
Problem Statement
In the realm of binary strings, a string is considered monotone increasing if it is composed of a section of 0
s, possibly none, followed by a section of 1
s, again, possibly none. Given a binary string s
, you have the capability to transform any character within this string from 0
to 1
or vice versa. The challenge here is to discern the least number of such transformations, or flips, required to convert the string into a monotone increasing sequence.
Examples
Example 1
Input:
s = "00110"
Output:
1
Explanation:
We flip the last digit to get 00111.
Example 2
Input:
s = "010110"
Output:
2
Explanation:
We flip to get 011111, or alternatively 000111.
Example 3
Input:
s = "00011000"
Output:
2
Explanation:
We flip to get 00000000.
Constraints
1 <= s.length <= 105
s[i]
is either'0'
or'1'
.
Approach and Intuition
Consider a binary string. To attain a monotone increasing sequence, the string should transition from a phase of 0
s to a phase of 1
s without any interruption. Each '1' before this transition and each '0' after can disrupt this sequence, necessitating a flip to maintain the expected order.
Estimate Flips for Various Cuts: The string can be sectioned into two parts at each index; the left part ideally should contain only
0
s, and the right part should ideally contain only1
s. Determine the flips needed for each breakpoint:- Calculate how many
1
s are present in the left part (which would need to be flipped to0
s) - Calculate how many
0
s are present in the right part (which would need to be flipped to1
s)
- Calculate how many
Prefix and Suffix Calculation: By utilizing prefix arrays that count
0
s and1
s up to each position, the flips needed can be dynamically calculated as the breakpoint is moved from the start to the end of the string.- Establish two arrays: one that maintains a cumulative count of
0
s up to each index and another for1
s. - For a given breakpoint, the flips can be derived using these arrays – the sum of turning all
1
s to0
s on the left and all remaining0
s to1
s on the right.
- Establish two arrays: one that maintains a cumulative count of
Optimize Breakpoint: Traverse through all possible breakpoints (from the start to just before the last character of the string). At every point, calculate the total flips needed if that point were the start of the
1
s, and update the minimum flips needed.- The optimized solution is the minimum of flips calculated over all these breakpoints.
Example walkthrough:
Example 1 ("00110"): Trace through possible breakpoints:
- Before index 0: (convert
0
,0
,1
,1
,0
into1
s) = 3 flips OR (convert1
,1
,0
into0
s) = 0 flips = 3 total - Between index 1 and 2: (convert
0
,0
into0
s) = 0 flips + (convert last0
into1
) = 1 flip = 1 total (Minimum)
Hence, flipping the last to get
00111
is the optimal solution.- Before index 0: (convert
Solutions
- C++
- Java
- Python
class Solution {
public:
int minimumFlipsMonotonic(string input) {
int result = 0, countOne = 0;
for (char ch : input) {
if (ch == '0') {
result = min(countOne, result + 1);
} else {
++countOne;
}
}
return result;
}
};
In the provided C++ code, a solution class has a method named minimumFlipsMonotonic
, which takes a string, input
, as parameter and returns an integer. This function aims to determine the minimum number of flips required to make the input string monotonic increasing. Monotonic increasing here implies that all the zeroes in the string come before all the ones.
An explanation of how the code accomplishes this:
Two integer variables,
result
andcountOne
, are initialized to zero. Theresult
will store the minimum number of flips needed, whilecountOne
will keep track of the number of '1's encountered as the string is processed.The function then loops through each character in the input string. For each character:
If it encounters a '0', it updates
result
by comparing the current value ofcountOne
(representing flips to change all previous '1' to '0') andresult + 1
(representing a flip of the current '0' to '1'), taking the minimum of these two values.If it encounters a '1', the
countOne
is incremented, updating the potential number of flipped ones.
Finally, the value of
result
is returned. This value represents the minimal flip operations required to make the string monotonic increasing.
class Solution {
public int minimumFlipsToMonotonic(String str) {
int flips = 0, countOnes = 0;
for (int index = 0; index < str.length(); ++index) {
if (str.charAt(index) == '0') {
flips = Math.min(countOnes, flips + 1);
} else {
++countOnes;
}
}
return flips;
}
}
The provided Java solution addresses the problem of converting a binary string into a monotone increasing string using the minimum number of flips. A monotone increasing string is one where every character is either the same as or greater than the one before it, which, in the case of a binary string, means the string could transition from any number of '0's to any number of '1's but not vice versa.
- The solution iterates through the input string to count flips needed to turn the string into a monotone increasing string.
- Maintains a
flips
counter to track the minimal flips required and acountOnes
counter to count the number of '1's encountered. - If a '0' is found in the string, the algorithm computes whether it is better to flip this '0' to '1' or to flip all previously encountered '1's to '0's.
- This decision is based on comparing the current number of flips required if flipping the '0' (
flips + 1
) versus the total '1's seen so far (countOnes
). - At the end, the function returns the computed minimum number of flips (
flips
) that would turn the input string into a monotone increasing string.
Ensure to provide a valid binary string input to avoid unexpected behaviors since the function assumes a valid input consisting only of '0's and '1's.
class Solution:
def minimumFlipsToMonotonic(self, binary_str: str) -> int:
flip_count = 0
one_count = 0
for char in binary_str:
if char == '0':
flip_count = min(one_count, flip_count + 1)
else:
one_count += 1
return flip_count
The given Python solution addresses the problem of flipping a binary string to make it monotone increasing. The approach utilizes dynamic programming to efficiently determine the minimal number of flips required. Below is an outline of the solution steps:
- Initialize two counters:
flip_count
to track the number of flips, andone_count
to track the number of '1's encountered. - Traverse the binary string character by character using a loop.
- If the current character is '0', update
flip_count
as the minimal value betweenone_count
(flipping all previous '1's to '0's) andflip_count + 1
(flipping the current '0' to '1'). - If the current character is '1', increment
one_count
.
- If the current character is '0', update
- Return the
flip_count
finally determined, which represents the minimum number of flips required to make the string monotone increasing.
This strategy ensures an optimal solution by considering flipping sequences dynamically, counting flips and ones separately to make decisions at each step. The time complexity of this solution is O(n), where n is the length of the string, due to the single traversal of the characters.
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