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 arraysint 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 Codeint 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 approachimport java.io.*;class GFG{// Function to count maximum number// of uncrossed lines between the// two given arraysstatic 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 Codepublic 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 arraysdef 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 Codeif __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 approachusing System;class GFG{// Function to count maximum number// of uncrossed lines between the// two given arraysstatic 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 Codepublic 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 arraysfunction 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 calldocument.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 arraysint 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 Codeint 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 approachimport 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 arraysdef 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 Codeif __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 approachusing 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 arraysfunction uncrossedLines(a, b, n, m) {// Stores the length of lcs// obtained upto every indexlet dp = new Array(m + 1).fill(0);// Iterate over first arrayfor (let i = 1; i <= n; i++) {// Initialize prev to 0let prev = 0;// Iterate over second arrayfor (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 answerreturn 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 Callconsole.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 arraysdef 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 codeif __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 LCSfunction 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 functionfunction 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
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