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 iff 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 intervalsvoid 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 codeint 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 intervalsclass GFG{ // Function to print intersecting intervalsstatic 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 codepublic 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 # intervalsdef 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 codearr1 = [ [ 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 intervalsusing System;class GFG{ // Function to print intersecting intervalsstatic 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 codepublic 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 |
{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
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
