Longest common subarray in the given two arrays
Last Updated :
12 Jul, 2025
Given two arrays A[] and B[] of N and M integers respectively, the task is to find the maximum length of an equal subarray or the longest common subarray between the two given array.
Examples:
Input: A[] = {1, 2, 8, 2, 1}, B[] = {8, 2, 1, 4, 7}
Output: 3
Explanation:
The subarray that is common to both arrays are {8, 2, 1} and the length of the subarray is 3.
Input: A[] = {1, 2, 3, 2, 1}, B[] = {8, 7, 6, 4, 7}
Output: 0
Explanation:
There is no such subarrays which are equal in the array A[] and B[].
Naive Approach: The idea is to generate all the subarrays of the two given array A[] and B[] and find the longest matching subarray. This solution is exponential in terms of time complexity.
C++
#include <iostream>
#include <vector>
using namespace std;
// Recursive function to find the longest common subarray (LCS)
int LCS(int i, int j, const vector<int>& A, const vector<int>& B, int count) {
// Base case: If either of the indices reaches the end of the array, return the count
if (i == A.size() || j == B.size())
return count;
// If the current elements are equal, recursively check the next elements
if (A[i] == B[j])
count = LCS(i + 1, j + 1, A, B, count + 1);
// Recursively check for the longest common subarray by considering two possibilities:
// 1. Exclude current element from array A and continue with array B
// 2. Exclude current element from array B and continue with array A
count = max(count, max(LCS(i + 1, j, A, B, 0), LCS(i, j + 1, A, B, 0)));
return count;
}
int main() {
// Example arrays
vector<int> A = {1, 2, 3, 2, 1};
vector<int> B = {3, 2, 1, 4, 7};
// Call the LCS function to find the maximum length of the common subarray
int maxLength = LCS(0, 0, A, B, 0);
// Print the result
cout << "Max length of common subarray: " << maxLength << endl;
return 0;
}
import java.util.Arrays;
import java.util.Vector;
public class Main {
// Recursive function to find the longest common subarray (LCS)
static int LCS(int i, int j, Vector<Integer> A,
Vector<Integer> B, int count) {
// Base case: If either of the indices reaches the end of the array, return the count
if (i == A.size() || j == B.size())
return count;
// If the current elements are equal, recursively check the next elements
if (A.get(i).equals(B.get(j)))
count = LCS(i + 1, j + 1, A, B, count + 1);
// Recursively check for the longest common subarray by considering two possibilities:
// 1. Exclude the current element from array A and continue with array B
// 2. Exclude the current element from array B and continue with array A
count = Math.max(count, Math.max(LCS(i + 1, j, A, B, 0),
LCS(i, j + 1, A, B, 0)));
return count;
}
public static void main(String[] args) {
// Example arrays
Vector<Integer> A = new Vector<>(Arrays.asList(1, 2, 3, 2, 1));
Vector<Integer> B = new Vector<>(Arrays.asList(3, 2, 1, 4, 7));
// Call the LCS function to find the maximum length of the common subarray
int maxLength = LCS(0, 0, A, B, 0);
// Print the result
System.out.println("Max length of common subarray: " + maxLength);
}
}
// This code is contributed by akshitaguprjz3
def LCS(i, j, A, B, count):
# Base case: If either of the indices reaches the end of the array,
# return the count
if i == len(A) or j == len(B):
return count
# If the current elements are equal, recursively check the
# next elements
if A[i] == B[j]:
count = LCS(i + 1, j + 1, A, B, count + 1)
# Recursively check for the longest common subarray by considering two possibilities:
# 1. Exclude current element from array A and continue with array B
# 2. Exclude current element from array B and continue with array A
return max(count, max(LCS(i + 1, j, A, B, 0), LCS(i, j + 1, A, B, 0)))
# Example arrays
A = [1, 2, 3, 2, 1]
B = [3, 2, 1, 4, 7]
# Call the LCS function to find the maximum length of the common subarray
maxLength = LCS(0, 0, A, B, 0)
# Print the result
print("Max length of common subarray:", maxLength)
using System;
class GFG {
// Recursive function to find the longest common
// subarray (LCS)
static int LCS(int i, int j, int[] A, int[] B,
int count)
{
// Base case: If either of the indices reaches the
// end of the array, return the count
if (i == A.Length || j == B.Length)
return count;
// If the current elements are equal, recursively
// check the next elements
if (A[i] == B[j])
