Minimum number of Water to Land conversion to make two islands connected in a Grid

Given a 2D grid arr[][] of ‘W’ and ‘L’ where ‘W’ denotes water and ‘L’ denotes land, the task is to find the minimum number of water components ‘W’ that must be changed to land component ‘L’ so that two islands becomes connected. 

An island is the set of connected ‘L’s. 

Note: There can be only two disjoint islands.

Examples:

Input: arr[][] = {{‘W’, ‘L’}, {‘L’, ‘W’}}; 
Output: 1 
Explanation:
For the given set of islands if we change arr[1][1] to ‘W’ then, set of all island are connected. 
Therefore, the minimum number of ‘W’ must be changed to ‘L’ is 1.

Input: arr[][] = {{‘W’, ‘L’, ‘W’}, {‘W’, ‘W’, ‘W’}, {‘W’, ‘W’, ‘L’}} 
Output: 2 

Approach: This problem can be solved using Floodfill algorithm. Below are the steps:  

1. Use Floodfill algorithm for the first set of the connected islands and make all the islands as visited and store the coordinates in a hash (say hash1).
2. Use Floodfill algorithm for the second set of the connected islands and make all the islands as visited and store the coordinates in a second hash(say hash2).
3. The minimum difference between coordinates stored in both the hash is the required result.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Determine the distance between two
// coordinates
int dist(pair<int, int>& p1,
         pair<int, int>& p2)
{

    return abs(p1.first - p2.first)
           + abs(p2.second - p1.second) - 1;
}

// Function to implement floodfill algorithm
void floodfill(set<pair<int, int> >& hash,
               int i, int j,
               vector<vector<char> >& A)
{

    // Base Case
    if (i < 0 || i >= A.size() || j < 0
        || j >= A[0].size() || A[i][j] != 'L') {
        return;
    }

    // Mark the current node visited
    A[i][j] = 'V';

    // Store the coordinates of in the
    // hash set
    hash.insert(make_pair(i, j));

    // Recursively iterate for the next
    // four directions
    floodfill(hash, i - 1, j, A);
    floodfill(hash, i + 1, j, A);
    floodfill(hash, i, j - 1, A);
    floodfill(hash, i, j + 1, A);
}

// Function to find the minimum 'W' to flipped
void findMinimumW(vector<vector<char> >& A)
{

    // Base Case
    if (A.size() == 0)
        return;

    // Two sets to store the coordinates of
    // connected island
    set<pair<int, int> > hash1, hash2;

    // Traversing the given grid[][]
    for (int i = 0; i < A.size(); i++) {

        for (int j = 0; j < A[0].size(); j++) {

            // If an island is found
            if (A[i][j] == 'L') {

                // Floodfill Algorithm for
                // the first island
                if (hash1.empty()) {
                    floodfill(hash1, i, j, A);
                }

                // Floodfill Algorithm for
                // the second island
                if (hash2.empty()
                    && !hash1.count({ i, j })) {
                    floodfill(hash2, i, j, A);
                }
            }
        }
    }

    // To store the minimum count of 'W'
    int ans = INT_MAX;

    // Traverse both pairs of hashes
    for (auto i : hash1) {
        for (auto j : hash2) {
            ans = min(ans, dist(i, j));
        }
    }

    // Print the final answer
    cout << ans << endl;
}

// Driver Code
int main()
{

    // Given grid of land and water
    vector<vector<char> > arr;
    arr = { { 'W', 'L' }, { 'L', 'W' } };

    // Function Call
    findMinimumW(arr);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG {
  // Determine the distance between two
  // coordinates
  static int dist(Pair p1, Pair p2)
  {
    return Math.abs(p1.first - p2.first)
      + Math.abs(p2.second - p1.second) - 1;
  }

  // class Pair to represent coordinates
  // of elements in grid
  static class Pair {
    int first;
    int second;
    Pair(int f, int s)
    {
      this.first = f;
      this.second = s;
    }
  }

  // Function to implement floodfill algorithm
  static void floodfill(HashSet<Pair> hash, int i, int j,
                        char[][] A)
  {

    // Base Case
    if (i < 0 || i >= A.length || j < 0
        || j >= A[0].length || A[i][j] != 'L') {
      return;
    }

    // Mark the current node visited
    A[i][j] = 'V';

    // Store the coordinates of in the
    // hash set
    hash.add(new Pair(i, j));

