CPMagical CP Library

Problem solution · Java

Apply Operations to Maximize Score

Apply Operations to Maximize Score: a Java 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 Java from the credited upstream file 2818.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup, work queue.
  • 12 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 · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maximumScore(List<Integer> nums, int k) {    final int n = nums.size();    final int mx = Collections.max(nums);    final int[] minPrimeFactors = sieveEratosthenes(mx + 1);    final 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]    int[] left = new int[n];    Arrays.fill(left, -1);    // right[i] := the next index on the right (if any)    //             s.t. primeScores[right[i]] > primeScores[i]    int[] right = new int[n];    Arrays.fill(right, n);    Deque<Integer> stack = new ArrayDeque<>();     // Find the next indices on the left where `primeScores` are greater or equal.    for (int i = n - 1; i >= 0; --i) {      while (!stack.isEmpty() && primeScores[stack.peek()] <= primeScores[i])        left[stack.pop()] = i;      stack.push(i);    }     stack.clear();     // Find the next indices on the right where `primeScores` are greater.    for (int i = 0; i < n; ++i) {      while (!stack.isEmpty() && primeScores[stack.peek()] < primeScores[i])        right[stack.pop()] = i;      stack.push(i);    }     Pair<Integer, Integer>[] numAndIndexes = new Pair[n];     for (int i = 0; i < n; ++i)      numAndIndexes[i] = new Pair<>(nums.get(i), i);     Arrays.sort(numAndIndexes,                Comparator.comparing(Pair<Integer, Integer>::getKey, Comparator.reverseOrder())                    .thenComparingInt(Pair<Integer, Integer>::getValue));     for (Pair<Integer, Integer> numAndIndex : numAndIndexes) {      final int num = numAndIndex.getKey();      final int i = numAndIndex.getValue();      // 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.      final long rangeCount = (long) (i - left[i]) * (right[i] - i);      final long actualCount = Math.min(rangeCount, (long) k);      k -= actualCount;      ans = (int) ((1L * ans * modPow(num, actualCount)) % MOD);    }     return ans;  }   private static final int MOD = 1_000_000_007;   private long modPow(long x, long n) {    if (n == 0)      return 1;    if (n % 2 == 1)      return x * modPow(x, n - 1) % MOD;    return modPow(x * x % MOD, n / 2);  }   // Gets the minimum prime factor of i, where 1 < i <= n.  private int[] sieveEratosthenes(int n) {    int[] minPrimeFactors = new int[n + 1];    for (int i = 2; i <= n; ++i)      minPrimeFactors[i] = i;    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] = Math.min(minPrimeFactors[j], i);    return minPrimeFactors;  }   private int[] getPrimeScores(List<Integer> nums, int[] minPrimeFactors) {    int[] primeScores = new int[nums.size()];    for (int i = 0; i < nums.size(); ++i)      primeScores[i] = getPrimeScore(nums.get(i), minPrimeFactors);    return primeScores;  }   private int getPrimeScore(int num, int[] minPrimeFactors) {    Set<Integer> primeFactors = new HashSet<>();    while (num > 1) {      final int divisor = minPrimeFactors[num];      primeFactors.add(divisor);      while (num % divisor == 0)        num /= divisor;    }    return primeFactors.size();  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