[LeetCode] Longest Consecutive Sequence

This problem is not very intuitive at first glance. However, the final idea should be very self-explanatory. You visit each element in nums, and then find its left and right neighbors and extend the length accordingly. The time complexity is O(n) if we guard from visiting the same element again.

You can implement it using an unordered_set or unordered_map.

The code is as follows. The last one is taken from this page which includes a nice explanation in the follong answers.

// O(n) solution using unordered_set
int longestConsecutive(vector<int>& nums) {
    unordered_set<int> copy(nums.begin(), nums.end());
    unordered_set<int> filter;
    int len = 0;
    for (int i = 0; i < nums.size(); i++) {
        if (filter.find(nums[i]) != filter.end()) continue;
        int l = nums[i] - 1, r = nums[i] + 1;
        while (copy.find(l) != copy.end()) filter.insert(l--);
        while (copy.find(r) != copy.end()) filter.insert(r++);
        len = max(len, r - l - 1);
    }
    return len;
}

// O(n) solution using unordered_map
int longestConsecutive(vector<int>& nums) {
    unordered_map<int, int> mp;
    int len = 0;
    for (int i = 0; i < nums.size(); i++) {
        if (mp[nums[i]]) continue;
        mp[nums[i]] = 1;
        if (mp.find(nums[i] - 1) != mp.end())
            mp[nums[i]] += mp[nums[i] - 1];
        if (mp.find(nums[i] + 1) != mp.end())
            mp[nums[i]] += mp[nums[i] + 1];
        mp[nums[i] - mp[nums[i] - 1]] = mp[nums[i]]; // left boundary
        mp[nums[i] + mp[nums[i] + 1]] = mp[nums[i]]; // right boundary
        len = max(len, mp[nums[i]]);
    }
    return len;
}

// O(n) super-concise solution (merging the above cases)
int longestConsecutive(vector<int>& nums) {
    unordered_map<int, int> mp;
    int len = 0;
    for (int i = 0; i < nums.size(); i++) {
        if (mp[nums[i]]) continue;
        len = max(len, mp[nums[i]] = mp[nums[i] - mp[nums[i] - 1]] = mp[nums[i] + mp[nums[i] + 1]] = mp[nums[i] - 1] + mp[nums[i] + 1] + 1);
    }
    return len;
}