Maximum number of uncrossed lines between two given arrays
Last Updated :
15 Jul, 2025
Given two arrays A[] and B[], the task is to find the maximum number of uncrossed lines between the elements of the two given arrays.
A straight line can be drawn between two array elements A[i] and B[j] only if:
- A[i] = B[j]
- The line does not intersect any other line.
Examples:
Input: A[] = {3, 9, 2}, B[] = {3, 2, 9}
Output: 2
Explanation:
The lines between A[0] to B[0] and A[1] to B[2] does not intersect each other.
Input: A[] = {1, 2, 3, 4, 5}, B[] = {1, 2, 3, 4, 5}
Output: 5
Naive Approach: The idea is to generate all the subsequences of array A[] and try to find them in array B[] so that the two subsequences can be connected by joining straight lines. The longest such subsequence found to be common in A[] and B[] would have the maximum number of uncrossed lines. So print the length of that subsequence.
Time Complexity: O(M * 2N)
Auxiliary Space: O(1)
Efficient Approach: From the above approach, it can be observed that the task is to find the longest subsequence common in both the arrays. Therefore, the above approach can be optimized by finding the Longest Common Subsequence between the two arrays using Dynamic Programming.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
int uncrossedLines(int* a, int* b,
int n, int m)
{
// Stores the length of lcs
// obtained upto every index
int dp[n + 1][m + 1];
// Iterate over first array
for (int i = 0; i <= n; i++) {
// Iterate over second array
for (int j = 0; j <= m; j++) {
if (i == 0 || j == 0)
// Update value in dp table
dp[i][j] = 0;
// If both characters
// are equal
else if (a[i - 1] == b[j - 1])
// Update the length of lcs
dp[i][j] = 1 + dp[i - 1][j - 1];
// If both characters
// are not equal
else
// Update the table
dp[i][j] = max(dp[i - 1][j],
dp[i][j - 1]);
}
}
// Return the answer
return dp[n][m];
}
// Driver Code
int main()
{
// Given array A[] and B[]
int A[] = { 3, 9, 2 };
int B[] = { 3, 2, 9 };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
// Function Call
cout << uncrossedLines(A, B, N, M);
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG{
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
static int uncrossedLines(int[] a, int[] b,
int n, int m)
{
// Stores the length of lcs
// obtained upto every index
int[][] dp = new int[n + 1][m + 1];
// Iterate over first array
for(int i = 0; i <= n; i++)
{
// Iterate over second array
for(int j = 0; j <= m; j++)
{
if (i == 0 || j == 0)
// Update value in dp table
dp[i][j] = 0;
// If both characters
// are equal
else if (a[i - 1] == b[j - 1])
// Update the length of lcs
dp[i][j] = 1 + dp[i - 1][j - 1];
// If both characters
// are not equal
else
// Update the table
dp[i][j] = Math.max(dp[i - 1][j],
dp[i][j - 1]);
}
}
// Return the answer
return dp[n][m];
}
// Driver Code
public static void main (String[] args)
{
// Given array A[] and B[]
int A[] = { 3, 9, 2 };
int B[] = { 3, 2, 9 };
int N = A.length;
int M = B.length;
// Function call
System.out.print(uncrossedLines(A, B, N, M));
}
}
// This code is contributed by code_hunt
# Python3 program for
# the above approach
# Function to count maximum number
# of uncrossed lines between the
# two given arrays
def uncrossedLines(a, b,
n, m):
# Stores the length of lcs
# obtained upto every index
dp = [[0 for x in range(m + 1)]
