Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Uncrossed Lines in C++
Suppose we have written the integers of A and B (in the order they are given) on two separate horizontal lines. Now, we may draw connecting lines: a straight line connecting two numbers A[i] and B[j] such that −
A[i] == B[j];
The line we draw that does not intersect any other connecting (non-horizontal) line.
We have to keep in mind that connecting lines cannot intersect even at the endpoints − each number can only belong to one connecting line. Find the maximum number of connecting lines. So if the input is like [1,4,2] and [1,2,4], then the output will be 2.
| 1 | 4 | 2 |
| 1 | 2 | 4 |
We can draw 2 uncrossed lines as in the diagram. We cannot draw 3 uncrossed lines, this is because the line from A[1]=4 to B[2]=4 will intersect the line from A[2]=2 to B[1]=2.
To solve this, we will follow these steps −
Define a method called solve(), this will take, i, j, the array A, array B and the matrix dp
if i is out of range of array A, return 0
if j is out of range of array B, then return 0
nj := j
while nj < size of B and B[nj] is not A[i]
increase nj by 1
temp := 1 when nj < size of B, otherwise 0
ret := max of (solve(i + 1, j, A, B, dp) and temp) + solve(i + 1, nj + 1, A, B, dp)
dp[i, j] := ret and return ret
From the main method
n := size of A, m := size of B
create a matrix dp of size n x m, and fill this with – 1
call solve(0, 0, A, , dp)
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int solve(int i, int j, vector <int>&A, vector <int>&B, vector <
vector <int> >& dp){
if(i >= A.size()) return 0;
if(j >= B.size()) return 0;
if(dp[i][j] != -1) return dp[i][j];
int nj = j;
while(nj < B.size() && B[nj] != A[i]) nj++;
int ret = max(solve(i + 1, j, A, B, dp), (nj < B.size() ? 1 :
0) + solve(i + 1, nj + 1, A, B, dp));
return dp[i][j] = ret;
}
int maxUncrossedLines(vector<int>& A, vector<int>& B) {
int n = A.size();
int m = B.size();
vector < vector <int > > dp(n, vector <int>(m, -1));
return solve(0, 0, A, B, dp);
}
};
main(){
vector<int> v1 = {1,4,2};
vector<int> v2 = {1,2,4};
Solution ob;
cout << (ob.maxUncrossedLines(v1, v2));
}Input
[1,4,2] [1,2,4]
Output
2
142 Views
