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Maximum number of uncrossed lines between two given arrays

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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:
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

// 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

# 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#

// 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
}

                    

Javascript

<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>

                    

Output
2







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

// 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));
  }
}

                    

Python3

# 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#

// 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

// 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;
}

                    

Java

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));
    }
}

                    

Python3

# 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))

                    

C#

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

                    

Javascript

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();

                    

Output
2







Time complexity: O(M*N),two loops iterations 

Auxiliary Space: O(M+N),Exta dp array required to store the desired results



Last Updated : 20 Oct, 2023
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