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Maximize Subarrays of 0s of Length X in Given Binary String after Flipping at Most One '1' Using C++
In computer programming, it is often necessary to manipulate binary strings and find optimal solutions for various problems. One such problem is to maximize the subarrays of 0s of a specific length, X, in a given binary string. Before we proceed further, it's essential to acknowledge that there exists a restriction - the string permits modification for just one '1'. Let's delve into an effective methodology to overcome this obstacle by applying C++.
Syntax
Before diving into the codes ahead its crucial to clarify the syntax of the method we'll be utilizing.
int maxSubarrays(string binaryString, int X);
Algorithm
Here is the step-by-step algorithm to maximize the subarrays of 0s of length X −
Initialize variables −
`n` as the length of the binary string.
`maxZeros` as 0, representing the maximum number of zeros found.
`currentZeros` as 0, representing the number of zeros in the current subarray.
`start` as 0, representing the starting index of the current subarray.
Iterate over the binary string −
For each index `i` from 0 to n-1, do the following −
In the event that the current character at index 'I' is '0', increase 'currentZeros' by 1.
In the event that 'currentZeros' surpasses X, decrement 'currentZeros' by 1 and move the 'start' index forward until 'currentZeros' becomes X once more.
If the current character at index `i` is '1', check if we can flip it −
In the event that the number of flips made is not exactly or equivalent to 1, increase 'currentZeros' by 1.
In the event that the number of flips made is equivalent to 1, compute the length of the current subarray by subtracting 'start' from 'I-1' and update 'maxZeros' if necessary.
Reset `currentZeros` to 0 and update `start` to the next index.
Finally, calculate the length of the last subarray and update `maxZeros` if necessary.
Return `maxZeros` as the maximum number of zeros in subarrays of length X.
Approach 1: Sliding Window Technique
In this approach, we use a sliding window to keep track of the subarrays.
We maintain a window of size X and move it along the binary string.
By keeping track of the number of zeros and the number of flips made, we can maximize the subarrays of 0s.
Example
#include <iostream>
#include <string>
using namespace std;
int maxSubarrays(string binaryString, int X) {
int n = binaryString.length();
int maxZeros = 0;
int currentZeros = 0;
int start = 0;
for (int i = 0; i < n; i++) {
if (binaryString[i] == '0')
currentZeros++;
else if (currentZeros < X) {
currentZeros++;
} else {
maxZeros = max(maxZeros, i - start);
while (binaryString[start] != '0')
start++;
start++;
}
}
maxZeros = max(maxZeros, n - start);
return maxZeros;
}
int main() {
string binaryString = "110101001011010011";
int X = 3;
int maxZeros = maxSubarrays(binaryString, X);
cout << "Maximum number of zeros in subarrays of length " << X << ": " << maxZeros << endl;
return 0;
}
Output
Maximum number of zeros in subarrays of length 3: 3
Approach 2: Two Pointers Technique
In this approach, we utilize two pointers to monitor the subarrays.
Our algorithm employs a pair of pointers- known as 'left' and 'right'- that are responsible for identifying the respective boundaries of the targeted subarray.
By making tactical adjustments to these pointers based upon both flip count and zero count present within our selected range, we are able to increase said range's size while maintaining accuracy.
Example
#include <iostream>
#include <string>
using namespace std;
int maxSubarrays(string binaryString, int X) {
int n = binaryString.length();
int maxZeros = 0;
int currentZeros = 0;
int flips = 0;
int left = 0;
int right = 0;
while (right < n) {
if (binaryString[right] == '0') {
currentZeros++;
right++;
} else if (flips < 1) {
currentZeros++;
right++;
flips++;
} else {
maxZeros = max(maxZeros, currentZeros);
while (binaryString[left] != '0') {
currentZeros--;
left++;
}
left++;
flips--;
}
}
maxZeros = max(maxZeros, currentZeros);
return maxZeros;
}
int main() {
string binaryString = "110100111010";
int X = 3;
int maxZeros = maxSubarrays(binaryString, X);
cout << "Maximum number of zeros in subarrays of length " << X << ": " << maxZeros << endl;
return 0;
}
Output
Maximum number of zeros in subarrays of length 3: 4
Conclusion
In this article, we discussed an efficient approach to maximize the subarrays of 0s of a specific length, X, in a given binary string. By using either the sliding window technique or the two pointers technique, we were able to manipulate the string to find the optimal solution. The provided C++ codes are fully executable and can be used to solve this problem efficiently. By understanding and implementing these approaches, you can handle similar problems that involve manipulating binary strings and optimizing subarrays.
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