Hard
Count the Number of Good Partitions — C++
Full explanation · Time O(n) · Space O(n)
// Time: O(n)
// Space: O(n)
// hash table, combinatorics
class Solution {
public:
int numberOfGoodPartitions(vector<int>& nums) {
static const int MOD = 1e9 + 7;
unordered_map<int, int> lookup;
for (int i = 0; i < size(nums); ++i) {
lookup[nums[i]] = i;
}
int result = 1;
for (int left = 0, right = 0; left < size(nums); ++left) {
if (left == right + 1) {
result = (result << 1) % MOD;
}
right = max(right, lookup[nums[left]]);
}
return result;
}
};