给定一个长度为N的二进制字符串S ，任务是找到需要从给定字符串S 中删除的最小0数以获得最长的1s子串。
例子：
Input: S = “010011”
Output: 2
Explanation:
Removing str[2] and str[3] modifies the string S to “0111”. Therefore, the minimum number of removals required is 2.
Input: S = “011111”
Output: 0
方法：这个想法是在字符串找到最左边和最右边的索引1 ，然后计算它们之间存在的0的数量。最后，打印获得的计数值。请按照以下步骤解决问题：
- 遍历字符串s找到1字符串中的第一个和最后发生和存储其指数中的变量，说左，右，分别。
- 使用变量i在范围[left, right] 上迭代，如果str[i] 的值等于0 ，则将count增加 1。
- 完成以上步骤后，打印count的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count the minimum number
// of 0s required to be removed to
// maximize the longest substring of 1s
void minNumZeros(string str)
{
// Stores leftmost and rightmost
// indices consisting of 1
int left = INT_MAX, right = INT_MIN;
// Stores the count of 0s
// between left and right
int count = 0;
// Traverse the string str
for (int i = 0; i < str.length(); i++) {
// If the current character is 1
if (str[i] == '1') {
// Update leftmost and rightmost
// index consisting of 1
left = min(i, left);
right = max(right, i);
}
}
// If string consists only of 0s
if (left == INT_MAX) {
cout << "0";
return;
}
// Count the number of 0s
// between left and right
for (int i = left; i < right; i++) {
if (str[i] == '0') {
count++;
}
}
// Print the result
cout << count;
}
// Driver Code
int main()
{
string str = "010011";
minNumZeros(str);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to count the minimum number
// of 0s required to be removed to
// maximize the longest substring of 1s
static void minNumZeros(String str)
{
// Stores leftmost and rightmost
// indices consisting of 1
int left = Integer.MAX_VALUE;
int right = Integer.MIN_VALUE;
// Stores the count of 0s
// between left and right
int count = 0;
// Traverse the string str
for(int i = 0; i < str.length(); i++)
{
// If the current character is 1
if (str.charAt(i) == '1')
{
// Update leftmost and rightmost
// index consisting of 1
left = Math.min(i, left);
right = Math.max(right, i);
}
}
// If string consists only of 0s
if (left == Integer.MAX_VALUE)
{
System.out.print("0");
return;
}
// Count the number of 0s
// between left and right
for(int i = left; i < right; i++)
{
if (str.charAt(i) == '0')
{
count++;
}
}
// Print the result
System.out.print(count);
}
// Driver Code
public static void main(String[] args)
{
String str = "010011";
minNumZeros(str);
}
}
// This code is contributed by Dharanendra L VPython3
# Python program for the above approach
import sys
# Function to count the minimum number
# of 0s required to be removed to
# maximize the longest substring of 1s
def minNumZeros(st) :
# Stores leftmost and rightmost
# indices consisting of 1
left = sys.maxsize
right = -sys.maxsize - 1
# Stores the count of 0s
# between left and right
count = 0
# Traverse the string str
for i in range(len(st)) :
#for (int i = 0; i < str.length(); i++) {
# If the current character is 1
if st[i] == '1' :
# Update leftmost and rightmost
# index consisting of 1
left = min(i, left)
right = max(right, i)
# If string consists only of 0s
if left == sys.maxsize :
print("0")
return
# Count the number of 0s
# between left and right
for i in range(left,right):
#for (int i = left; i < right; i++) {
if st[i] == '0' :
count += 1
# Print the result
print(count)
# Driver Code
if __name__ == "__main__" :
st = "010011";
minNumZeros(st)
# This code is contributed by jana_sayantan.C#
// C# program for the above approach
using System;
class GFG{
// Function to count the minimum number
// of 0s required to be removed to
// maximize the longest substring of 1s
static void minNumZeros(string str)
{
// Stores leftmost and rightmost
// indices consisting of 1
int left = int.MaxValue;
int right = int.MinValue;
// Stores the count of 0s
// between left and right
int count = 0;
// Traverse the string str
for(int i = 0; i < str.Length; i++)
{
// If the current character is 1
if (str[i] == '1')
{
// Update leftmost and rightmost
// index consisting of 1
left = Math.Min(i, left);
right = Math.Max(right, i);
}
}
// If string consists only of 0s
if (left == int.MaxValue)
{
Console.WriteLine("0");
return;
}
// Count the number of 0s
// between left and right
for(int i = left; i < right; i++)
{
if (str[i] == '0')
{
count++;
}
}
// Print the result
Console.WriteLine(count);
}
// Driver Code
static public void Main()
{
string str = "010011";
minNumZeros(str);
}
}
// This code is contributed by Dharanendra L V输出：
2时间复杂度： O(N)
辅助空间： O(1)
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