count = LCS(i + 1, j + 1, A, B, count + 1);
// Recursively check for the longest common subarray
// by considering two possibilities:
// 1. Exclude the current element from array A and
// continue with array B
// 2. Exclude the current element from array B and
// continue with array A
count = Math.Max(count,
Math.Max(LCS(i + 1, j, A, B, 0),
LCS(i, j + 1, A, B, 0)));
return count;
}
static void Main()
{
// Example arrays
int[] A = { 1, 2, 3, 2, 1 };
int[] B = { 3, 2, 1, 4, 7 };
// Call the LCS function to find the maximum length
// of the common subarray
int maxLength = LCS(0, 0, A, B, 0);
// Print the result
Console.WriteLine("Max length of common subarray: "
+ maxLength);
}
}
// Recursive function to find the longest common subarray (LCS)
function LCS(i, j, A, B, count) {
// Base case: If either of the indices reaches the end of the array, return the count
if (i === A.length || j === B.length) {
return count;
}
// If the current elements are equal, recursively check the next elements
if (A[i] === B[j]) {
count = LCS(i + 1, j + 1, A, B, count + 1);
}
// Recursively check for the longest common subarray by considering two possibilities:
// 1. Exclude the current element from array A and continue with array B
// 2. Exclude the current element from array B and continue with array A
count = Math.max(count, Math.max(LCS(i + 1, j, A, B, 0), LCS(i, j + 1, A, B, 0)));
return count;
}
// Example arrays
const A = [1, 2, 3, 2, 1];
const B = [3, 2, 1, 4, 7];
// Call the LCS function to find the maximum length of the common subarray
const maxLength = LCS(0, 0, A, B, 0);
// Print the result
console.log("Max length of common subarray: " + maxLength);
OutputMax length of common subarray: 3
Time Complexity: O(2N+M), where N is the length of the array A[] and M is the length of the array B[].
Efficient Approach:
The efficient approach is to use Dynamic Programming(DP). This problem is the variation of the Longest Common Subsequence(LCS).
Let the input sequences are A[0..n-1] and B[0..m-1] of lengths m & n respectively. Following is the recursive implementation of the equal subarrays:
- Since common subarray of A[] and B[] must start at some index i and j such that A[i] is equals to B[j]. Let dp[i][j] be the longest common subarray of A[i...] and B[j...].
- Therefore, for any index i and j, if A[i] is equals to B[j], then dp[i][j] = dp[i+1][j+1] + 1.
- The maximum of all the elements in the array dp[][] will give the maximum length of equal subarrays.
For Example:
If the given array A[] = {1, 2, 8, 2, 1} and B[] = {8, 2, 1, 4, 7}.
If the characters match at index i and j for the array A[] and B[] respectively, then dp[i][j] will be updated as 1 + dp[i+1][j+1].
Below is the updated dp[][] table for the given array A[] and B[].
Below is the implementation of the above approach:
C++
// C++ program to DP approach
// to above solution
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum
// length of equal subarray
int FindMaxLength(int A[], int B[], int n, int m)
{
// Auxiliary dp[][] array
int dp[n + 1][m + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
dp[i][j] = 0;
// Updating the dp[][] table
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--) {
for (int j = m - 1; j >= 0; j--) {
// If A[i] is equal to B[i]
// then dp[j][i] = dp[j + 1][i + 1] + 1
if (A[i] == B[j])
dp[i][j] = dp[i + 1][j + 1] + 1;
}
}
int maxm = 0;
// Find maximum of all the values
// in dp[][] array to get the
// maximum length
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Update the length
maxm = max(maxm, dp[i][j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
int main()
{
int A[] = { 1, 2, 8, 2, 1 };
int B[] = { 8, 2, 1, 4, 7 };
int n = sizeof(A) / sizeof(A[0]);
int m = sizeof(B) / sizeof(B[0]);
// Function call to find
// maximum length of subarray
cout << (FindMaxLength(A, B, n, m));
}
// This code is contributed by chitranayal
// Java program to DP approach
// to above solution
class GFG
{
// Function to find the maximum
// length of equal subarray
static int FindMaxLength(int A[], int B[], int n, int m)
{
// Auxiliary dp[][] array
int[][] dp = new int[n + 1][m + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
dp[i][j] = 0;
// Updating the dp[][] table
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--)
{
for (int j = m - 1; j >= 0; j--)
{
// If A[i] is equal to B[i]
// then dp[j][i] = dp[j + 1][i + 1] + 1
