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.


Input: arr[][] = {{‘W’, ‘L’}, {‘L’, ‘W’}};
Output: 1
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++ 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') {
    // 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);
// Funtion to find the minimum 'W' to flipped
void findMinimumW(vector<vector<char> >& A)
    // Base Case
    if (A.size() == 0)
    // 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
    return 0;




Time Complexity: O(N2)

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