Problem
Given a set of N people (numbered 1, 2, …, N), we would like to split everyone into two groups of any size.
Each person may dislike some other people, and they should not go into the same group.
Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.
Return true if and only if it is possible to split everyone into two groups in this way.
example 1
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Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]
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example 2
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Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false
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example 3
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Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false
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example 4
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Input: N = 10, dislikes = [[4,7],[4,8],[5,6],[1,6],[3,7],[2,5],[5,8],[1,2],[4,9],[6,10],[8,10],[3,6],[2,10],[9,10],[3,9],[2,3],[1,9],[4,6],[5,7],[3,8],[1,8],[1,7],[2,4]]
Output: true
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Constraints
- 1 <=
N <= 2000
- 0 <=
dislikes.length <= 10000
- 1 <=
dislikes[i][j] <= N
dislikes[i][0] < dislikes[i][1]- There does not exist
i != j for which dislikes[i] == dislikes[j].
Solution
Hash map + DFS
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func possibleBipartition(N int, dislikes [][]int) bool {
m := make(map[int][]int, N+1)
c := make(map[int]int, N+1)
for idx := range dislikes {
if _, ok := m[dislikes[idx][0]]; !ok {
m[dislikes[idx][0]] = make([]int, 0)
}
m[dislikes[idx][0]] = append(m[dislikes[idx][0]], dislikes[idx][1])
m[dislikes[idx][1]] = append(m[dislikes[idx][1]], dislikes[idx][0])
}
for idx := range m {
if _, ok := c[idx]; !ok && !clr(m, c, idx, 0) {
return false
}
}
return true
}
func clr(m map[int][]int, c map[int]int, i, color int) bool {
if _, ok := c[i]; ok {
return c[i] == color
}
c[i] = color
for idx := range m[i] {
if !clr(m, c, m[i][idx], color^1) {
return false
}
}
return true
}
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