Description
You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].
You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:
- The first element in
s[k] starts with the selection of the element nums[k] of index = k.
- The next element in
s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.
- We stop adding right before a duplicate element occurs in
s[k].
Return the longest length of a set s[k].
Example 1:
Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation:
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}
Example 2:
Input: nums = [0,1,2]
Output: 1
Constraints:
1 <= nums.length <= 1050 <= nums[i] < nums.length- All the values of
nums are unique.
Solutions
Solution 1
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16 | class Solution:
def arrayNesting(self, nums: List[int]) -> int:
n = len(nums)
vis = [False] * n
res = 0
for i in range(n):
if vis[i]:
continue
cur, m = nums[i], 1
vis[cur] = True
while nums[cur] != nums[i]:
cur = nums[cur]
m += 1
vis[cur] = True
res = max(res, m)
return res
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21 | class Solution {
public int arrayNesting(int[] nums) {
int n = nums.length;
boolean[] vis = new boolean[n];
int res = 0;
for (int i = 0; i < n; i++) {
if (vis[i]) {
continue;
}
int cur = nums[i], m = 1;
vis[cur] = true;
while (nums[cur] != nums[i]) {
cur = nums[cur];
m++;
vis[cur] = true;
}
res = Math.max(res, m);
}
return res;
}
}
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20 | class Solution {
public:
int arrayNesting(vector<int>& nums) {
int n = nums.size();
vector<bool> vis(n);
int res = 0;
for (int i = 0; i < n; ++i) {
if (vis[i]) continue;
int cur = nums[i], m = 1;
vis[cur] = true;
while (nums[cur] != nums[i]) {
cur = nums[cur];
++m;
vis[cur] = true;
}
res = max(res, m);
}
return res;
}
};
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21 | func arrayNesting(nums []int) int {
n := len(nums)
vis := make([]bool, n)
ans := 0
for i := 0; i < n; i++ {
if vis[i] {
continue
}
cur, m := nums[i], 1
vis[cur] = true
for nums[cur] != nums[i] {
cur = nums[cur]
m++
vis[cur] = true
}
if m > ans {
ans = m
}
}
return ans
}
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Solution 2
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12 | class Solution:
def arrayNesting(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
for i in range(n):
cnt = 0
while nums[i] != n:
j = nums[i]
nums[i] = n
i = j
cnt += 1
ans = max(ans, cnt)
return ans
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17 | class Solution {
public int arrayNesting(int[] nums) {
int ans = 0, n = nums.length;
for (int i = 0; i < n; ++i) {
int cnt = 0;
int j = i;
while (nums[j] < n) {
int k = nums[j];
nums[j] = n;
j = k;
++cnt;
}
ans = Math.max(ans, cnt);
}
return ans;
}
}
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18 | class Solution {
public:
int arrayNesting(vector<int>& nums) {
int ans = 0, n = nums.size();
for (int i = 0; i < n; ++i) {
int cnt = 0;
int j = i;
while (nums[j] < n) {
int k = nums[j];
nums[j] = n;
j = k;
++cnt;
}
ans = max(ans, cnt);
}
return ans;
}
};
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16 | func arrayNesting(nums []int) int {
ans, n := 0, len(nums)
for i := range nums {
cnt, j := 0, i
for nums[j] != n {
k := nums[j]
nums[j] = n
j = k
cnt++
}
if ans < cnt {
ans = cnt
}
}
return ans
}
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