Class Solution
java.lang.Object
g3401_3500.s3499_maximize_active_section_with_trade_i.Solution
3499 - Maximize Active Section with Trade I.
Medium
You are given a binary string s of length n, where:
'1'represents an active section.'0'represents an inactive section.
You can perform at most one trade to maximize the number of active sections in s. In a trade, you:
- Convert a contiguous block of
'1's that is surrounded by'0's to all'0's. - Afterward, convert a contiguous block of
'0's that is surrounded by'1's to all'1's.
Return the maximum number of active sections in s after making the optimal trade.
Note: Treat s as if it is augmented with a '1' at both ends, forming t = '1' + s + '1'. The augmented '1's do not contribute to the final count.
Example 1:
Input: s = “01”
Output: 1
Explanation:
Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 1.
Example 2:
Input: s = “0100”
Output: 4
Explanation:
- String
"0100"→ Augmented to"101001". - Choose
"0100", convert“10 1 001”→“1 0000 1”→“1 1111 1”. - The final string without augmentation is
"1111". The maximum number of active sections is 4.
Example 3:
Input: s = “1000100”
Output: 7
Explanation:
- String
"1000100"→ Augmented to"110001001". - Choose
"000100", convert“11000 1 001”→“11 000000 1”→“11 111111 1”. - The final string without augmentation is
"1111111". The maximum number of active sections is 7.
Example 4:
Input: s = “01010”
Output: 4
Explanation:
- String
"01010"→ Augmented to"1010101". - Choose
"010", convert“10 1 0101”→“1 000 101”→“1 111 101”. - The final string without augmentation is
"11110". The maximum number of active sections is 4.
Constraints:
1 <= n == s.length <= 105s[i]is either'0'or'1'
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Constructor Summary
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Solution
public Solution()
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Method Details
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maxActiveSectionsAfterTrade
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