Intersection of intervals given by two lists
Last Updated :
15 Jul, 2025
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.
Note: Disjoint means no element is common in a list
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} is the overlap of {0, 4} and {1, 5}. Similarly, {24, 24} is the overlap of {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} is the overlap of {0, 2} and {1, 4}. Similarly, {12, 12} is the overlap of {12, 22} and {9, 12}.
We use two-pointer to solve by following these steps:
Use two pointers, i and j, to traverse arr1 and arr2. For each pair, find the overlap:
l = max(arr1[i][0],arr2[j][0]),
r = min(arr1[i][1],arr2[j][1])
If l ≤ r, add [l, r] to the result. Move the pointer with the smaller endpoint to continue.
C++
#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++;
}
}
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 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++;
}
}
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);
}
}
# 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)
// 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++;
}
}
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);
}
}
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)
console.log("[" + 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++;
}
}
var arr1 = [ [ 0, 4 ], [ 5, 10 ], [ 13, 20 ], [ 24, 25 ] ];
var arr2 = [ [ 1, 5 ], [ 8, 12 ], [ 15, 24 ], [ 25, 26 ] ];
printIntervals(arr1, arr2);
Output[1, 4]
[5, 5]
[8, 10]
[15, 20]
[24, 24]
[25, 25]
Time Complexity O(n + m)
Auxiliary Space O(1)
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