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) —
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