for y in range(n + 1)]
# Iterate over first array
for i in range (n + 1):
# Iterate over second array
for j in range (m + 1):
if (i == 0 or j == 0):
# Update value in dp table
dp[i][j] = 0
# If both characters
# are equal
elif (a[i - 1] == b[j - 1]):
# Update the length of lcs
dp[i][j] = 1 + dp[i - 1][j - 1]
# If both characters
# are not equal
else:
# Update the table
dp[i][j] = max(dp[i - 1][j],
dp[i][j - 1])
# Return the answer
return dp[n][m]
# Driver Code
if __name__ == "__main__":
# Given array A[] and B[]
A = [3, 9, 2]
B = [3, 2, 9]
N = len(A)
M = len(B)
# Function Call
print (uncrossedLines(A, B, N, M))
# This code is contributed by Chitranayal
// C# program for the above approach
using System;
class GFG{
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
static int uncrossedLines(int[] a, int[] b,
int n, int m)
{
// Stores the length of lcs
// obtained upto every index
int[,] dp = new int[n + 1, m + 1];
// Iterate over first array
for(int i = 0; i <= n; i++)
{
// Iterate over second array
for(int j = 0; j <= m; j++)
{
if (i == 0 || j == 0)
// Update value in dp table
dp[i, j] = 0;
// If both characters
// are equal
else if (a[i - 1] == b[j - 1])
// Update the length of lcs
dp[i, j] = 1 + dp[i - 1, j - 1];
// If both characters
// are not equal
else
// Update the table
dp[i, j] = Math.Max(dp[i - 1, j],
dp[i, j - 1]);
}
}
// Return the answer
return dp[n, m];
}
// Driver Code
public static void Main (String[] args)
{
// Given array A[] and B[]
int[] A = { 3, 9, 2 };
int[] B = { 3, 2, 9 };
int N = A.Length;
int M = B.Length;
// Function call
Console.Write(uncrossedLines(A, B, N, M));
}
}
// This code is contributed by code_hunt
}
<script>
// Javascript program for the above approach
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
function uncrossedLines(a, b, n, m)
{
// Stores the length of lcs
// obtained upto every index
let dp = new Array(n + 1);
for(let i = 0; i< (n + 1); i++)
{
dp[i] = new Array(m + 1);
for(let j = 0; j < (m + 1); j++)
{
dp[i][j] = 0;
}
}
// Iterate over first array
for(let i = 0; i <= n; i++)
{
// Iterate over second array
for(let j = 0; j <= m; j++)
{
if (i == 0 || j == 0)
// Update value in dp table
dp[i][j] = 0;
// If both characters
// are equal
else if (a[i - 1] == b[j - 1])
// Update the length of lcs
dp[i][j] = 1 + dp[i - 1][j - 1];
// If both characters
// are not equal
else
// Update the table
dp[i][j] = Math.max(dp[i - 1][j],
dp[i][j - 1]);
}
}
// Return the answer
return dp[n][m];
}
// Driver Code
// Given array A[] and B[]
let A = [ 3, 9, 2 ];
let B = [3, 2, 9];
let N = A.length;
let M = B.length;
// Function call
document.write(uncrossedLines(A, B, N, M));
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(N*M)
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 space we use a 1D vectors dp to store previous value and use prev to store the previous diagonal element and get the current computation.
Implementation Steps:
- Define a vector dp of size m+1 and initialize its first element to 0.
- For each element j in b[], iterate in reverse order from n to 1 and update dp[i] as follows:
a. If a[i - 1] == b[j - 1], set dp[j] to the previous value of dp[i-1] + 1 (diagonal element).
b. If a[i-1] != b[j-1], set dp[j] to the maximum value between dp[j] and dp[j-1] (value on the left). - Finally, return dp[m].
