Intersection of Two Sorted Arrays with Distinct Elements
Last Updated :
23 Jul, 2025
Given two sorted arrays a[] and b[] with distinct elements of size n and m respectively, the task is to find intersection (or common elements) of the two arrays. We need to return the intersection in sorted order.
Note: Intersection of two arrays can be defined as a set containing distinct common elements between the two arrays.
Examples:
Input: a[] = { 1, 2, 4, 5, 6 }, b[] = { 2, 4, 7, 9 }
Output: { 2, 4 }
Explanation: The common elements in both arrays are 2 and 4.
Input: a[] = { 2, 3, 4, 5 }, b[] = { 1, 7 }
Output: { }
Explanation: There are no common elements in array a[] and b[].
Approaches Same as Unsorted Arrays with distinct elements – Best Time O(n+m) and O(n) Space
The idea is to use the same approaches as discussed in Intersection of Two Arrays with Distinct Elements. The most optimal approach among them is to use Hash Set which has a time complexity of O(n+m) and Auxiliary Space of O(n).
[Expected Approach] Using Merge Step of Merge Sort - O(n+m) Time and O(1) Space
The idea is to find the intersection of two sorted arrays using merge step of merge sort. We maintain two pointers to traverse both arrays simultaneously.
- If the element in first array is smaller, move the pointer of first array forward because this element cannot be part of the intersection.
- If the element in second array is smaller, move the pointer of second array forward.
- If both elements are equal, add one of them and move both the pointers forward.
This continues until one of the pointers reaches the end of its array.
C++
// C++ program for intersection of
// two sorted arrays with distinct elements
#include <iostream>
#include <vector>
using namespace std;
vector<int> intersection(vector<int>& a, vector<int>& b) {
vector<int> res;
int i=0;
int j=0;
// Start simultaneous traversal on both arrays
while (i < a.size() && j < b.size()) {
// if a[i] is smaller, then move in a[]
// towards larger value
if(a[i] < b[j]) {
i++;
}
// if b[j] is smaller, then move in b[]
// towards larger value
else if (a[i] > b[j]) {
j++;
}
// if a[i] == b[j], then this element is common
// add it to result array and move in both arrays
else {
res.push_back(a[i]);
i++;
j++;
}
}
return res;
}
int main() {
vector<int> a = {1, 2, 4, 5, 6};
vector<int> b = {2, 4, 7, 9};
vector<int> res = intersection(a, b);
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
return 0;
}
// C program for intersection of
// two sorted arrays with distinct elements
#include <stdio.h>
int* intersection(int a[], int b[], int n, int m, int *size) {
int* res = (int*) malloc(n * sizeof(int));
*size = 0;
int i = 0;
int j = 0;
// Start simultaneous traversal on both arrays
while (i < n && j < m) {
// if a[i] is smaller, then move in a[]
// towards larger value
if (a[i] < b[j]) {
i++;
}
// if b[j] is smaller, then move in b[]
// towards larger value
else if (a[i] > b[j]) {
j++;
}
// if a[i] == b[j], then this element is common
// add it to result array and move in both arrays
else {
res[(*size)++] = a[i];
i++;
j++;
}
}
return res;
}
int main() {
int a[] = {1, 2, 4, 5, 6};
int b[] = {2, 4, 7, 9};
int size;
int* res = intersection(a, b, 5, 4, &size);
for (int i = 0; i < size; i++)
printf("%d ", res[i]);
return 0;
}
// Java program for intersection of
// two sorted arrays with distinct elements
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> intersection(int[] a, int[] b) {
ArrayList<Integer> res = new ArrayList<>();
int i = 0;
int j = 0;
// Start simultaneous traversal on both arrays
while (i < a.length && j < b.length) {
// if a[i] is smaller, then move in a[]
// towards larger value
if (a[i] < b[j]) {
i++;
}
// if b[j] is smaller, then move in b[]
// towards larger value
else if (a[i] > b[j]) {
j++;
}
// if a[i] == b[j], then this element is common
// add it to result array and move in both arrays
else {
res.add(a[i]);
i++;
j++;
}
}
return res;
}
public static void main(String[] args) {
int[] a = {1, 2, 4, 5, 6};
