CPMagical CP Library

Problem solution · Java

Find X-Sum of All K-Long Subarrays II

Find X-Sum of All K-Long Subarrays II: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
59 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

Hash-based lookup

For Find X-Sum of All K-Long Subarrays II, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 59 lines of Java from the credited upstream file 3321.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 3 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeFind X-Sum of All K-Long Subarrays II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  // Same as 3318. Find X-Sum of All K-Long Subarrays I  public long[] findXSum(int[] nums, int k, int x) {    long[] ans = new long[nums.length - k + 1];    Map<Integer, Integer> count = new HashMap<>();    TreeSet<Pair<Integer, Integer>> top =        new TreeSet<>(Comparator.comparingInt(Pair<Integer, Integer>::getKey)                          .thenComparingInt(Pair<Integer, Integer>::getValue));    TreeSet<Pair<Integer, Integer>> bot =        new TreeSet<>(Comparator.comparingInt(Pair<Integer, Integer>::getKey)                          .thenComparingInt(Pair<Integer, Integer>::getValue));     for (int i = 0; i < nums.length; ++i) {      update(nums[i], 1, count, top, bot);      if (i >= k)        update(nums[i - k], -1, count, top, bot);      // Move the bottom elements to the top if needed.      while (!bot.isEmpty() && top.size() < x) {        Pair<Integer, Integer> pair = bot.pollLast();        top.add(pair);        windowSum += (long) pair.getValue() * pair.getKey();      }      // Swap the bottom and top elements if needed.      while (!bot.isEmpty() && (bot.last().getKey() > top.first().getKey() ||                                bot.last().getKey().equals(top.first().getKey()) &&                                    bot.last().getValue() > top.first().getValue())) {        Pair<Integer, Integer> pairB = bot.pollLast();        Pair<Integer, Integer> pairT = top.pollFirst();        top.add(pairB);        bot.add(pairT);        windowSum += (long) pairB.getValue() * pairB.getKey();        windowSum -= (long) pairT.getValue() * pairT.getKey();      }      if (i >= k - 1)        ans[i - k + 1] = windowSum;    }    return ans;  }   private long windowSum = 0;   // Updates the count of num by freq and the window sum accordingly.  private void update(int num, int freq, Map<Integer, Integer> count,                      TreeSet<Pair<Integer, Integer>> top, TreeSet<Pair<Integer, Integer>> bot) {    if (count.getOrDefault(num, 0) > 0) { // Clean up the old count.      Pair<Integer, Integer> pair = new Pair<>(count.get(num), num);      if (bot.remove(pair)) {        // Do nothing.      } else {        top.remove(pair);        windowSum -= (long) num * count.get(num);      }    }    count.merge(num, freq, Integer::sum);    if (count.get(num) > 0)      bot.add(new Pair<>(count.get(num), num));  }}

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 ↗