How to Find the Longest Harmonious Subsequence?

  • 时间：2020-09-18 17:01:02
  • 分类：网络文摘
  • 阅读：140 次

We define a harmounious array as an array where the difference between its maximum value and its minimum value is exactly 1. Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible subsequences.

Example 1:
Input: [1,3,2,2,5,2,3,7]
Output: 5
Explanation: The longest harmonious subsequence is [3,2,2,2,3].

Note: The length of the input array will not exceed 20,000.

Bruteforce All Subsequences

One solution is to bruteforce all the subsequences, which is exponential. The total subsequences is C(n, 1) + C(n, 2) + C(n, 3) + … C(n, n). We can use two nested loops, the outer loop is from 0 to (2^n)-1 where n is the size of the array, and the inner loop is from 0 to n-1.

By generating all the subsequences, we can increment the counter once we find the Harmonious subsequence where the min and max value is exactly one difference.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

class Solution {
public:
    int findLHS(vector<int>& nums) {
        int ans = 0;
        for (int i = 0; i < (1 << nums.size()); ++ i) {
            int count = 0, minv = INT_MAX, maxv = INT_MIN;
            for (int j = 0; j < nums.size(); ++ j) {
                if ((i & (1 << j)) != 0) {
                    minv = min(minv, nums[j]);
                    maxv = max(maxv, nums[j]);
                    count ++;
                }
            }
            if ((minv != INT_MAX) && (maxv == minv + 1)) {
                ans = max(ans, count);
            }
        }        
        return ans;
    }
};
class Solution {
public:
    int findLHS(vector<int>& nums) {
        int ans = 0;
        for (int i = 0; i < (1 << nums.size()); ++ i) {
            int count = 0, minv = INT_MAX, maxv = INT_MIN;
            for (int j = 0; j < nums.size(); ++ j) {
                if ((i & (1 << j)) != 0) {
                    minv = min(minv, nums[j]);
                    maxv = max(maxv, nums[j]);
                    count ++;
                }
            }
            if ((minv != INT_MAX) && (maxv == minv + 1)) {
                ans = max(ans, count);
            }
        }        
        return ans;
    }
};

Apparently, this bruteforce is extremely inefficient.

Using a Hashmap

We can do one-pass scan to count the occurences for each number in a hash map. Then, by going through the numbers in the hash map, then the Harmonious subsequence can be checked if the next number is in the hash map – the length will be the sum of the counters for both numbers.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

class Solution {
public:
    int findLHS(vector<int>& nums) {        
        unordered_map<int, int> count;
        for (const auto &n: nums) {
            count[n] ++;
        }
        int ans = 0;
        for (auto it = begin(count); it != end(count); ++ it) {
            int next = it->first + 1;
            if (count.find(next) != count.end()) {
                ans = max(ans, it->second + count[next]);
            }
        }
        return ans;
    }
};
class Solution {
public:
    int findLHS(vector<int>& nums) {        
        unordered_map<int, int> count;
        for (const auto &n: nums) {
            count[n] ++;
        }
        int ans = 0;
        for (auto it = begin(count); it != end(count); ++ it) {
            int next = it->first + 1;
            if (count.find(next) != count.end()) {
                ans = max(ans, it->second + count[next]);
            }
        }
        return ans;
    }
};

We can also do this one-pass, by checking the previous and next number in the hash map as we going through each number.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

class Solution {
public:
    int findLHS(vector<int>& nums) {
        int ans = 0;
        unordered_map<int, int> data;
        for (const auto &n: nums) {
            data[n] ++;
            if (data.find(n + 1) != data.end()) {
                ans = max(ans, data[n] + data[n + 1]);
            }
            if (data.find(n - 1) != data.end()) {
                ans = max(ans, data[n] + data[n - 1]);
            }
        }        
        return ans;
    }
};
class Solution {
public:
    int findLHS(vector<int>& nums) {
        int ans = 0;
        unordered_map<int, int> data;
        for (const auto &n: nums) {
            data[n] ++;
            if (data.find(n + 1) != data.end()) {
                ans = max(ans, data[n] + data[n + 1]);
            }
            if (data.find(n - 1) != data.end()) {
                ans = max(ans, data[n] + data[n - 1]);
            }
        }        
        return ans;
    }
};

Both approaches are using hash map, thus O(N) space and O(N) time complexity.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
数学题：把一个棱长为4分米的正方体填满水　 数学题：一批工人到甲乙两个工地进行清理工作　 数学题:问乙等了甲多长时间？　 数学题:一个商人骑一头驴要穿越1000公里长的沙漠　 数学题：有一个底面周长为9.42厘米的圆柱体，斜着截去一段　 百度算法经常更新要怎么解决？　 讲一讲我这10年的站长经历　 为什么企业网站不要用模板建站 模板建站有哪些弊端　 网站安全渗透测试难度系数有多大　 什么内容才是被百度肯定的优质内容？优质内容应该这样做！　
评论列表
添加评论