Implementation:
C++
// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
int uncrossedLines(int* a, int* b, int n, int m)
{
// Stores the length of lcs
// obtained upto every index
vector<int> dp(m + 1, 0);
// Iterate over first array
for (int i = 1; i <= n; i++) {
// Initialize prev to 0
int prev = 0;
// Iterate over second array
for (int j = 1; j <= m; j++) {
// Store the current dp[j]
int curr = dp[j];
if (a[i - 1] == b[j - 1])
dp[j] = prev + 1;
else
dp[j] = max(dp[j], dp[j - 1]);
// Update prev
prev = curr;
}
}
// Return the answer
return dp[m];
}
// Driver Code
int main()
{
// Given array A[] and B[]
int A[] = { 3, 9, 2 };
int B[] = { 3, 2, 9 };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
// Function Call
cout << uncrossedLines(A, B, N, M);
return 0;
}
// this code is contributed by bhardwajji
// Java code for above approach
import java.io.*;
class Main {
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
static int uncrossedLines(int[] a, int[] b, int n, int m)
{
// Stores the length of lcs
// obtained upto every index
int[] dp = new int[m + 1];
// Iterate over first array
for (int i = 1; i <= n; i++) {
// Initialize prev to 0
int prev = 0;
// Iterate over second array
for (int j = 1; j <= m; j++) {
// Store the current dp[j]
int curr = dp[j];
if (a[i - 1] == b[j - 1])
dp[j] = prev + 1;
else
dp[j] = Math.max(dp[j], dp[j - 1]);
// Update prev
prev = curr;
}
}
// Return the answer
return dp[m];
}
// Driver Code
public static void main(String args[]) {
// Given array A[] and B[]
int[] A = { 3, 9, 2 };
int[] B = { 3, 2, 9 };
int N = A.length;
int M = B.length;
// Function Call
System.out.print(uncrossedLines(A, B, N, M));
}
}
# Function to count maximum number
# of uncrossed lines between the
# two given arrays
def uncrossedLines(a, b, n, m):
# Stores the length of lcs
# obtained upto every index
dp = [0] * (m + 1)
# Iterate over first array
for i in range(1, n + 1):
# Initialize prev to 0
prev = 0
# Iterate over second array
for j in range(1, m + 1):
# Store the current dp[j]
curr = dp[j]
if a[i - 1] == b[j - 1]:
dp[j] = prev + 1
else:
dp[j] = max(dp[j], dp[j - 1])
# Update prev
prev = curr
# Return the answer
return dp[m]
# Driver Code
if __name__ == '__main__':
# Given array A[] and B[]
A = [3, 9, 2]
B = [3, 2, 9]
N = len(A)
M = len(B)
# Function Call
print(uncrossedLines(A, B, N, M))
// C# code for above approach
using System;
class GFG {
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
static int UncrossedLines(int[] a, int[] b, int n,
int m)
{
// Stores the length of lcs
// obtained upto every index
int[] dp = new int[m + 1];
Array.Fill(dp, 0);
// Iterate over first array
for (int i = 1; i <= n; i++) {
// Initialize prev to 0
int prev = 0;
// Iterate over second array
for (int j = 1; j <= m; j++) {
// Store the current dp[j]
int curr = dp[j];
if (a[i - 1] == b[j - 1])
dp[j] = prev + 1;
else
dp[j] = Math.Max(dp[j], dp[j - 1]);
// Update prev
prev = curr;
}
}
// Return the answer
return dp[m];
}
// Driver Code
public static void Main()
{
// Given array A[] and B[]
int[] A = { 3, 9, 2 };
int[] B = { 3, 2, 9 };
int N = A.Length;
int M = B.Length;
// Function Call
Console.WriteLine(UncrossedLines(A, B, N, M));
}
}
// JavaScript code for above approach
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
function uncrossedLines(a, b, n, m)
{
// Stores the length of lcs
// obtained upto every index
let dp = new Array(m + 1).fill(0);
// Iterate over first array
for (let i = 1; i <= n; i++)
{
// Initialize prev to 0
let prev = 0;
// Iterate over second array
for (let j = 1; j <= m; j++)
{
// Store the current dp[j]
let curr = dp[j];
if (a[i - 1] == b[j - 1]) dp[j] = prev + 1;
else dp[j] = Math.max(dp[j], dp[j - 1]);
// Update prev
prev = curr;
}
}
// Return the answer
return dp[m];
}
// Driver Code
// Given array A[] and B[]
let A = [3, 9, 2];
let B = [3, 2, 9];
let N = A.length;
let M = B.length;
// Function Call
console.log(uncrossedLines(A, B, N, M));
Output
2
Time Complexity: O(N*M)
Auxiliary Space: O(M)
Memoization(Top Down) Approach:
0 1 2 3
0 +--+--+--+
| | | |
1 +--+--+--+
| |? |? |
2 +--+? | + |
| | + |? |
3 +--+--+? |
| | | + |
+--+--+--+
0 1 2 3
0 + 0 0 0
| | | |
1 + 0 1 1
| |?|?|
2 + 0 1+1+
| |+|?|
3 + 0 1 2+
| | |+|
+ + + +
Hint:
First, add one dummy -1 to A and B to represent empty list
Then, we define the notation DP[ y ][ x ].
