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Problem solution · Java

Maximum Number of Non-overlapping Palindrome Substrings

Maximum Number of Non-overlapping Palindrome Substrings: a Java solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
35 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

Dynamic programming

For Maximum Number of Non-overlapping Palindrome Substrings, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 35 lines of Java from the credited upstream file 2472.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeMaximum Number of Non-overlapping Palindrome Substrings · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxPalindromes(String s, int k) {    final int n = s.length();    // dp[i] := the maximum number of substrings in the first i chars of s    int[] dp = new int[n + 1];     // If a palindrome is a subString of another palindrome, then considering    // the longer palindrome won't increase the number of non-overlapping    // palindromes. So, we only need to consider the shorter one. Also,    // considering palindromes with both k length and k + 1 length ensures that    // we look for both even and odd length palindromes.    for (int i = k; i <= n; ++i) {      dp[i] = dp[i - 1];      // Consider palindrome with length k.      if (isPalindrome(s, i - k, i - 1))        dp[i] = Math.max(dp[i], 1 + dp[i - k]);      // Consider palindrome with length k + 1.      if (isPalindrome(s, i - k - 1, i - 1))        dp[i] = Math.max(dp[i], 1 + dp[i - k - 1]);    }     return dp[n];  }   // Returns true is s[i..j) is a palindrome.  private boolean isPalindrome(String s, int l, int r) {    if (l < 0)      return false;    while (l < r)      if (s.charAt(l++) != s.charAt(r--))        return false;    return true;  }}

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