if (A[i] == B[j])
dp[i][j] = dp[i + 1][j + 1] + 1;
}
}
int maxm = 0;
// Find maximum of all the values
// in dp[][] array to get the
// maximum length
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Update the length
maxm = Math.max(maxm, dp[i][j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 2, 8, 2, 1 };
int B[] = { 8, 2, 1, 4, 7 };
int n = A.length;
int m = B.length;
// Function call to find
// maximum length of subarray
System.out.print(FindMaxLength(A, B, n, m));
}
}
// This code is contributed by PrinciRaj1992
# Python program to DP approach
# to above solution
# Function to find the maximum
# length of equal subarray
def FindMaxLength(A, B):
n = len(A)
m = len(B)
# Auxiliary dp[][] array
dp = [[0 for i in range(n + 1)] for i in range(m + 1)]
# Updating the dp[][] table
# in Bottom Up approach
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
# If A[i] is equal to B[i]
# then dp[j][i] = dp[j + 1][i + 1] + 1
if A[i] == B[j]:
dp[i][j] = dp[i + 1][j + 1] + 1
maxm = 0
# Find maximum of all the values
# in dp[][] array to get the
# maximum length
for i in dp:
for j in i:
# Update the length
maxm = max(maxm, j)
# Return the maximum length
return maxm
# Driver Code
if __name__ == '__main__':
A = [1, 2, 8, 2, 1]
B = [8, 2, 1, 4, 7]
# Function call to find
# maximum length of subarray
print(FindMaxLength(A, B))
// C# program to DP approach
// to above solution
using System;
class GFG
{
// Function to find the maximum
// length of equal subarray
static int FindMaxLength(int[] A, int[] B, int n, int m)
{
// Auxiliary [,]dp array
int[, ] dp = new int[n + 1, m + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
dp[i, j] = 0;
// Updating the [,]dp table
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--)
{
for (int j = m - 1; j >= 0; j--)
{
// If A[i] is equal to B[i]
// then dp[j, i] = dp[j + 1, i + 1] + 1
if (A[i] == B[j])
dp[i, j] = dp[i + 1, j + 1] + 1;
}
}
int maxm = 0;
// Find maximum of all the values
// in [,]dp array to get the
// maximum length
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Update the length
maxm = Math.Max(maxm, dp[i, j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
public static void Main(String[] args)
{
int[] A = { 1, 2, 8, 2, 1 };
int[] B = { 8, 2, 1, 4, 7 };
int n = A.Length;
int m = B.Length;
// Function call to find
// maximum length of subarray
Console.Write(FindMaxLength(A, B, n, m));
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript program to DP approach
// to above solution
// Function to find the maximum
// length of equal subarray
function FindMaxLength(A,B,n,m)
{
// Auxiliary dp[][] array
let dp = new Array(n + 1);
for (let i = 0; i <= n; i++)
{
dp[i]=new Array(m+1);
for (let j = 0; j <= m; j++)
dp[i][j] = 0;
}
// Updating the dp[][] table
// in Bottom Up approach
for (let i = n - 1; i >= 0; i--)
{
for (let j = m - 1; j >= 0; j--)
{
// If A[i] is equal to B[i]
// then dp[i][j] = dp[i + 1][j + 1] + 1
if (A[i] == B[j])
dp[j][i] = dp[j + 1][i + 1] + 1;
}
}
let maxm = 0;
// Find maximum of all the values
// in dp[][] array to get the
// maximum length
for (let i = 0; i < n; i++)
{
for (let j = 0; j < m; j++)
{
// Update the length
maxm = Math.max(maxm, dp[i][j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
let A=[1, 2, 8, 2, 1 ];
let B=[8, 2, 1, 4, 7];
let n = A.length;
let m = B.length;
// Function call to find
// maximum length of subarray
document.write(FindMaxLength(A, B, n, m));
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(N*M), where N is the length of array A[] and M is the length of array B[].
Auxiliary Space: O(N*M)
Efficient approach: space optimization
In previous approach the dp[i][j] is depend upon the current and previous row of 2D matrix. So to optimize the space complexity we use only single vector dp for current row and a variable prev to get the previous computation.
Implementation Steps :
- Create a vector DP of size M+1 , that stores the computation of subproblems and initialize it with 0.
- Initialize a variable maxm with 0 used to store the maximum length.
- Now iterative over subproblems and update the DP vector in bottom up approach.
- Initialize variable prev and temp. prev is used to get the previous row value of Dp and temp is used to get the value of prev in another iteration.
- While iterating update the maxm with respect to the value stored in DP.
