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LeetCode Solutions

1042. Flower Planting With No Adjacent

Time: $O(n)$

Space: $O(n)$

			


class Solution {
 public:
  vector<int> gardenNoAdj(int n, vector<vector<int>>& paths) {
    vector<int> ans(n);  // ans[i] := 1, 2, 3, or 4
    vector<vector<int>> graph(n);

    for (const vector<int>& p : paths) {
      const int u = p[0] - 1;
      const int v = p[1] - 1;
      graph[u].push_back(v);
      graph[v].push_back(u);
    }

    for (int i = 0; i < n; ++i) {
      vector<bool> used(5);
      for (const int v : graph[i])
        used[ans[v]] = true;
      for (int type = 1; type < 5; ++type)
        if (!used[type]) {
          ans[i] = type;
          break;
        }
    }

    return ans;
  }
};




			


class Solution {
  public int[] gardenNoAdj(int n, int[][] paths) {
    int[] ans = new int[n]; // ans[i] := 1, 2, 3, or 4
    List<Integer>[] graph = new List[n];

    for (int i = 0; i < n; ++i)
      graph[i] = new ArrayList<>();

    for (int[] p : paths) {
      final int u = p[0] - 1;
      final int v = p[1] - 1;
      graph[u].add(v);
      graph[v].add(u);
    }

    for (int i = 0; i < n; ++i) {
      boolean[] used = new boolean[5];
      for (final int v : graph[i])
        used[ans[v]] = true;
      for (int type = 1; type < 5; ++type)
        if (!used[type]) {
          ans[i] = type;
          break;
        }
    }

    return ans;
  }
}




			


class Solution:
  def gardenNoAdj(self, n: int, paths: List[List[int]]) -> List[int]:
    ans = [0] * n  # ans[i] := 1, 2, 3, or 4
    graph = [[] for _ in range(n)]

    for a, b in paths:
      u = a - 1
      v = b - 1
      graph[u].append(v)
      graph[v].append(u)

    for i in range(n):
      used = [False] * 5
      for v in graph[i]:
        used[ans[v]] = True
      for type in range(1, 5):
        if not used[type]:
          ans[i] = type
          break

    return ans