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:
- 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).
- 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).
- 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// coordinatesint 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 algorithmvoid 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 flippedvoid 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 Codeint main(){ // Given grid of land and water vector<vector<char> > arr; arr = { { 'W', 'L' }, { 'L', 'W' } }; // Function Call findMinimumW(arr); return 0;} |
Python3
# Python3 program for the above approachimport sys# Determine the distance between two# coordinatesdef dist(p1, p2): return (abs(p1[0] - p2[0]) + abs(p2[1] - p1[1]) - 1)# Function to implement floodfill algorithmdef 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 flippeddef 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 Codeif __name__=='__main__': # Given grid of land and water arr = [] arr = [ [ 'W', 'L' ], [ 'L', 'W' ] ] # Function Call findMinimumW(arr)# This code is contributed by pratham76 |
1
Time Complexity: O(N2)
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