Let DP[y][x] denote the maximal number of uncrossed lines between A[ 1 ... y ] and B[ 1 ... x ]
We have optimal substructure as following:
Base case:
Any sequence with empty list yield no uncrossed lines.
If y = 0 or x = 0:
DP[ y ][ x ] = 0
General case:
If A[ y ] == B[ x ]:
DP[ y ][ x ] = DP[ y-1 ][ x-1 ] + 1
Current last number is matched, therefore, add one more uncrossed line
If A[ y ] =/= B[ x ]:
DP[ y ][ x ] = Max( DP[ y ][ x-1 ], DP[ y-1 ][ x ] )
Current last number is not matched,
backtrack to A[ 1...y ]B[ 1...x-1 ], A[ 1...y-1 ]B[ 1...x ]
to find maximal number of uncrossed line
Top-down DP; for each step we can decide to draw the line from the current pointer i (if possible, add this line to the result), or skip this position. Maximize the result of these two choices.
This is a simplified solution when we just scan the other array to find the matching value; we can use some faster lockup method instead. However, the memoisation helps and the simplified solution has the same runtime as the optimized solution with hast set + set.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
vector<vector<int>>dp;
// Stores the length of lcs
// obtained upto every index
int helper(int i,int j,vector<int>&nums1,vector<int>&nums2){
//Check for the base condition
if(i==-1||j==-1)return 0;
//Check if the value already exist in the dp array
if(dp[i][j]!=-1)return dp[i][j];
//check for equality
if(nums1[i]==nums2[j])return dp[i][j]=1+helper(i-1,j-1,nums1,nums2);
//return the max value of the uncrossed lines
return dp[i][j]=max(helper(i-1,j,nums1,nums2),helper(i,j-1,nums1,nums2));
}
int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {
int n1=nums1.size();
int n2=nums2.size();
//make the dp array size according to the inputs
dp.resize(n1,vector<int>(n2,-1));
//return the resultant answer
return helper(n1-1,n2-1,nums1,nums2);
}
int main() {
//Declare two vectors
vector<int> A{ 3, 9, 2 };
vector<int> B{ 3, 2, 9 };
// Function Call
cout << maxUncrossedLines(A, B);
return 0;
}
import java.util.Arrays;
public class GFG {
// Function to count maximum number
// of uncrossed lines between the
// two given arrays
static int[][] dp;
// Stores the length of lcs
// obtained up to every index
static int helper(int i, int j, int[] nums1, int[] nums2) {
// Check for the base condition
if (i == -1 || j == -1)
return 0;
// Check if the value already exists in the dp array
if (dp[i][j] != -1)
return dp[i][j];
// Check for equality
if (nums1[i] == nums2[j])
return dp[i][j] = 1 + helper(i - 1, j - 1, nums1, nums2);
// Return the max value of the uncrossed lines
return dp[i][j] = Math.max(helper(i - 1, j, nums1, nums2),
helper(i, j - 1, nums1, nums2));
}
static int maxUncrossedLines(int[] nums1, int[] nums2) {
int n1 = nums1.length;
int n2 = nums2.length;
// Make the dp array size according to the inputs
dp = new int[n1][n2];
for (int[] row : dp) {
Arrays.fill(row, -1);
}
// Return the resultant answer
return helper(n1 - 1, n2 - 1, nums1, nums2);
}
public static void main(String[] args) {
// Declare two arrays
int[] A = {3, 9, 2};
int[] B = {3, 2, 9};
// Function Call
System.out.println(maxUncrossedLines(A, B));
}
}
# Function to count maximum number
# of uncrossed lines between the
# two given arrays
def max_uncrossed_lines(nums1, nums2):
