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"
\u2192 Augmented to"101001"
. - Choose
"0100"
, convert“10 1 001”
\u2192“1 0000 1”
\u2192“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"
\u2192 Augmented to"110001001"
. - Choose
"000100"
, convert“11000 1 001”
\u2192“11 000000 1”
\u2192“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"
\u2192 Augmented to"1010101"
. - Choose
"010"
, convert“10 1 0101”
\u2192“1 000 101”
\u2192“1 111 101”
. - The final string without augmentation is
"11110"
. The maximum number of active sections is 4.
Constraints:
1 <= n == s.length <= 105
s[i]
is either'0'
or'1'
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Constructor Summary
Constructors -
Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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maxActiveSectionsAfterTrade
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