    // Recursively iterate for the next
    // four directions
    floodfill(hash, i - 1, j, A);
    floodfill(hash, i + 1, j, A);
    floodfill(hash, i, j - 1, A);
    floodfill(hash, i, j + 1, A);
  }

  // Function to find the minimum 'W' to flipped
  static void findMinimumW(char[][] A)
  {

    // Base Case
    if (A.length == 0)
      return;

    // Two sets to store the coordinates of
    // connected island
    HashSet<Pair> hash1 = new HashSet<>();
    HashSet<Pair> hash2 = new HashSet<>();

    // Traversing the given grid[][]
    for (int i = 0; i < A.length; i++) {

      for (int j = 0; j < A[0].length; j++) {

        // If an island is found
        if (A[i][j] == 'L') {

          // Floodfill Algorithm for
          // the first island
          if (hash1.isEmpty()) {
            floodfill(hash1, i, j, A);
          }

          // Floodfill Algorithm for
          // the second island
          if (hash2.isEmpty()
              && !hash1.contains(
                new Pair(i, j))) {
            floodfill(hash2, i, j, A);
          }
        }
      }
    }

    // To store the minimum count of 'W'
    int ans = Integer.MAX_VALUE;

    // Traverse both pairs of hashes
    for (Pair i : hash1) {
      for (Pair j : hash2) {
        ans = Math.min(ans, dist(i, j));
      }
    }

    // Print the final answer
    System.out.println(ans);
  }

  // Driver Code
  public static void main(String[] args)
  {
    // Given grid of land and water
    char[][] arr = { { 'W', 'L' }, { 'L', 'W' } };

    // Function Call
    findMinimumW(arr);
  }
}

// This code is contributed by kdeepsingh2002

Python3

# Python3 program for the above approach
import sys

# Determine the distance between two
# coordinates
def dist(p1, p2):

    return (abs(p1[0] - p2[0]) +
            abs(p2[1] - p1[1]) - 1)

# Function to implement floodfill algorithm
def floodfill(hash, i, j, A):

    # Base Case
    if (i < 0 or i >= len(A) or j < 0 or
        j >= len(A[0]) or A[i][j] != 'L'):
        return hash, A

    # Mark the current node visited
    A[i][j] = 'V'

    # Store the coordinates of in the
    # hash set
    hash.add((i, j))

    # Recursively iterate for the next
    # four directions
    hash, A = floodfill(hash, i - 1, j, A)
    hash, A = floodfill(hash, i + 1, j, A)
    hash, A = floodfill(hash, i, j - 1, A)
    hash, A = floodfill(hash, i, j + 1, A)
    return hash, A

# Function to find the minimum 'W' to flipped
def findMinimumW(A):

    # Base Case
    if (len(A) == 0):
        return set(), A

    # Two sets to store the coordinates of
    # connected island
    hash1 = set()
    hash2 = set()

    # Traversing the given grid[][]
    for i in range(len(A)):
        for j in range(len(A[0])):

            # If an island is found
            if (A[i][j] == 'L'):

                # Floodfill Algorithm for
                # the first island
                if (len(hash1) == 0):
                    hash1, A = floodfill(hash1, i, j, A)

                # Floodfill Algorithm for
                # the second island
                if (len(hash2) == 0 and
                   (i, j) not in hash1):
                    hash2, A = floodfill(hash2, i, j, A)

    # To store the minimum count of 'W'
    ans = sys.maxsize

    # Traverse both pairs of hashes
    for i in hash1:
        for j in hash2:
            ans = min(ans, dist(i, j))

    # Print the final answer
    print(ans)

# Driver Code
if __name__=='__main__':

    # Given grid of land and water
    arr = []
    arr = [ [ 'W', 'L' ], [ 'L', 'W' ] ]

    # Function Call
    findMinimumW(arr)

# This code is contributed by pratham76

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Determine the distance between two
  // coordinates
  static int dist(Pair p1, Pair p2)
  {
    return Math.Abs(p1.first - p2.first)
      + Math.Abs(p2.second - p1.second) - 1;
  }

  // class Pair to represent coordinates
  // of elements in grid
  class Pair {
    public int first;
    public int second;
    public Pair(int f, int s)
    {
      this.first = f;
      this.second = s;
    }
  }

  // Function to implement floodfill algorithm
  static void floodfill(HashSet<Pair> hash, int i, int j,
                        char[,] A)
  {

