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Making A Large Island in C++
Suppose we have a 2D grid of binary values (0s and 1s), we change at most one 0 to a 1. After that we have to find what is the size of the largest island? Here an island is a 4-directionally (top, bottom, left, right) connected group of 1s.
So, if the input is like [[1, 0], [0, 1]], then the output will be 3, this is because if we change one 0 to 1 and connect two 1s, then we will get an island with area = 3.
To solve this, we will follow these steps −
Define an array dir of size: 4 x 2, dir := {{1, 0}, { - 1, 0}, {0, 1}, {0, - 1}}
Define a function dfs(), this will take idx, i, j, the grid,
if (i,j) are inside the grid region and grid[i, j] is not equal to 1, then −
return 0
ret := 1
grid[i, j] := idx
for initialize k := 0, when k < 4, update (increase k by 1), do −
ni := dir[k, 0] + i, nj := dir[k, 1] + j
ret := ret + dfs(grid, ni, nj, idx)
return ret
From the main method, do the following −
ret := 0, idx := 2
Define an array area of size 2
n := Row count of grid, m := column count of grid
for initialize i := 0, when i < n, update (increase i by 1), do −
for initialize j := 0, when j < m, update (increase j by 1), do −
if grid[i, j] is same as 1, then −
insert dfs(grid, i, j, idx) at the end of area
ret := maximum of ret and last element of area
(increase idx by 1)
for initialize i := 0, when i < n, update (increase i by 1), do −
if grid[i, j] is same as 0, then −
Define one set idxs
for initialize k := 0, when k < 4, update (increase k by 1), do −
ni := i + dir[k, 0],nj := j + dir[k, 1]
if ni,nj in the range of grid, then −
if grid[ni, nj] is non-zero, then −
insert grid[ni, nj] into idxs
temp := 1
for all elements it in idxs, do −
temp := temp + area[it]
(increase it by 1)p + area[it]
ret := maximum of ret and temp
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
class Solution {
public:
int dfs(vector < vector <int> >& grid, int i, int j, int idx){
if(i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size()
|| grid[i][j] != 1) return 0;
int ret = 1;
grid[i][j] = idx;
for(int k = 0; k < 4; k++){
int ni = dir[k][0] + i;
int nj = dir[k][1] + j;
ret += dfs(grid, ni, nj, idx);
}
return ret;
}
int largestIsland(vector<vector<int>>& grid) {
int ret = 0;
int idx = 2;
vector <int > area(2);
int n = grid.size();
int m = grid[0].size();
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(grid[i][j] == 1){
area.push_back(dfs(grid, i, j, idx));
ret = max(ret, area.back());
idx++;
}
}
}
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(grid[i][j] == 0){
set <int> idxs;
for(int k = 0; k < 4; k++){
int ni = i + dir[k][0];
int nj = j + dir[k][1];
if(ni < 0 || nj < 0 || ni >= grid.size() ||
nj >= grid[0].size()) continue;
if(grid[ni][nj]){
idxs.insert(grid[ni][nj]);
}
}
int temp = 1;
set <int> :: iterator it = idxs.begin();
while(it != idxs.end()){
temp += area[*it];
it++;
}
ret = max(ret, temp);
}
}
}
return ret;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{1,0},{0,1}};
cout << (ob.largestIsland(v));
}Input
{{1,0},{0,1}}Output
3
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