1653. Minimum Deletions to Make String Balanced
Medium
You are given a string s
consisting only of characters 'a'
and 'b'
.
You can delete any number of characters in s
to make s
balanced. s
is balanced if there is no pair of indices (i,j)
such that i < j
and s[i] = 'b'
and s[j]= 'a'
.
Return the minimum number of deletions needed to make s
balanced.
Example 1:
- Input: s = "aababbab"
- Output: 2
-
Explanation: You can either:
- Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
- Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
Example 2:
- Input: s = "bbaaaaabb"
- Output: 2
- Explanation: The only solution is to delete the first two characters.
Constraints:
1 <= s.length <= 105
-
s[i]
isa
orb
.
Hint:
- You need to find for every index the number of Bs before it and the number of A's after it
- You can speed up the finding of A's and B's in suffix and prefix using preprocessing
Solution:
To solve this problem, we can follow these steps:
- Preprocess the counts of 'b's up to each index: This helps in knowing how many 'b's are present before any given index.
- Preprocess the counts of 'a's from each index: This helps in knowing how many 'a's are present after any given index.
- Calculate the minimum deletions needed: For each index, you can calculate the deletions needed by considering deleting all 'b's before it and all 'a's after it.
Let's implement this solution in PHP: 1653. Minimum Deletions to Make String Balanced
<?php
// Test cases
echo minimumDeletions("aababbab") . "\n"; // Output: 2
echo minimumDeletions("bbaaaaabb") . "\n"; // Output: 2
?>
Explanation:
-
Prefix Array for 'b's: This array keeps track of the cumulative count of 'b's up to each index. For example, if the string is "aababbab", the prefix array for 'b's will be
[0, 0, 1, 2, 2, 3, 3, 4]
. -
Suffix Array for 'a's: This array keeps track of the cumulative count of 'a's from each index to the end. For the same string, the suffix array for 'a's will be
[4, 3, 3, 2, 2, 1, 1, 0]
. -
Calculating Minimum Deletions: For each index
i
, the total deletions needed to make the string balanced if we split ati
can be calculated as the sum of 'b's beforei
and 'a's afteri
. The minimum of these values across all indices gives the result.
This approach ensures that the solution is efficient with a time complexity of (O(n)), which is suitable given the constraints.
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