给定长度为N的二进制字符串str ，任务是找到从给定的二进制字符串删除的最小数目的字符，以使子字符串为0 s，然后子字符串为1 s。
例子：
Input: str = “00101101”
Output: 2
Explanation: Removing str[2] and str[6] or removing str[3] and str[6] modifies the given binary string to “000111” or “001111” respectively. The number of removals required in both the cases is 2, which is the minimum possible.
Input: str = “111000001111”
Output: 3
天真的方法：解决此问题的最简单方法是遍历字符串，对于遇到的每个“ 1” ，通过删除0 s或删除1 s计算所需的最小删除次数。最后，打印所需的最少删除内容。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效的方法：可以通过有一个辅助空间来优化上述方法，该辅助空间可以使1s之后的0s计数保持不变。使用这种预计算，可以将时间复杂度提高N倍。步骤如下：
- 初始化一个变量，例如ans ，以存储需要删除的最少字符数。
- 初始化一个数组，比如zeroCount [] ，以计算在给定索引之后出现的0数。
- 从端部超过[N – 2,0]的范围内遍历字符串str并且如果当前字符是0，则更新zeroCount [I]作为(zeroCount [I + 1] + 1)。否则，将zeroCount [i]更新为zeroCount [i + 1] 。
- 初始化一个变量，例如oneCount ，以计算1s的数量。
- 再次遍历给定的字符串。为每个字符发现是‘1’，更新ANS作为最小ANS的和(oneCount + zeroCount [I])。
- 完成上述步骤后，如果ans的值与其初始值相同，则需要删除0个字符。否则， ans是要删除的所需字符数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
#include
using namespace std;
// Function to count minimum removals
// required to make a given string
// concatenation of substring of 0s
// followed by substring of 1s
int minimumDeletions(string s)
{
// Stores the length of the string
int n = s.size();
// Precompute the count of 0s
int zeroCount[n];
// Check for the last character
zeroCount[n - 1] = (s[n - 1] == '0') ? 1 : 0;
// Traverse and update zeroCount array
for(int i = n - 2; i >= 0; i--)
// If current character is 0,
zeroCount[i] = (s[i] == '0') ?
// Update aCount[i] as
// aCount[i + 1] + 1
zeroCount[i + 1] + 1 :
// Update as aCount[i + 1]
zeroCount[i + 1];
// Keeps track of deleted 1s
int oneCount = 0;
// Stores the count of removals
int ans = INT_MAX;
// Traverse the array
for(int i = 0; i < n; i++)
{
// If current character is 1
if (s[i] == '1')
{
// Update ans
ans = min(ans,
oneCount +
zeroCount[i]);
oneCount++;
}
}
// If all 1s are deleted
ans = min(ans, oneCount);
// Return the minimum
// number of deletions
return (ans == INT_MAX) ? 0 : ans;
}
// Driver Code
int main()
{
string stri = "00101101";
cout << minimumDeletions(stri) << endl;
return 0;
}
// This code is contributed by AnkThon Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to count minimum removals
// required to make a given string
// concatenation of substring of 0s
// followed by substring of 1s
public static int minimumDeletions(String s)
{
// Stores the length of the string
int n = s.length();
// Precompute the count of 0s
int zeroCount[] = new int[n];
// Check for the last character
zeroCount[n - 1] = (s.charAt(n - 1)
== '0')
? 1
: 0;
// Traverse and update zeroCount array
for (int i = n - 2; i >= 0; i--)
// If current character is 0,
zeroCount[i] = (s.charAt(i) == '0')
// Update aCount[i] as
// aCount[i + 1] + 1
? zeroCount[i + 1] + 1
// Update as aCount[i + 1]
: zeroCount[i + 1];
// Keeps track of deleted 1s
int oneCount = 0;
// Stores the count of removals
int ans = Integer.MAX_VALUE;
// Traverse the array
for (int i = 0; i < n; i++) {
// If current character is 1
if (s.charAt(i) == '1') {
// Update ans
ans = Math.min(ans,
oneCount
+ zeroCount[i]);
oneCount++;
}
}
// If all 1s are deleted
ans = Math.min(ans, oneCount);
// Return the minimum
// number of deletions
return (ans == Integer.MAX_VALUE)
? 0
: ans;
}
// Driver Code
public static void main(String[] args)
{
String str = "00101101";
System.out.println(
minimumDeletions(str));
}
}Python3
# Python3 program for the above approach
# Function to count minimum removals
# required to make a given string
# concatenation of substring of 0s
# followed by substring of 1s
def minimumDeletions(s):
# Stores the length of the string
n = len(s)
# Precompute the count of 0s
zeroCount = [ 0 for i in range(n)]
# Check for the last character
zeroCount[n - 1] = 1 if s[n - 1] == '0' else 0
# Traverse and update zeroCount array
for i in range(n - 2, -1, -1):
# If current character is 0,
zeroCount[i] = zeroCount[i + 1] + 1 if (s[i] == '0') else zeroCount[i + 1]
# Keeps track of deleted 1s
oneCount = 0
# Stores the count of removals
ans = 10**9
# Traverse the array
for i in range(n):
# If current character is 1
if (s[i] == '1'):
# Update ans
ans = min(ans,oneCount + zeroCount[i])
oneCount += 1
# If all 1s are deleted
ans = min(ans, oneCount)
# Return the minimum
# number of deletions
return 0 if ans == 10**18 else ans
# Driver Code
if __name__ == '__main__':
str = "00101101"
print(minimumDeletions(str))
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to count minimum removals
// required to make a given string
// concatenation of substring of 0s
// followed by substring of 1s
public static int minimumDeletions(String s)
{
// Stores the length of the string
int n = s.Length;
// Precompute the count of 0s
int []zeroCount = new int[n];
// Check for the last character
zeroCount[n - 1] = (s[n - 1]
== '0')
? 1
: 0;
// Traverse and update zeroCount array
for (int i = n - 2; i >= 0; i--)
// If current character is 0,
zeroCount[i] = (s[i] == '0')
// Update aCount[i] as
// aCount[i + 1] + 1
? zeroCount[i + 1] + 1
// Update as aCount[i + 1]
: zeroCount[i + 1];
// Keeps track of deleted 1s
int oneCount = 0;
// Stores the count of removals
int ans = int.MaxValue;
// Traverse the array
for (int i = 0; i < n; i++) {
// If current character is 1
if (s[i] == '1') {
// Update ans
ans = Math.Min(ans,
oneCount
+ zeroCount[i]);
oneCount++;
}
}
// If all 1s are deleted
ans = Math.Min(ans, oneCount);
// Return the minimum
// number of deletions
return (ans == int.MaxValue)
? 0
: ans;
}
// Driver Code
public static void Main(String[] args)
{
String str = "00101101";
Console.WriteLine(
minimumDeletions(str));
}
}
// This code is contributed by 29AjayKumar输出：
2时间复杂度： O(N)
辅助空间： O(N)