# Helper function to calculate the LCS and maximum uncrossed lines
def helper(i, j, nums1, nums2):
# Check for the base condition
if i == -1 or j == -1:
return 0
# Check if the value already exists in the dp array
if dp[i][j] != -1:
return dp[i][j]
# Check for equality
if nums1[i] == nums2[j]:
dp[i][j] = 1 + helper(i - 1, j - 1, nums1, nums2)
else:
# Return the max value of the uncrossed lines
dp[i][j] = max(helper(i - 1, j, nums1, nums2), helper(i, j - 1, nums1, nums2))
return dp[i][j]
n1 = len(nums1)
n2 = len(nums2)
# Initialize the dp array with -1
dp = [[-1 for _ in range(n2)] for _ in range(n1)]
# Return the result using helper function
return helper(n1 - 1, n2 - 1, nums1, nums2)
# Driver code
if __name__ == "__main__":
# Declare two lists
A = [3, 9, 2]
B = [3, 2, 9]
# Function Call
print(max_uncrossed_lines(A, B))
using System;
using System.Collections.Generic;
class MainClass
{
static List<List<int>> dp;
static int Helper(int i, int j, List<int> nums1, List<int> nums2)
{
// Check for the base condition
if (i == -1 || j == -1) return 0;
// Check if the value already exists in the dp array
if (dp[i][j] != -1) return dp[i][j];
// Check for equality
if (nums1[i] == nums2[j]) return dp[i][j] = 1 + Helper(i - 1, j - 1, nums1, nums2);
// Return the max value of the uncrossed lines
return dp[i][j] = Math.Max(Helper(i - 1, j, nums1, nums2), Helper(i, j - 1, nums1, nums2));
}
static int MaxUncrossedLines(List<int> nums1, List<int> nums2)
{
int n1 = nums1.Count;
int n2 = nums2.Count;
// Make the dp array size according to the inputs
dp = new List<List<int>>();
for (int i = 0; i < n1; i++)
{
dp.Add(new List<int>());
for (int j = 0; j < n2; j++)
{
dp[i].Add(-1);
}
}
// Return the resultant answer
return Helper(n1 - 1, n2 - 1, nums1, nums2);
}
public static void Main(string[] args)
{
// Declare two lists
List<int> A = new List<int> { 3, 9, 2 };
List<int> B = new List<int> { 3, 2, 9 };
// Function Call
Console.WriteLine(MaxUncrossedLines(A, B));
}
}
// This code is contributed by rambabuguphka
let dp = [];
// Stores the length of LCS
function GFG(i, j, nums1, nums2) {
// Check for the base condition
if (i === -1 || j === -1) return 0;
// Check if the value already exists in the
// dp array
if (dp[i][j] !== undefined) return dp[i][j];
// Check for equality
if (nums1[i] === nums2[j]) return (dp[i][j] = 1 + GFG(i - 1, j - 1, nums1, nums2));
// Return the max value of the uncrossed lines
return (dp[i][j] = Math.max(GFG(i - 1, j, nums1, nums2), GFG(i, j - 1, nums1, nums2)));
}
function maxUncrossedLines(nums1, nums2) {
const n1 = nums1.length;
const n2 = nums2.length;
// Make the dp array size according to the inputs
dp = new Array(n1).fill(null).map(() => new Array(n2).fill(undefined));
// Return the resultant answer
return GFG(n1 - 1, n2 - 1, nums1, nums2);
}
// Main function
function main() {
// Declare two arrays
const A = [3, 9, 2];
const B = [3, 2, 9];
// Function Call
console.log(maxUncrossedLines(A, B));
}
main();
Time complexity: O(M*N),two loops iterations
Auxiliary Space: O(M+N),Exta dp array required to store the desired results
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Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
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