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Minimize count of 0s required to be removed to maximize length of longest substring of 1s
In this article, we will delve into an intriguing problem that involves string manipulation. The problem we're examining today is how to "Minimize the count of 0s required to be removed to maximize the length of the longest substring of 1s". This problem is a great way to hone your skills in string manipulation and dynamic programming in various programming languages.
Problem Statement
Given a binary string, the task is to minimize the count of 0s required to be removed in order to maximize the length of the longest substring of 1s.
Solution Approach
To solve this problem, we can use a sliding window approach. We'll maintain two pointers, left and right. Initially, both pointers point to the first element. Then, we'll keep moving the right pointer to the right. If we encounter a '0', we increment a counter. If the counter becomes larger than the number of allowed zero removals, we move the left pointer to the right until we encounter a '0' and decrement the counter.
We'll also maintain a variable maxLen to store the maximum length of the substring of 1s we've seen so far.
Example
Following are the programs that solves the problem −
#include <stdio.h>
#include <string.h>
int maxSubstring(char str[], int k) {
int zeroCount = 0;
int left = 0;
int maxLen = 0;
int strLen = strlen(str);
for (int right = 0; right < strLen; right++) {
if (str[right] == '0') {
zeroCount++;
}
while (zeroCount > k) {
if (str[left] == '0') {
zeroCount--;
}
left++;
}
maxLen = (maxLen > right - left + 1) ? maxLen : right - left + 1;
}
return maxLen;
}
int main() {
char str[] = "110100110";
int k = 2; // number of zeros that can be removed
int result = maxSubstring(str, k);
printf("The maximum length of the substring of 1s is: %d\n", result);
return 0;
}
Output
The maximum length of the substring of 1s is: 5
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int maxSubstring(string str, int k) {
int zeroCount = 0;
int left = 0;
int maxLen = 0;
for (int right = 0; right < str.size(); right++) {
if (str[right] == '0') {
zeroCount++;
}
while (zeroCount > k) {
if (str[left] == '0') {
zeroCount--;
}
left++;
}
maxLen = max(maxLen, right - left + 1);
}
return maxLen;
}
int main() {
string str = "110100110";
int k = 2; // number of zeros that can be removed
int result = maxSubstring(str, k);
cout << "The maximum length of the substring of 1s is: " << result << endl;
return 0;
}
Output
The maximum length of the substring of 1s is: 5
public class MaxSubstring {
public static int maxSubstring(String str, int k) {
int zeroCount = 0;
int left = 0;
int maxLen = 0; // Stores the maximum length of the substring of 1s
for (int right = 0; right < str.length(); right++) {
if (str.charAt(right) == '0') {
zeroCount++;
}
while (zeroCount > k) {
if (str.charAt(left) == '0') {
zeroCount--;
}
left++;
}
// Update maxLen to store the maximum length of the substring of 1s found so far
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen; // Return the maximum length of the substring of 1s
}
public static void main(String[] args) {
String str = "110100110";
int k = 2;
int result = maxSubstring(str, k); // Call the maxSubstring function
System.out.println("The maximum length of the substring of 1s is: " + result);
}
}
Output
The maximum length of the substring of 1s is: 5
def maxSubstring(s, k):
zeroCount = 0
left = 0
maxLen = 0
for right in range(len(s)):
if s[right] == '0':
zeroCount += 1
while zeroCount > k:
if s[left] == '0':
zeroCount -= 1
left += 1
# Update maxLen to store the maximum length of the substring of 1s found so far
maxLen = max(maxLen, right - left + 1)
return maxLen
if __name__ == "__main__":
s = "110100110"
k = 2
result = maxSubstring(s, k) # Call the maxSubstring function
print("The maximum length of the substring of 1s is:", result)
Output
The maximum length of the substring of 1s is: 5
Explanation with a Test Case
Let's take the binary string "110100110" and we're allowed to remove 2 zeros.
When we pass this string and the value of k to the maxSubstring function, it starts scanning from the left. Every time it encounters a '0', it increments the zeroCount. When zeroCount exceeds k, it starts moving the left pointer right until it encounters a '0' and decrements the zeroCount.
During this process, it keeps updating maxLen which is the maximum length of the substring of 1s. For the given string, the maximum length of the substring of 1s after removing at most 2 zeros is 5, which is the substring "11111" after removing the second and third '0'.
So, the function will return 5.
Conclusion
This problem demonstrates how the sliding window technique can be effectively used for solving complex string manipulation problems. It's an excellent problem for understanding and practicing dynamic programming and string handling techniques. Keep practicing such problems to enhance your coding skills.
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