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565. Array Nesting

DifficultyMedium

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 <= 105
  • 0 <= nums[i] < nums.length
  • All the values of nums are unique.

Solutions

Solution 1

Thinking

\(A[i]\) points to \(A[A[i]]\), partitioning the array into disjoint cycles. A nest is a cycle. Walking from every index would revisit the same cycle.

A visited array marks seen indices; start a walk only from an unseen index and count the cycle. Disjointness visits each index once. The longest cycle is the answer.

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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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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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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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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
}

Solution 2

Thinking

Solution 1 uses an \(O(n)\) visited array. Values already lie in \([0,n-1]\), so the sentinel \(n\) can mark a visited cell in place.

Walk the cycle, write \(n\) into each cell, and count. A stored \(n\) means this index was already processed. Extra space becomes constant; the cycles are the same.

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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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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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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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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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