Hard
Maximum Frequency Score of a Subarray — C++
Full explanation · Time O(n) · Space O(n)
// Time: O(n)
// Space: O(n)
// two pointers, sliding window, freq table, hash table
class Solution {
public:
int maxFrequencyScore(vector<int>& nums, int k) {
static const int MOD = 1e9 + 7;
unordered_map<int, unordered_map<int, int>> lookup;
const auto& powmod = [&](int n, int p) {
if (!lookup.count(n) || !lookup[n].count(p)) {
lookup[n][p] = p >= 2 ? (static_cast<int64_t>(lookup[n][p - 1]) * n) % MOD : n % MOD; // assumed powmod(n, p-1) was accessed before powmod(n, p)
}
return lookup[n][p];
};
int result = 0, curr = 0;
unordered_map<int, int> cnt;
for (int i = 0; i < size(nums); ++i) {
if (i >= k) {
curr = ((curr - powmod(nums[i - k], cnt[nums[i - k]])) % MOD + MOD) % MOD;
--cnt[nums[i - k]];
if (cnt[nums[i - k]]) {
curr = (curr + powmod(nums[i - k], cnt[nums[i - k]])) % MOD;
}
}
if (cnt[nums[i]]) {
curr = ((curr - powmod(nums[i], cnt[nums[i]])) % MOD + MOD) % MOD;
}
++cnt[nums[i]];
curr = (curr + powmod(nums[i], cnt[nums[i]])) % MOD;
if (i >= k - 1) {
result = max(result, curr);
}
}
return result;
}
};
// Time: O(n)
// Space: O(n)
// two pointers, sliding window, freq table
class Solution2 {
public:
int maxFrequencyScore(vector<int>& nums, int k) {
static const int MOD = 1e9 + 7;
const auto& powmod = [&](int a, int b) {
a %= MOD;
int64_t result = 1;
while (b) {
if (b & 1) {
result = result * a % MOD;
}
a = int64_t(a) * a % MOD;
b >>= 1;
}
return result;
};
int result = 0, curr = 0;
unordered_map<int, int> cnt;
for (int i = 0; i < size(nums); ++i) {
if (i >= k) {
curr = ((curr - powmod(nums[i - k], cnt[nums[i - k]])) % MOD + MOD) % MOD;
--cnt[nums[i - k]];
if (cnt[nums[i - k]]) {
curr = (curr + powmod(nums[i - k], cnt[nums[i - k]])) % MOD;
}
}
if (cnt[nums[i]]) {
curr = ((curr - powmod(nums[i], cnt[nums[i]])) % MOD + MOD) % MOD;
}
++cnt[nums[i]];
curr = (curr + powmod(nums[i], cnt[nums[i]])) % MOD;
if (i >= k - 1) {
result = max(result, curr);
}
}
return result;
}
};