int[] b = {2, 4, 7, 9};
ArrayList<Integer> res = intersection(a, b);
for (int i = 0; i < res.size(); i++)
System.out.print(res.get(i) + " ");
}
}
# Python program for intersection of
# two sorted arrays with distinct elements
def intersection(a, b):
res = []
i = 0
j = 0
# Start simultaneous traversal on both arrays
while i < len(a) and j < len(b):
# if a[i] is smaller, then move in a[]
# towards larger value
if a[i] < b[j]:
i += 1
# if b[j] is smaller, then move in b[]
# towards larger value
elif a[i] > b[j]:
j += 1
# if a[i] == b[j], then this element is common
# add it to result array and move in both arrays
else:
res.append(a[i])
i += 1
j += 1
return res
if __name__ == "__main__":
a = [1, 2, 4, 5, 6]
b = [2, 4, 7, 9]
res = intersection(a, b)
for i in range(len(res)):
print(res[i], end=" ")
// C# program for intersection of
// two sorted arrays with distinct elements
using System;
using System.Collections.Generic;
class GfG {
static List<int> intersection(int[] a, int[] b) {
List<int> res = new List<int>();
int i = 0;
int j = 0;
// Start simultaneous traversal on both arrays
while (i < a.Length && j < b.Length) {
// if a[i] is smaller, then move in a[]
// towards larger value
if (a[i] < b[j]) {
i++;
}
// if b[j] is smaller, then move in b[]
// towards larger value
else if (a[i] > b[j]) {
j++;
}
// if a[i] == b[j], then this element is common
// add it to result array and move in both arrays
else {
res.Add(a[i]);
i++;
j++;
}
}
return res;
}
static void Main() {
int[] a = {1, 2, 4, 5, 6};
int[] b = {2, 4, 7, 9};
List<int> res = intersection(a, b);
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
}
// JavaScript program for intersection of
// two sorted arrays with distinct elements
function intersection(a, b) {
const res = [];
let i = 0;
let j = 0;
// Start simultaneous traversal on both arrays
while (i < a.length && j < b.length) {
// if a[i] is smaller, then move in a[]
// towards larger value
if (a[i] < b[j]) {
i++;
}
// if b[j] is smaller, then move in b[]
// towards larger value
else if (a[i] > b[j]) {
j++;
}
// if a[i] == b[j], then this element is common
// add it to result array and move in both arrays
else {
res.push(a[i]);
i++;
j++;
}
}
return res;
}
const a = [1, 2, 4, 5, 6];
const b = [2, 4, 7, 9];
const res = intersection(a, b);
console.log(res.join(' '));
Time Complexity: O(n + m), where n and m are size of array a[] and b[] respectively.
Auxiliary Space: O(1)
[Alternate Approach 1] Using Binary Search - O(n * log (m)) Time and O(1) Space
The idea is to check each element of array a[] and see if it is in array b[]. If it is, we add it to the result array. Since b[] is already sorted, we can use binary search to find the elements in log(n) time.
C++
// C++ program for intersection of two sorted arrays
// with distinct elements using Binary Search
#include <iostream>
#include <vector>
using namespace std;
// Function to perform binary search
int binarySearch(vector<int> &arr, int lo, int hi, int target) {
while (lo <= hi){
int mid = lo + (hi - lo) / 2;
if (arr[mid] == target)
return mid;
if (arr[mid] < target)
lo = mid + 1;
else
hi = mid - 1;
}
return -1;
}
vector<int> intersection(vector<int>& a, vector<int>& b) {
vector<int> res;
// Traverse through array a[] and search every
// element a[i] in array b[]
for (int i = 0; i < a.size(); i++) {
// If found in b[], then add this
// element to result array
if (binarySearch(b, 0, b.size()-1, a[i]) != -1) {
res.push_back(a[i]);
}
}
return res;
}
int main() {
vector<int> a = {1, 2, 4, 5, 6};
vector<int> b = {2, 4, 7, 9};
vector<int> res = intersection(a, b);
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
return 0;
}
// C program for intersection of two sorted arrays
// with distinct elements using Binary Search
#include <stdio.h>
int binarySearch(int arr[], int lo, int hi, int target) {
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (arr[mid] == target)
return mid;
if (arr[mid] < target)
lo = mid + 1;
else
hi = mid - 1;
}
return -1;
}
