Find intersection of intervals given by two lists

Given two 2-D arrays which represent intervals. Each 2-D array represents a list of intervals. Each list of intervals is disjoint and sorted in increasing order. Find the intersection or set of ranges that are common to both the lists.

Disjoint means no element is common in a list. Example: {1, 4} and {5, 6} are disjoint while {1, 4} and {2, 5} are not as 2, 3 and 4 are common to both intervals. 

Examples:

Input: arr1[][] = {{0, 4}, {5, 10}, {13, 20}, {24, 25}} 
arr2[][] = {{1, 5}, {8, 12}, {15, 24}, {25, 26}} 
Output: {{1, 4}, {5, 5}, {8, 10}, {15, 20}, {24, 24}, {25, 25}}
Explanation: 
{1, 4} lies completely within range {0, 4} and {1, 5}. Hence, {1, 4} is the desired intersection. Similarly, {24, 24} lies completely within two intervals {24, 25} and {15, 24}.
Input: arr1[][] = {{0, 2}, {5, 10}, {12, 22}, {24, 25}} 
arr2[][] = {{1, 4}, {9, 12}, {15, 24}, {25, 26}} 
Output: {{1, 2}, {9, 10}, {12, 12}, {15, 22}, {24, 24}, {25, 25}} 
Explanation:
{1, 2} lies completely within range {0, 2} and {1, 4}. Hence, {1, 2} is the desired intersection. Similarly, {12, 12} lies completely within two intervals {12, 22} and {9, 12}. 

Approach:
To solve the problem mentioned above, two pointer technique can be used, as per the steps given below:

• Maintain two pointers i and j to traverse the two interval lists, arr1 and arr2 respectively.
• Now, if arr1[i] has smallest endpoint, it can only intersect with arr2[j]. Similarly, if arr2[j] has smallest endpoint, it can only intersect with arr1[i]. If intersection occurs, find the intersecting segment.
• [l, r] will be the intersecting segment if l <= r, where l = max(arr1[i][0], arr2[j][0]) and r = min(arr1[i][1], arr2[j][1]).
• Increment the i and j pointers accordingly to move ahead.

Below is the implementation of the approach: 

C++

// C++ implementation to find the
// intersection of the two intervals

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

// Function to print intersecting intervals
void printIntervals(vector<vector<int> > arr1,
                    vector<vector<int> > arr2)
{

    // i and j pointers for
    // arr1 and arr2 respectively
    int i = 0, j = 0;

    // Size of the two lists
    int n = arr1.size(), m = arr2.size();

    // Loop through all intervals unless
    // one of the interval gets exhausted
    while (i < n && j < m) {
        // Left bound for intersecting segment
        int l = max(arr1[i][0], arr2[j][0]);

        // Right bound for intersecting segment
        int r = min(arr1[i][1], arr2[j][1]);

        // If segment is valid print it
        if (l <= r)
            cout << "{" << l << ", "
                 << r << "}\n";

        // If i-th interval's right
        // bound is smaller
        // increment i else
        // increment j
        if (arr1[i][1] < arr2[j][1])
            i++;
        else
            j++;
    }
}

// Driver code
int main()
{

    vector<vector<int> > arr1
        = { { 0, 4 }, { 5, 10 },
            { 13, 20 }, { 24, 25 } };

    vector<vector<int> > arr2
        = { { 1, 5 }, { 8, 12 },
            { 15, 24 }, { 25, 26 } };

    printIntervals(arr1, arr2);

    return 0;
}

Java

// Java implementation to find
// intersecting intervals
class GFG{

// Function to print intersecting intervals
static void printIntervals(int arr1[][],
                           int arr2[][])
{

    // i and j pointers for arr1 and
    // arr2 respectively
    int i = 0, j = 0;

    int n = arr1.length, m = arr2.length;

    // Loop through all intervals unless
    // one of the interval gets exhausted
    while (i < n && j < m)
    {

        // Left bound for intersecting segment
        int l = Math.max(arr1[i][0], arr2[j][0]);

        // Right bound for intersecting segment
        int r = Math.min(arr1[i][1], arr2[j][1]);

        // If segment is valid print it
        if (l <= r)
            System.out.println("{" + l + ", " +
                                 r + "}");

        // If i-th interval's right bound is
        // smaller increment i else increment j
        if (arr1[i][1] < arr2[j][1])
            i++;
        else
            j++;
    }
}

