        // Skip the similar ranges
        // of size 1
        if (last[0] == current[0]
            && last[1] == current[1]
            && current[1] == current[0])
            continue;

        // Find the range of integer
        // available to be assigned
        int start = max(prev[0], current[0] - 1);
        int end = max(prev[1], current[1]);

        // Check if an integer is
        // available to be assigned
        if (end - start > 0) {

            // Update the previously
            // assigned range
            prev[0] = 1 + start;
            prev[1] = end;

            // Update the count of range
            count++;
        }
    }

    // Return the maximum count of
    // ranges
    return count;
}

// Driver Code
int main()
{
    vector<vector<int> > range = {
        { 1, 4 }, { 4, 4 },
        { 2, 2 }, { 3, 4 },
        { 1, 1 }
    };
    int N = range.size();
    cout << maxRanges(range, N);

    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

public class GFG {

    // Function to find the maximum number
    // of ranges where each range can be
    // uniquely represented by an integer
    static int maxRanges(Integer arr[][], int N)
    {
        // Sort the ranges in ascending order
        Arrays.sort(arr, (a, b) -> {
            if (a[0].equals(b[0]))
                return Integer.compare(a[1], b[1]);
            return Integer.compare(a[0], b[0]);
        });

        // Stores the count of ranges
        int count = 1;

        // Stores previously assigned range
        Integer prev[] = arr[0];

        // Traverse the vector arr[]
        for (int i = 1; i < N; i++) {

            Integer last[] = arr[i - 1];
            Integer current[] = arr[i];

            // Skip the similar ranges
            // of size 1
            if (last[0] == current[0]
                && last[1] == current[1]
                && current[1] == current[0])
                continue;

            // Find the range of integer
            // available to be assigned
            int start = Math.max(prev[0], current[0] - 1);
            int end = Math.max(prev[1], current[1]);

            // Check if an integer is
            // available to be assigned
            if (end - start > 0) {

                // Update the previously
                // assigned range
                prev[0] = 1 + start;
                prev[1] = end;

                // Update the count of range
                count++;
            }
        }

        // Return the maximum count of
        // ranges
        return count;
    }

    // Driver Code
    public static void main(String[] args)
    {

        Integer range[][] = {
            { 1, 4 }, { 4, 4 }, { 2, 2 }, { 3, 4 }, { 1, 1 }
        };
        int N = range.length;
        System.out.print(maxRanges(range, N));
    }
}

// This code is contributed by Kingash.

Python3

# Python 3 program for the above approach

# Function to find the maximum number
# of ranges where each range can be
# uniquely represented by an integer
def maxRanges(arr, N):

    # Sort the ranges in ascending order
    arr.sort()

    # Stores the count of ranges
    count = 1

    # Stores previously assigned range
    prev = arr[0]

    # Traverse the vector arr[]
    for i in range(1, N):

        last = arr[i - 1]
        current = arr[i]

        # Skip the similar ranges
        # of size 1
        if (last[0] == current[0]
            and last[1] == current[1]
                and current[1] == current[0]):
            continue

        # Find the range of integer
        # available to be assigned
        start = max(prev[0], current[0] - 1)
        end = max(prev[1], current[1])

        # Check if an integer is
        # available to be assigned
        if (end - start > 0):

            # Update the previously
            # assigned range
            prev[0] = 1 + start
            prev[1] = end

            # Update the count of range
            count += 1

    # Return the maximum count of
    # ranges
    return count


# Driver Code
if __name__ == "__main__":

    arr = [
        [1, 4], [4, 4],
        [2, 2], [3, 4],
        [1, 1]]
    N = len(arr)
    print(maxRanges(arr, N))

    # This code is contributed by ukasp.

Javascript

<script>
// Javascript program for the above approach

// Function to find the maximum number
// of ranges where each range can be
// uniquely represented by an integer
function maxRanges(arr, N) {
    // Sort the ranges in ascending order
    arr.sort((a, b) => {
        return b[0] - a[0]
    })

    // Stores the count of ranges
    let count = 1;

    // Stores previously assigned range
    let prev = arr[0];

    // Traverse the vector arr[]
    for (let i = 1; i < N; i++) {

        let last = arr[i - 1];
        let current = arr[i];

        // Skip the similar ranges
        // of size 1
        if (last[0] == current[0]
            && last[1] == current[1]
            && current[1] == current[0]) {
            continue;
        }
        // Find the range of integer
        // available to be assigned
        let start = Math.max(prev[0], current[0] - 1);
        let end = Math.max(prev[1], current[1]);

        // Check if an integer is
        // available to be assigned
        if ((end - start) > 0) {

            // Update the previously
            // assigned range
            prev[0] = 1 + start;
            prev[1] = end;

            // Update the count of range
            count++;
        }
    }

    // Return the maximum count of
    // ranges
    return count;
}

// Driver Code

let range = [
    [1, 4], [4, 4],
    [2, 2], [3, 4],
    [1, 1]];
let N = range.length;
document.write(maxRanges(range, N));


// This code is contributed by _Saurabh_jaiswal
</script>

Output:

4

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