int* intersection(int a[], int b[], int n, int m, int *size) {
*size = 0;
int* res = (int*) malloc(n * sizeof(int));
// Traverse through array a[] and search every
// element a[i] in array b[]
for (int i = 0; i < n; i++) {
// If found in b[], then add this
// element to result array
if (binarySearch(b, 0, m - 1, a[i]) != -1) {
res[(*size)++] = a[i];
}
}
return res;
}
int main() {
int a[] = {1, 2, 4, 5, 6};
int b[] = {2, 4, 7, 9};
int size;
int* res = intersection(a, b, 5, 4, &size);
for (int i = 0; i < size; i++)
printf("%d ", res[i]);
return 0;
}
// Java program for intersection of two sorted arrays
// with distinct elements using Binary Search
import java.util.ArrayList;
class GfG {
// Function to perform binary search
static int binarySearch(int[] arr, int lo, int hi, int target) {
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (arr[mid] == target)
return mid;
if (arr[mid] < target)
lo = mid + 1;
else
hi = mid - 1;
}
return -1;
}
static ArrayList<Integer> intersection(int[] a, int[] b) {
ArrayList<Integer> res = new ArrayList<>();
// Traverse through array a[] and search every
// element a[i] in array b[]
for (int i = 0; i < a.length; i++) {
// If found in b[], then add this
// element to result array
if (binarySearch(b, 0, b.length - 1, a[i]) != -1) {
res.add(a[i]);
}
}
return res;
}
public static void main(String[] args) {
int[] a = {1, 2, 4, 5, 6};
int[] b = {2, 4, 7, 9};
ArrayList<Integer> res = intersection(a, b);
for (int i = 0; i < res.size(); i++)
System.out.print(res.get(i) + " ");
}
}
# Python program for intersection of two sorted arrays
# with distinct elements using Binary Search
def binarySearch(arr, lo, hi, target):
while lo <= hi:
mid = lo + (hi - lo) // 2
if arr[mid] == target:
return mid
if arr[mid] < target:
lo = mid + 1
else:
hi = mid - 1
return -1
def intersection(a, b):
res = []
# Traverse through array a[] and search every
# element a[i] in array b[]
for i in range(len(a)):
# If found in b[], then add this
# element to result array
if binarySearch(b, 0, len(b) - 1, a[i]) != -1:
res.append(a[i])
return res
if __name__ == "__main__":
a = [1, 2, 4, 5, 6]
b = [2, 4, 7, 9]
res = intersection(a, b)
for i in range(len(res)):
print(res[i], end=" ")
// C# program for intersection of two sorted arrays
// with distinct elements using Binary Search
using System;
using System.Collections.Generic;
class GfG {
// Function to perform binary search
static int binarySearch(int[] arr, int lo, int hi, int target) {
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (arr[mid] == target)
return mid;
if (arr[mid] < target)
lo = mid + 1;
else
hi = mid - 1;
}
return -1;
}
static List<int> intersection(int[] a, int[] b) {
List<int> res = new List<int>();
// Traverse through array a[] and search every
// element a[i] in array b[]
for (int i = 0; i < a.Length; i++) {
// If found in b[], then add this
// element to result array
if (binarySearch(b, 0, b.Length - 1, a[i]) != -1) {
res.Add(a[i]);
}
}
return res;
}
static void Main() {
int[] a = {1, 2, 4, 5, 6};
int[] b = {2, 4, 7, 9};
List<int> res = intersection(a, b);
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
}
// JavaScript program for intersection of two sorted arrays
// with distinct elements using Binary Search
function binarySearch(arr, lo, hi, target) {
while (lo <= hi) {
const mid = lo + Math.floor((hi - lo) / 2);
if (arr[mid] === target)
return mid;
if (arr[mid] < target)
lo = mid + 1;
else
hi = mid - 1;
}
return -1;
}
function intersection(a, b) {
const res = [];
// Traverse through array a[] and search every
// element a[i] in array b[]
for (let i = 0; i < a.length; i++) {
// If found in b[], then add this
// element to result array
if (binarySearch(b, 0, b.length - 1, a[i]) !== -1) {
res.push(a[i]);
}
}
return res;
}
const a = [1, 2, 4, 5, 6];
const b = [2, 4, 7, 9];
const res = intersection(a, b);
console.log(res.join(' '));
Time Complexity: O(n * log (m)), where n and m are size of array a[] and b[] respectively.
Auxiliary Space: O(1)
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