Making A Large Island Problem
Description
LeetCode Problem 827.
You are given an n x n binary matrix grid. You are allowed to change at most one 0 to be 1. Return the size of the largest island in grid after applying this operation.
An island is a 4-directionally connected group of 1s.
Example 1:
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Input: grid = [[1,0],[0,1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
Example 2:
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Input: grid = [[1,1],[1,0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 4.
Example 3:
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Input: grid = [[1,1],[1,1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 4.
Constraints:
- n == grid.length
- n == grid[i].length
- 1 <= n <= 500
- grid[i][j] is either 0 or 1.
Sample C++ Code
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class Solution {
public:
const int DIR[5] = {0, 1, 0, -1, 0};
int m, n;
unordered_map<int, int> componentSize;
int largestIsland(vector<vector<int>>& grid) {
m = grid.size(); n = grid[0].size();
int ans = 0, nextColor = 2;
for (int r = 0; r < m; ++r) {
for (int c = 0; c < n; ++c) {
if (grid[r][c] != 1) continue; // Only paint when it's an island cell
paint(grid, r, c, nextColor++);
ans = max(ans, componentSize[nextColor - 1]);
}
}
for (int r = 0; r < m; ++r) {
for (int c = 0; c < n; ++c) {
if (grid[r][c] != 0) continue; // Skip non-empty cell
unordered_set<int> neighborColors;
for (int i = 0; i < 4; ++i) {
int nr = r + DIR[i], nc = c + DIR[i+1];
if (nr < 0 || nr == m || nc < 0 || nc == n || grid[nr][nc] == 0)
continue;
neighborColors.insert(grid[nr][nc]);
}
int sizeFormed = 1;
for (int color : neighborColors)
sizeFormed += componentSize[color];
ans = max(ans, sizeFormed);
}
}
return ans;
}
void paint(vector<vector<int>>& grid, int r, int c, int color) {
if (r < 0 || r == m || c < 0 || c == n || grid[r][c] != 1)
return;
grid[r][c] = color;
componentSize[color] += 1;
for (int i = 0; i < 4; ++i)
paint(grid, r + DIR[i], c + DIR[i+1], color);
}
};