// Driver code
public static void main(String[] args)
{
    int arr1[][] = { { 0, 4 }, { 5, 10 },
                     { 13, 20 }, { 24, 25 } };

    int arr2[][] = { { 1, 5 }, { 8, 12 },
                     { 15, 24 }, { 25, 26 } };

    printIntervals(arr1, arr2);
}
}

// This code is contributed by sarthak_eddy

Python3

# Python3 implementation to find
# intersecting intervals

# Function to print intersecting
# intervals
def printIntervals(arr1, arr2):

    # i and j pointers for arr1
    # and arr2 respectively
    i = j = 0

    n = len(arr1)
    m = len(arr2)

    # Loop through all intervals unless one
    # of the interval gets exhausted
    while i < n and j < m:

        # Left bound for intersecting segment
        l = max(arr1[i][0], arr2[j][0])

        # Right bound for intersecting segment
        r = min(arr1[i][1], arr2[j][1])

        # If segment is valid print it
        if l <= r:
            print('{', l, ',', r, '}')

        # If i-th interval's right bound is
        # smaller increment i else increment j
        if arr1[i][1] < arr2[j][1]:
            i += 1
        else:
            j += 1

# Driver code
arr1 = [ [ 0, 4 ], [ 5, 10 ],
         [ 13, 20 ], [ 24, 25 ] ]

arr2 = [ [ 1, 5 ], [ 8, 12 ],
         [ 15, 24 ], [ 25, 26 ] ]

printIntervals(arr1, arr2)

# This code is contributed by sarthak_eddy

C#

// C# implementation to find
// intersecting intervals
using System;
class GFG{

// Function to print intersecting intervals
static void printIntervals(int [,]arr1,
                           int [,]arr2)
{

    // i and j pointers for arr1 and
    // arr2 respectively
    int i = 0, j = 0;

    int n = arr1.GetLength(0),
        m = arr2.GetLength(0);

    // Loop through all intervals unless
    // one of the interval gets exhausted
    while (i < n && j < m)
    {

        // Left bound for intersecting segment
        int l = Math.Max(arr1[i, 0], arr2[j, 0]);

        // Right bound for intersecting segment
        int r = Math.Min(arr1[i, 1], arr2[j, 1]);

        // If segment is valid print it
        if (l <= r)
        Console.WriteLine("{" + l + ", " +
                            r + "}");

        // If i-th interval's right bound is
        // smaller increment i else increment j
        if (arr1[i, 1] < arr2[j, 1])
            i++;
        else
            j++;
    }
}

// Driver code
public static void Main(String[] args)
{
    int [,]arr1 = { { 0, 4 }, { 5, 10 },
                    { 13, 20 }, { 24, 25 } };

    int [,]arr2 = { { 1, 5 }, { 8, 12 },
                    { 15, 24 }, { 25, 26 } };

    printIntervals(arr1, arr2);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript implementation to find
// intersecting intervals
// Function to print intersecting intervals
function printIntervals(arr1, arr2)
{

    // i and j pointers for arr1 and
    // arr2 respectively
    var i = 0, j = 0;

    var n = arr1.length, m = arr2.length;

    // Loop through all intervals unless
    // one of the interval gets exhausted
    while (i < n && j < m)
    {

        // Left bound for intersecting segment
        var l = Math.max(arr1[i][0], arr2[j][0]);

        // Right bound for intersecting segment
        var r = Math.min(arr1[i][1], arr2[j][1]);

        // If segment is valid print it
        if (l <= r)
            document.write("{" + l + ", " +
                                 r + "}" + "<br>");

        // If i-th interval's right bound is
        // smaller increment i else increment j
        if (arr1[i][1] < arr2[j][1])
            i++;
        else
            j++;
    }
}

// Driver code
    var arr1 = [ [ 0, 4 ], [ 5, 10 ],
                     [ 13, 20 ], [ 24, 25 ] ];

    var arr2 = [ [ 1, 5 ], [ 8, 12 ],
                     [ 15, 24 ], [ 25, 26 ] ];

    printIntervals(arr1, arr2);

// This code is contributed by shivanisinghss2110

</script>

Output:

{1, 4}
{5, 5}
{8, 10}
{15, 20}
{24, 24}
{25, 25}

Time Complexity: O(N + M), where N and M are lengths of the 2-D arrays
Auxiliary Space: O(1)