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Problem solution · Java

Find Maximum Removals From Source String

Find Maximum Removals From Source String: 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
25 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 Find Maximum Removals From Source String, 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

  • 25 lines of Java from the credited upstream file 3316-2.java.
  • The implementation visibly relies on sequence storage, ordered lookup, 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 codeFind Maximum Removals From Source String · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxRemovals(String source, String pattern, int[] targetIndices) {    final int m = source.length();    final int n = pattern.length();    Set<Integer> target = Arrays.stream(targetIndices).boxed().collect(Collectors.toSet());    // dp[i][j] := the maximum number of operations that can be performed for    // source[i..m) and pattern[j..n)    int[][] dp = new int[m + 1][n + 1];    Arrays.stream(dp).forEach(A -> Arrays.fill(A, Integer.MIN_VALUE));    dp[m][n] = 0;     for (int i = m - 1; i >= 0; --i) {      dp[i][n] = (target.contains(i) ? 1 : 0) + dp[i + 1][n];      for (int j = n - 1; j >= 0; --j) {        final int pick =            source.charAt(i) == pattern.charAt(j) ? dp[i + 1][j + 1] : Integer.MIN_VALUE;        final int skip = (target.contains(i) ? 1 : 0) + dp[i + 1][j];        dp[i][j] = Math.max(pick, skip);      }    }     return dp[0][0] == Integer.MIN_VALUE ? 0 : dp[0][0];  }}

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