Consider a 2-D map with a horizontal river passing through its center. There are n cities on the southern bank with x-coordinates a(1) … a(n) and n cities on the northern bank with x-coordinates b(1) … b(n). You want to connect as many north-south pairs of cities as possible with bridges such that no two bridges cross. When connecting cities, you can only connect city a(i) on the northern bank to city b(i) on the southern bank. Maximum number of bridges that can be built to connect north-south pairs with the aforementioned constraints.

The values in the upper bank can be considered as the northern x-coordinates of the cities and the values in the bottom bank can be considered as the corresponding southern x-coordinates of the cities to which the northern x-coordinate city can be connected.

Examples:

Input : 6 4 2 1 2 3 6 5 Output :Maximum number of bridges= 2Explanation:Let the north-south x-coordinates be written in increasing order. 1 2 3 4 5 6 \ \ \ \ For the north-south pairs \ \(2, 6)and(1, 5)\ \ the bridges can be built. \ \ We can consider other pairs also, \ \ but then only one bridge can be built \ \ because more than one bridge built will \ \ then cross each other. \ \ 1 2 3 4 5 6 Input : 8 1 4 3 5 2 6 7 1 2 3 4 5 6 7 8 Output :Maximum number of bridges= 2

**Approach:** It is a variation of LIS problem. The following are the steps to solve the problem.

- Sort the north-south pairs on the basis of increasing order of south x-coordinates.
- If two south x-coordinates are same, then sort on the basis of increasing order of north x-coordinates.
- Now find the Longest Increasing Subsequence of the north x-coordinates.
- One thing to note that in the increasing subsequence a value can be greater as well as can be equal to its previous value.

We can also sort on the basis of north x-coordinates and find the LIS on the south x-coordinates.

`// C++ implementation of building bridges ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `// north-south coodinates ` `// of each City Pair ` `struct` `CityPairs ` `{ ` ` ` `int` `north, south; ` `}; ` ` ` `// comparison function to sort ` `// the given set of CityPairs ` `bool` `compare(` `struct` `CityPairs a, ` `struct` `CityPairs b) ` `{ ` ` ` `if` `(a.south == b.south) ` ` ` `return` `a.north < b.north; ` ` ` `return` `a.south < b.south; ` `} ` ` ` `// function to find the maximum number ` `// of bridges that can be built ` `int` `maxBridges(` `struct` `CityPairs values[], ` `int` `n) ` `{ ` ` ` `int` `lis[n]; ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `lis[i] = 1; ` ` ` ` ` `sort(values, values+n, compare); ` ` ` ` ` `// logic of longest increasing subsequence ` ` ` `// applied on the northern coordinates ` ` ` `for` `(` `int` `i=1; i<n; i++) ` ` ` `for` `(` `int` `j=0; j<i; j++) ` ` ` `if` `(values[i].north >= values[j].north ` ` ` `&& lis[i] < 1 + lis[j]) ` ` ` `lis[i] = 1 + lis[j]; ` ` ` ` ` ` ` `int` `max = lis[0]; ` ` ` `for` `(` `int` `i=1; i<n; i++) ` ` ` `if` `(max < lis[i]) ` ` ` `max = lis[i]; ` ` ` ` ` `// required number of bridges ` ` ` `// that can be built ` ` ` `return` `max; ` `} ` ` ` `// Driver program to test above ` `int` `main() ` `{ ` ` ` `struct` `CityPairs values[] = {{6, 2}, {4, 3}, {2, 6}, {1, 5}}; ` ` ` `int` `n = 4; ` ` ` `cout << ` `"Maximum number of bridges = "` ` ` `<< maxBridges(values, n); ` ` ` `return` `0; ` `} ` |

Output:

Maximum number of bridges = 2

Time Complexity: O(n^{2})

Problem References:

https://www.geeksforgeeks.org/dynamic-programming-set-14-variations-of-lis/

Solution References:

https://www.youtube.com/watch?v=w6tSmS86C4w

This article is contributed by **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Longest Increasing Path in Matrix
- Minimum insertions to sort an array
- Shortest Uncommon Subsequence
- Count number of ways to jump to reach end
- Subset Sum Problem | DP-25
- Optimal Binary Search Tree | DP-24
- Box Stacking Problem | DP-22
- Variations of LIS | DP-21
- Maximum Length Chain of Pairs | DP-20
- Partition problem | DP-18
- Palindrome Partitioning | DP-17
- Maximum Sum Increasing Subsequence | DP-14
- 0-1 Knapsack Problem | DP-10
- Longest Common Subsequence | DP-4
- Longest Increasing Subsequence | DP-3