Implement minimum-deletions-to-make-string-balanced

[yamcodes]

Implement minimum-deletions-to-make-string-balanced

Summary by CodeRabbit

Release Notes

• New Features

  • Added fully-functional algorithm implementation with intelligent cost evaluation mechanism that analyzes multiple partitioning points to determine the optimal solution.

• Tests

  • Substantially expanded test suite with additional comprehensive test cases covering various input patterns and boundary conditions for robust validation.

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Walkthrough

This PR implements the core algorithm for calculating minimum character deletions to balance a string of 'a' and 'b' characters. A new helper method counts 'b' characters before and 'a' characters after each position, enabling the main method to compute the minimum deletion count across all possible split points. Test coverage expands with five additional test cases.

Changes

Cohort / File(s)	Summary
Algorithm Implementation src/main/java/.../minimumdeletionstomakestringbalanced/Solution.java	Added getSliceInfo(String, int) helper method returning counts of 'b' before and 'a' after a given index. Replaced placeholder logic in minimumDeletions() with full O(n²) scan that iterates over split indices, retrieves info from helper, and tracks minimum deletion cost.
Test Cases & Imports src/test/java/.../minimumdeletionstomakestringbalanced/SolutionTest.java	Reordered imports; expanded test data with five new test cases (b→0, ba→1, aabbbaa→3, abababab→3, bbbbaaaa→0) alongside existing cases.

Estimated code review effort

🎯 2 (Simple) | ⏱️ ~12 minutes

Poem

🐰 A bunny hops through strings of 'a' and 'b',
Counting left and right with glee,
Balancing characters, slice by slice,
Finding deletions? Oh how nice!
Now tests validate our logic tight,
The algorithm shines so bright! ✨

🚥 Pre-merge checks | ✅ 2 | ❌ 1

❌ Failed checks (1 warning)

Check name	Status	Explanation	Resolution
Docstring Coverage	⚠️ Warning	Docstring coverage is 25.00% which is insufficient. The required threshold is 80.00%.	Write docstrings for the functions missing them to satisfy the coverage threshold.

✅ Passed checks (2 passed)

Check name	Status	Explanation
Description Check	✅ Passed	Check skipped - CodeRabbit’s high-level summary is enabled.
Title check	✅ Passed	The title directly and clearly describes the main change: implementing the minimum-deletions-to-make-string-balanced solution.

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✨ Finishing touches

🧪 Generate unit tests (beta)

• Create PR with unit tests
• Post copyable unit tests in a comment
• Commit unit tests in branch minimum-deletions-to-make-string-balanced

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[coderabbitai]

Actionable comments posted: 2

🤖 Fix all issues with AI agents

In
`@src/main/java/codes/yam/leetcode/minimumdeletionstomakestringbalanced/Solution.java`:
- Around line 5-24: Add HTML-formatted Javadoc to both methods: document the
public method minimumDeletions(String s) with a descriptive summary and include
"Time Complexity: <code>...</code>" and "Space Complexity: <code>...</code>"
tags, and update getSliceInfo(String s, int index)'s Javadoc to use the same
HTML formatting (summary, param tags, return tag) plus explicit Time Complexity
and Space Complexity in <code>...</code> tags; ensure the param names and return
description reference getSliceInfo and minimumDeletions exactly as named so
reviewers can find the methods.

In
`@src/test/java/codes/yam/leetcode/minimumdeletionstomakestringbalanced/SolutionTest.java`:
- Around line 14-21: Update the incorrect expected values in the parameterized
test data: in the Stream.of arguments for the test method (the arguments(...)
entries), change the expected value for the input "aabbbaa" from 3 to 2 and
change the expected value for "bbbbaaaa" from 0 to 4 so the assertions use the
correct minimal deletion counts.

🧹 Nitpick comments (1)

src/main/java/codes/yam/leetcode/minimumdeletionstomakestringbalanced/Solution.java (1)

26-34: Consider an O(n) single-pass approach instead of O(n²).

The current implementation calls getSliceInfo (an O(n) scan) for every index, yielding O(n²) overall. This will TLE on LeetCode's larger inputs (n up to 10⁵). A single-pass O(n) solution is straightforward: track a running count of 'b's seen so far, and at each 'a', decide whether to "delete" it or pair it against a previous 'b'.

♻️ O(n) implementation

  int minimumDeletions(String s) {
-   int min = Integer.MAX_VALUE;
-   for (int i = 0; i < s.length(); i++) {
-     int[] info = getSliceInfo(s, i);
-     int computed = info[0] + info[1];
-     min = Math.min(min, computed);
-   }
-   return min;
+   int deletions = 0;
+   int bCount = 0;
+   for (int i = 0; i < s.length(); i++) {
+     if (s.charAt(i) == 'b') {
+       bCount++;
+     } else if (bCount > 0) {
+       deletions++;
+       bCount--;
+     }
+   }
+   return deletions;
  }