- At last return the maxm.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum
// length of equal subarray
int FindMaxLength(int A[], int B[], int n, int m)
{
// Auxiliary dp[] vector
vector<int> dp(m + 1, 0);
int maxm = 0;
// Updating the dp[] vector
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--) {
int prev = 0;
for (int j = m - 1; j >= 0; j--) {
int temp = dp[j];
if (A[i] == B[j]) {
dp[j] = prev + 1;
maxm = max(maxm, dp[j]);
}
else {
dp[j] = 0;
}
prev = temp;
}
}
// Return the maximum length
return maxm;
}
// Driver Code
int main()
{
int A[] = { 1, 2, 8, 2, 1 };
int B[] = { 8, 2, 1, 4, 7 };
int n = sizeof(A) / sizeof(A[0]);
int m = sizeof(B) / sizeof(B[0]);
// Function call to find
// maximum length of subarray
cout << (FindMaxLength(A, B, n, m));
}
import java.util.Arrays;
public class Main {
// Function to find the maximum length of equal subarray
static int findMaxLength(int[] A, int[] B, int n, int m) {
// Auxiliary dp[] array
int[] dp = new int[m + 1];
int maxm = 0;
// Updating the dp[] array in Bottom Up approach
for (int i = n - 1; i >= 0; i--) {
int prev = 0;
for (int j = m - 1; j >= 0; j--) {
int temp = dp[j];
if (A[i] == B[j]) {
dp[j] = prev + 1;
maxm = Math.max(maxm, dp[j]);
} else {
dp[j] = 0;
}
prev = temp;
}
}
// Return the maximum length
return maxm;
}
public static void main(String[] args) {
int[] A = { 1, 2, 8, 2, 1 };
int[] B = { 8, 2, 1, 4, 7 };
int n = A.length;
int m = B.length;
// Function call to find maximum length of subarray
System.out.println(findMaxLength(A, B, n, m));
}
}
# code
def find_max_length(A, B, n, m):
# Auxiliary dp[] list
dp = [0] * (m + 1)
maxm = 0
# Updating the dp[] list
# in Bottom Up approach
for i in range(n - 1, -1, -1):
prev = 0
for j in range(m - 1, -1, -1):
temp = dp[j]
if A[i] == B[j]:
dp[j] = prev + 1
maxm = max(maxm, dp[j])
else:
dp[j] = 0
prev = temp
# Return the maximum length
return maxm
# Driver Code
if __name__ == '__main__':
A = [1, 2, 8, 2, 1]
B = [8, 2, 1, 4, 7]
n = len(A)
m = len(B)
print(find_max_length(A, B, n, m))
using System;
class Program
{
// Function to find the maximum
// length of equal subarray
static int FindMaxLength(int[] A, int[] B, int n, int m)
{
// Auxiliary dp[] array
int[] dp = new int[m + 1];
int maxm = 0;
// Updating the dp[] array
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--)
{
int prev = 0;
for (int j = m - 1; j >= 0; j--)
{
int temp = dp[j];
if (A[i] == B[j])
{
dp[j] = prev + 1;
maxm = Math.Max(maxm, dp[j]);
}
else
{
dp[j] = 0;
}
prev = temp;
}
}
// Return the maximum length
return maxm;
}
// Driver Code
static void Main()
{
int[] A = { 1, 2, 8, 2, 1 };
int[] B = { 8, 2, 1, 4, 7 };
int n = A.Length;
int m = B.Length;
// Function call to find
// maximum length of subarray
Console.WriteLine(FindMaxLength(A, B, n, m));
}
}
// Function to find the maximum length of equal subarray
function FindMaxLength(A, B, n, m) {
// Auxiliary dp[] vector
let dp = new Array(m + 1).fill(0);
let maxm = 0;
// Updating the dp[] vector in Bottom Up approach
for (let i = n - 1; i >= 0; i--) {
let prev = 0;
for (let j = m - 1; j >= 0; j--) {
let temp = dp[j];
if (A[i] == B[j]) {
dp[j] = prev + 1;
maxm = Math.max(maxm, dp[j]);
} else {
dp[j] = 0;
}
prev = temp;
}
}
// Return the maximum length
return maxm;
}
let A = [1, 2, 8, 2, 1];
let B = [8, 2, 1, 4, 7];
let n = A.length;
let m = B.length;
// Function call to find maximum length of subarray
console.log(FindMaxLength(A, B, n, m));
Time Complexity: O(N*M)
Auxiliary Space: O(M) , only use one vector dp of size M
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