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# Maximum number of uncrossed lines between two given arrays

• Difficulty Level : Hard
• Last Updated : 29 Mar, 2023

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 ` `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

 ``

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 ` `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));` `  ``}` `}`

Output

`2`

Time Complexity: O(N*M)
Auxiliary Space: O(M)

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