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128. Longest Consecutive Sequence

Description

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

 

Example 1:

Input: nums = [100,4,200,1,3,2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

Example 2:

Input: nums = [0,3,7,2,5,8,4,6,0,1]
Output: 9

 

Constraints:

  • 0 <= nums.length <= 105
  • -109 <= nums[i] <= 109

Solutions

Solution 1: Sorting

First, we sort the array, then use a variable $t$ to record the current length of the consecutive sequence, and a variable $ans$ to record the length of the longest consecutive sequence.

Next, we start traversing the array from index $i=1$. For the current element $nums[i]$:

  • If $nums[i] = nums[i-1]$, it means the current element is repeated and does not need to be considered.
  • If $nums[i] = nums[i-1] + 1$, it means the current element can be appended to the previous consecutive sequence to form a longer consecutive sequence. We update $t = t + 1$, and then update the answer $ans = \max(ans, t)$.
  • Otherwise, it means the current element cannot be appended to the previous consecutive sequence, and we reset $t$ to $1$.

Finally, we return the answer $ans$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

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class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        n = len(nums)
        if n < 2:
            return n
        nums.sort()
        ans = t = 1
        for a, b in pairwise(nums):
            if a == b:
                continue
            if a + 1 == b:
                t += 1
                ans = max(ans, t)
            else:
                t = 1
        return ans

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class Solution {
    public int longestConsecutive(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return n;
        }
        Arrays.sort(nums);
        int ans = 1, t = 1;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == nums[i - 1]) {
                continue;
            }
            if (nums[i] == nums[i - 1] + 1) {
                ans = Math.max(ans, ++t);
            } else {
                t = 1;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        int n = nums.size();
        if (n < 2) {
            return n;
        }
        sort(nums.begin(), nums.end());
        int ans = 1, t = 1;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == nums[i - 1]) {
                continue;
            }
            if (nums[i] == nums[i - 1] + 1) {
                ans = max(ans, ++t);
            } else {
                t = 1;
            }
        }
        return ans;
    }
};

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func longestConsecutive(nums []int) int {
    n := len(nums)
    if n < 2 {
        return n
    }
    sort.Ints(nums)
    ans, t := 1, 1
    for i, x := range nums[1:] {
        if x == nums[i] {
            continue
        }
        if x == nums[i]+1 {
            t++
            ans = max(ans, t)
        } else {
            t = 1
        }
    }
    return ans
}

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function longestConsecutive(nums: number[]): number {
    const n = nums.length;
    if (n < 2) {
        return n;
    }
    let ans = 1;
    let t = 1;
    nums.sort((a, b) => a - b);
    for (let i = 1; i < n; ++i) {
        if (nums[i] === nums[i - 1]) {
            continue;
        }
        if (nums[i] === nums[i - 1] + 1) {
            ans = Math.max(ans, ++t);
        } else {
            t = 1;
        }
    }
    return ans;
}

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use std::collections::HashSet;

impl Solution {
    #[allow(dead_code)]
    pub fn longest_consecutive(nums: Vec<i32>) -> i32 {
        let mut s = HashSet::new();
        let mut ret = 0;

        // Initialize the set
        for num in &nums {
            s.insert(*num);
        }

        for num in &nums {
            if s.contains(&(*num - 1)) {
                continue;
            }
            let mut cur_num = num.clone();
            while s.contains(&cur_num) {
                cur_num += 1;
            }
            // Update the answer
            ret = std::cmp::max(ret, cur_num - num);
        }

        ret
    }
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function (nums) {
    const n = nums.length;
    if (n < 2) {
        return n;
    }
    nums.sort((a, b) => a - b);
    let ans = 1;
    let t = 1;
    for (let i = 1; i < n; ++i) {
        if (nums[i] === nums[i - 1]) {
            continue;
        }
        if (nums[i] === nums[i - 1] + 1) {
            ans = Math.max(ans, ++t);
        } else {
            t = 1;
        }
    }
    return ans;
};

Solution 2: Hash Table

We use a hash table to store all elements in the array, and then traverse each element $x$ in the array. If the predecessor $x-1$ of the current element is not in the hash table, then we start with the current element and continuously try to match $x+1, x+2, x+3, \dots$, until no match is found. The length of the match at this time is the longest consecutive sequence length starting with $x$, and we update the answer accordingly.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

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class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        s = set(nums)
        ans = 0
        for x in nums:
            if x - 1 not in s:
                y = x + 1
                while y in s:
                    y += 1
                ans = max(ans, y - x)
        return ans

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class Solution {
    public int longestConsecutive(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int x : nums) {
            s.add(x);
        }
        int ans = 0;
        for (int x : nums) {
            if (!s.contains(x - 1)) {
                int y = x + 1;
                while (s.contains(y)) {
                    ++y;
                }
                ans = Math.max(ans, y - x);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        int ans = 0;
        for (int x : nums) {
            if (!s.count(x - 1)) {
                int y = x + 1;
                while (s.count(y)) {
                    y++;
                }
                ans = max(ans, y - x);
            }
        }
        return ans;
    }
};

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func longestConsecutive(nums []int) (ans int) {
    s := map[int]bool{}
    for _, x := range nums {
        s[x] = true
    }
    for _, x := range nums {
        if !s[x-1] {
            y := x + 1
            for s[y] {
                y++
            }
            ans = max(ans, y-x)
        }
    }
    return
}

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function longestConsecutive(nums: number[]): number {
    const s: Set<number> = new Set(nums);
    let ans = 0;
    for (const x of s) {
        if (!s.has(x - 1)) {
            let y = x + 1;
            while (s.has(y)) {
                y++;
            }
            ans = Math.max(ans, y - x);
        }
    }
    return ans;
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function (nums) {
    const s = new Set(nums);
    let ans = 0;
    for (const x of nums) {
        if (!s.has(x - 1)) {
            let y = x + 1;
            while (s.has(y)) {
                y++;
            }
            ans = Math.max(ans, y - x);
        }
    }
    return ans;
};

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