    // Base Case
    if (i < 0 || i >= A.GetLength(0) || j < 0
        || j >= A.GetLength(1) || A[i,j] != 'L') {
      return;
    }

    // Mark the current node visited
    A[i,j] = 'V';

    // Store the coordinates of in the
    // hash set
    hash.Add(new Pair(i, j));

    // Recursively iterate for the next
    // four directions
    floodfill(hash, i - 1, j, A);
    floodfill(hash, i + 1, j, A);
    floodfill(hash, i, j - 1, A);
    floodfill(hash, i, j + 1, A);
  }

  // Function to find the minimum 'W' to flipped
  static void findMinimumW(char[,] A)
  {

    // Base Case
    if (A.GetLength(0) == 0)
      return;

    // Two sets to store the coordinates of
    // connected island
    HashSet<Pair> hash1 = new HashSet<Pair>();
    HashSet<Pair> hash2 = new HashSet<Pair>();

    // Traversing the given grid[,]
    for (int i = 0; i < A.GetLength(0); i++) {

      for (int j = 0; j < A.GetLength(1); j++) {

        // If an island is found
        if (A[i,j] == 'L') {

          // Floodfill Algorithm for
          // the first island
          if (hash1.Count==0) {
            floodfill(hash1, i, j, A);
          }

          // Floodfill Algorithm for
          // the second island
          if (hash2.Count==0
              && !hash1.Contains(
                new Pair(i, j))) {
            floodfill(hash2, i, j, A);
          }
        }
      }
    }

    // To store the minimum count of 'W'
    int ans = int.MaxValue;

    // Traverse both pairs of hashes
    foreach (Pair i in hash1) {
      foreach (Pair j in hash2) {
        ans = Math.Min(ans, dist(i, j));
      }
    }

    // Print the readonly answer
    Console.WriteLine(ans);
  }

  // Driver Code
  public static void Main(String[] args)
  {

    // Given grid of land and water
    char[,] arr = { { 'W', 'L' }, { 'L', 'W' } };

    // Function Call
    findMinimumW(arr);
  }
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// JavaScript program for the above approach


// Determine the distance between two
// coordinates
function dist(p1,p2)
{

    return Math.abs(p1[0] - p2[0])
           + Math.abs(p2[1] - p1[1]) - 1;
}

// Function to implement floodfill algorithm
function floodfill(hash,i,j, A)
{

    // Base Case
    if (i < 0 || i >= A.length || j < 0
        || j >= A[0].length || A[i][j] != 'L') {
        return;
    }

    // Mark the current node visited
    A[i][j] = 'V';

    // Store the coordinates of in the
    // hash set
    hash.add([i, j]);

    // Recursively iterate for the next
    // four directions
    floodfill(hash, i - 1, j, A);
    floodfill(hash, i + 1, j, A);
    floodfill(hash, i, j - 1, A);
    floodfill(hash, i, j + 1, A);
}

// Function to find the minimum 'W' to flipped
function findMinimumW(A)
{

    // Base Case
    if (A.length == 0)
        return;

    // Two sets to store the coordinates of
    // connected island
    let hash1 = new Set(), hash2 = new Set();

    // Traversing the given grid[][]
    for (let i = 0; i < A.length; i++) {

        for (let j = 0; j < A[0].length; j++) {

            // If an island is found
            if (A[i][j] == 'L') {

                // Floodfill Algorithm for
                // the first island
                if (hash1.size == 0) {
                    floodfill(hash1, i, j, A);
                }

                // Floodfill Algorithm for
                // the second island
                if (hash2.size == 0 && !hash1.has([ i, j ])) {
                    floodfill(hash2, i, j, A);
                }
            }
        }
    }

    // To store the minimum count of 'W'
    let ans = Number.MAX_VALUE;

    // Traverse both pairs of hashes
    for (let i of hash1) {
        for (let j of hash2) {
            ans = Math.min(ans, dist(i, j));
        }
    }

    // Print the final answer
    document.write(ans,"</br>");
}

// Driver Code

// Given grid of land and water
let arr = [ [ 'W', 'L' ], [ 'L', 'W' ] ];

// Function Call
findMinimumW(arr);

// This code is contributed by shinjanpatra

</script>

Output:

1

Time Complexity: O(N⁴)
Auxiliary Space: O(N²)