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Problem solution · C++

Apply Operations to Maximize Score

Apply Operations to Maximize Score: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
98 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Apply Operations to Maximize Score, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 98 lines of C++ from the credited upstream file 2818.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 11 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeApply Operations to Maximize Score · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int maximumScore(vector<int>& nums, int k) {    const int n = nums.size();    const int mx = ranges::max(nums);    const vector<int> minPrimeFactors = sieveEratosthenes(mx + 1);    const vector<int> primeScores = getPrimeScores(nums, minPrimeFactors);    int ans = 1;    // left[i] := the next index on the left (if any) s.t.    // primeScores[left[i]] >= primeScores[i]    vector<int> left(n, -1);    // right[i] := the next index on the right (if any) s.t.    // primeScores[right[i]] > primeScores[i]    vector<int> right(n, n);    stack<int> stack;     // Find the next indices on the left where `primeScores` are greater or    // equal.    for (int i = n - 1; i >= 0; --i) {      while (!stack.empty() && primeScores[stack.top()] <= primeScores[i])        left[stack.top()] = i, stack.pop();      stack.push(i);    }     stack = std::stack<int>();     // Find the next indices on the right where `primeScores` are greater.    for (int i = 0; i < n; ++i) {      while (!stack.empty() && primeScores[stack.top()] < primeScores[i])        right[stack.top()] = i, stack.pop();      stack.push(i);    }     vector<pair<int, int>> numAndIndexes;     for (int i = 0; i < n; ++i)      numAndIndexes.emplace_back(nums[i], i);     ranges::sort(numAndIndexes,                 [&](const pair<int, int>& a, const pair<int, int>& b) {      return a.first == b.first ? a.second < b.second : a.first > b.first;    });     for (const auto& [num, i] : numAndIndexes) {      // nums[i] is the maximum value in the range [left[i] + 1, right[i] - 1]      // So, there are (i - left[i]) * (right[i] - 1) ranges where nums[i] will      // be chosen.      const long rangeCount = static_cast<long>(i - left[i]) * (right[i] - i);      const long actualCount = min(rangeCount, static_cast<long>(k));      k -= actualCount;      ans = static_cast<long>(ans) * modPow(num, actualCount) % kMod;    }     return ans;  }  private:  static constexpr int kMod = 1'000'000'007;   long modPow(long x, long n) {    if (n == 0)      return 1;    if (n % 2 == 1)      return x * modPow(x % kMod, (n - 1)) % kMod;    return modPow(x * x % kMod, (n / 2)) % kMod;  }   // Gets the minimum prime factor of i, where 1 < i <= n.  vector<int> sieveEratosthenes(int n) {    vector<int> minPrimeFactors(n + 1);    iota(minPrimeFactors.begin() + 2, minPrimeFactors.end(), 2);    for (int i = 2; i * i < n; ++i)      if (minPrimeFactors[i] == i)  // `i` is prime.        for (int j = i * i; j < n; j += i)          minPrimeFactors[j] = min(minPrimeFactors[j], i);    return minPrimeFactors;  }   vector<int> getPrimeScores(const vector<int>& nums,                             const vector<int>& minPrimeFactors) {    vector<int> primeScores;    for (const int num : nums)      primeScores.push_back(getPrimeScore(num, minPrimeFactors));    return primeScores;  }   int getPrimeScore(int num, const vector<int>& minPrimeFactors) {    unordered_set<int> primeFactors;    while (num > 1) {      const int divisor = minPrimeFactors[num];      primeFactors.insert(divisor);      while (num % divisor == 0)        num /= divisor;    }    return primeFactors.size();  }};

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