# 2524. Maximum Frequency Score of a Subarray

## Description

You are given an integer array nums and a positive integer k.

The frequency score of an array is the sum of the distinct values in the array raised to the power of their frequencies, taking the sum modulo 109 + 7.

• For example, the frequency score of the array [5,4,5,7,4,4] is (43 + 52 + 71) modulo (109 + 7) = 96.

Return the maximum frequency score of a subarray of size k in nums. You should maximize the value under the modulo and not the actual value.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,1,1,2,1,2], k = 3
Output: 5
Explanation: The subarray [2,1,2] has a frequency score equal to 5. It can be shown that it is the maximum frequency score we can have.


Example 2:

Input: nums = [1,1,1,1,1,1], k = 4
Output: 1
Explanation: All the subarrays of length 4 have a frequency score equal to 1.


Constraints:

• 1 <= k <= nums.length <= 105
• 1 <= nums[i] <= 106

## Solutions

• class Solution {
private final int mod = (int) 1e9 + 7;

public int maxFrequencyScore(int[] nums, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
for (int i = 0; i < k; ++i) {
cnt.merge(nums[i], 1, Integer::sum);
}
long cur = 0;
for (var e : cnt.entrySet()) {
cur = (cur + qpow(e.getKey(), e.getValue())) % mod;
}
long ans = cur;
for (int i = k; i < nums.length; ++i) {
int a = nums[i - k];
int b = nums[i];
if (a != b) {
if (cnt.getOrDefault(b, 0) > 0) {
cur += (b - 1) * qpow(b, cnt.get(b)) % mod;
} else {
cur += b;
}
if (cnt.getOrDefault(a, 0) > 1) {
cur -= (a - 1) * qpow(a, cnt.get(a) - 1) % mod;
} else {
cur -= a;
}
cur = (cur + mod) % mod;
cnt.put(b, cnt.getOrDefault(b, 0) + 1);
cnt.put(a, cnt.getOrDefault(a, 0) - 1);
ans = Math.max(ans, cur);
}
}
return (int) ans;
}

private long qpow(long a, long n) {
long ans = 1;
for (; n > 0; n >>= 1) {
if ((n & 1) == 1) {
ans = ans * a % mod;
}
a = a * a % mod;
}
return ans;
}
}

• class Solution {
public:
int maxFrequencyScore(vector<int>& nums, int k) {
using ll = long long;
const int mod = 1e9 + 7;
auto qpow = [&](ll a, ll n) {
ll ans = 1;
for (; n; n >>= 1) {
if (n & 1) {
ans = ans * a % mod;
}
a = a * a % mod;
}
return ans;
};
unordered_map<int, int> cnt;
for (int i = 0; i < k; ++i) {
cnt[nums[i]]++;
}
ll cur = 0;
for (auto& [k, v] : cnt) {
cur = (cur + qpow(k, v)) % mod;
}
ll ans = cur;
for (int i = k; i < nums.size(); ++i) {
int a = nums[i - k], b = nums[i];
if (a != b) {
cur += cnt[b] ? (b - 1) * qpow(b, cnt[b]) % mod : b;
cur -= cnt[a] > 1 ? (a - 1) * qpow(a, cnt[a] - 1) % mod : a;
cur = (cur + mod) % mod;
ans = max(ans, cur);
cnt[b]++;
cnt[a]--;
}
}
return ans;
}
};

• class Solution:
def maxFrequencyScore(self, nums: List[int], k: int) -> int:
mod = 10**9 + 7
cnt = Counter(nums[:k])
ans = cur = sum(pow(k, v, mod) for k, v in cnt.items()) % mod
i = k
while i < len(nums):
a, b = nums[i - k], nums[i]
if a != b:
cur += (b - 1) * pow(b, cnt[b], mod) if cnt[b] else b
cur -= (a - 1) * pow(a, cnt[a] - 1, mod) if cnt[a] > 1 else a
cur %= mod
cnt[b] += 1
cnt[a] -= 1
ans = max(ans, cur)
i += 1
return ans


• func maxFrequencyScore(nums []int, k int) int {
cnt := map[int]int{}
for _, v := range nums[:k] {
cnt[v]++
}
cur := 0
const mod int = 1e9 + 7
qpow := func(a, n int) int {
ans := 1
for ; n > 0; n >>= 1 {
if n&1 == 1 {
ans = ans * a % mod
}
a = a * a % mod
}
return ans
}
for k, v := range cnt {
cur = (cur + qpow(k, v)) % mod
}
ans := cur
for i := k; i < len(nums); i++ {
a, b := nums[i-k], nums[i]
if a != b {
if cnt[b] > 0 {
cur += (b - 1) * qpow(b, cnt[b]) % mod
} else {
cur += b
}
if cnt[a] > 1 {
cur -= (a - 1) * qpow(a, cnt[a]-1) % mod
} else {
cur -= a
}
cur = (cur + mod) % mod
ans = max(ans, cur)
cnt[b]++
cnt[a]--
}
}
return ans
}