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Problem solution · Java

Count the Number of Good Partitions

Count the Number of Good Partitions: 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
27 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 Count the Number of Good Partitions, 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

  • 27 lines of Java from the credited upstream file 2963.java.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 2 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 codeCount the Number of Good Partitions · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int numberOfGoodPartitions(int[] nums) {    final int MOD = 1_000_000_007;    int ans = 1;     // lastSeen[num] := the index of the last time `num` appeared    HashMap<Integer, Integer> lastSeen = new HashMap<>();     for (int i = 0; i < nums.length; ++i)      lastSeen.put(nums[i], i);     // Track the maximum right index of each running partition by ensuring that    // the first and last occurrences of a number fall within the same    // partition.    int maxRight = 0;    for (int i = 0; i < nums.length; ++i) {      if (i > maxRight)        // Start a new partition that starts from nums[i].        // Each partition doubles the total number of good partitions.        ans = (int) ((ans * 2L) % MOD);      maxRight = Math.max(maxRight, lastSeen.get(nums[i]));    }     return ans;  }}

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