# Posts tagged as “graph”

You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string.

You can swap the characters at any pair of indices in the given pairs any number of times.

Return the lexicographically smallest string that s can be changed to after using the swaps.

Example 1:

Input: s = "dcab", pairs = [[0,3],[1,2]]
Output: "bacd"
Explaination:
Swap s[0] and s[3], s = "bcad"
Swap s[1] and s[2], s = "bacd"


Example 2:

Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
Output: "abcd"
Explaination:
Swap s[0] and s[3], s = "bcad"
Swap s[0] and s[2], s = "acbd"
Swap s[1] and s[2], s = "abcd"

Example 3:

Input: s = "cba", pairs = [[0,1],[1,2]]
Output: "abc"
Explaination:
Swap s[0] and s[1], s = "bca"
Swap s[1] and s[2], s = "bac"
Swap s[0] and s[1], s = "abc"



Constraints:

• 1 <= s.length <= 10^5
• 0 <= pairs.length <= 10^5
• 0 <= pairs[i][0], pairs[i][1] < s.length
• s only contains lower case English letters.

## Solution: Connected Components

Use DFS / Union-Find to find all the connected components of swapable indices. For each connected components (index group), extract the subsequence of corresponding chars as a string, sort it and put it back to the original string in the same location.

e.g. s = “dcab”, pairs = [[0,3],[1,2]]
There are two connected components: {0,3}, {1,2}
subsequences:
1. 0,3 “db”, sorted: “bd”
2. 1,2 “ca”, sorted: “ac”
0 => b
1 => a
2 => c
3 => d
final = “bacd”

Time complexity: DFS: O(nlogn + k*(V+E)), Union-Find: O(nlogn + V+E)
Space complexity: O(n)

## C++/Union-Find

There are n servers numbered from 0 to n-1 connected by undirected server-to-server connections forming a network where connections[i] = [a, b] represents a connection between servers a and b. Any server can reach any other server directly or indirectly through the network.

critical connection is a connection that, if removed, will make some server unable to reach some other server.

Return all critical connections in the network in any order.

Example 1:

Input: n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
Output: [[1,3]]
Explanation: [[3,1]] is also accepted.


Constraints:

• 1 <= n <= 10^5
• n-1 <= connections.length <= 10^5
• connections[i][0] != connections[i][1]
• There are no repeated connections.

## Solution: Tarjan

Time complexity: O(v+e)
Space complexity: O(v+e)

## C++

A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n.

The bus goes along both directions i.e. clockwise and counterclockwise.

Return the shortest distance between the given start and destination stops.

Example 1:

Input: distance = [1,2,3,4], start = 0, destination = 1
Output: 1
Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.

Example 2:

Input: distance = [1,2,3,4], start = 0, destination = 2
Output: 3
Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.


Example 3:

Input: distance = [1,2,3,4], start = 0, destination = 3
Output: 4
Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.


Constraints:

• 1 <= n <= 10^4
• distance.length == n
• 0 <= start, destination < n
• 0 <= distance[i] <= 10^4

Solution: Summation

1. compute the total sum
2. compute the sum from s to d, c
3. ans = min(c, sum – c)

Time complexity: O(d-s)
Space complexity: O(1)

## C++

Given an N x N grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized and return the distance.

The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1)is |x0 - x1| + |y0 - y1|.

If no land or water exists in the grid, return -1.

Example 1:

Input: [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation:
The cell (1, 1) is as far as possible from all the land with distance 2.


Example 2:

Input: [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation:
The cell (2, 2) is as far as possible from all the land with distance 4.


Note:

1. 1 <= grid.length == grid[0].length <= 100
2. grid[i][j] is 0 or 1

## Solution: BFS

Put all land cells into a queue as source nodes and BFS for water cells, the last expanded one will be the farthest.

Time compleixty: O(n^2)
Space complexity: O(n^2)

## C++

Consider a directed graph, with nodes labelled 0, 1, ..., n-1.  In this graph, each edge is either red or blue, and there could be self-edges or parallel edges.

Each [i, j] in red_edges denotes a red directed edge from node i to node j.  Similarly, each [i, j] in blue_edges denotes a blue directed edge from node i to node j.

Return an array answer of length n, where each answer[X] is the length of the shortest path from node 0 to node X such that the edge colors alternate along the path (or -1 if such a path doesn’t exist).

Example 1:

Input: n = 3, red_edges = [[0,1],[1,2]], blue_edges = []
Output: [0,1,-1]


Example 2:

Input: n = 3, red_edges = [[0,1]], blue_edges = [[2,1]]
Output: [0,1,-1]


Example 3:

Input: n = 3, red_edges = [[1,0]], blue_edges = [[2,1]]
Output: [0,-1,-1]


Example 4:

Input: n = 3, red_edges = [[0,1]], blue_edges = [[1,2]]
Output: [0,1,2]


Example 5:

Input: n = 3, red_edges = [[0,1],[0,2]], blue_edges = [[1,0]]
Output: [0,1,1]


Constraints:

• 1 <= n <= 100
• red_edges.length <= 400
• blue_edges.length <= 400
• red_edges[i].length == blue_edges[i].length == 2
• 0 <= red_edges[i][j], blue_edges[i][j] < n

Solution: BFS

Time complexity: O(|V| + |E|)
Space complexity: O(|V| + |E|)

## C++

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