使字符串满足给定条件所需的最小更改

给定二进制字符串str 。在单个操作中，我们可以将任何‘1’更改为‘0’或将任何‘0’更改为‘1’ 。任务是在字符串进行最少数量的更改，以便如果我们采用字符串的任何前缀，则1 的数量应大于或等于0 的数量。
例子：

Input: str = “10001” 
Output: 1 
We can change str[2] from ‘0’ to ‘1’.
Input: str = “0000” 
Output: 2 
 

处理方法：贪心的可以解决问题。字符串的第一个字符必须是1 。然后，该字符串的其余部分，我们通过字符通过字符串字符遍历和检查，如果需要的条件满足与否，如果没有的话，我们增加所需的更改的数量。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum
// changes required
int minChanges(string str, int n)
{
 
    // To store the count of minimum changes,
    // number of ones and the number of zeroes
    int count = 0, zeros = 0, ones = 0;
 
    // First character has to be '1'
    if (str[0] != '1') {
        count++;
        ones++;
    }
 
    for (int i = 1; i < n; i++) {
        if (str[i] == '0')
            zeros++;
        else
            ones++;
 
        // If condition fails
        // changes need to be made
        if (zeros > ones) {
            zeros--;
            ones++;
            count++;
        }
    }
 
    // Return the required count
    return count;
}
 
// Driver code
int main()
{
    string str = "0000";
    int n = str.length();
    cout << minChanges(str, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the minimum
// changes required
static int minChanges(char[] str, int n)
{
 
    // To store the count of minimum changes,
    // number of ones and the number of zeroes
    int count = 0, zeros = 0, ones = 0;
 
    // First character has to be '1'
    if (str[0] != '1')
    {
        count++;
        ones++;
    }
 
    for (int i = 1; i < n; i++)
    {
        if (str[i] == '0')
            zeros++;
        else
            ones++;
 
        // If condition fails
        // changes need to be made
        if (zeros > ones)
        {
            zeros--;
            ones++;
            count++;
        }
    }
 
    // Return the required count
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    char []str = "0000".toCharArray();
    int n = str.length;
    System.out.print(minChanges(str, n));
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the minimum
# changes required
def minChanges(str, n):
     
    # To store the count of minimum changes,
    # number of ones and the number of zeroes
    count, zeros, ones = 0, 0, 0
 
    # First character has to be '1'
    if (ord(str[0])!= ord('1')):
        count += 1
        ones += 1
 
    for i in range(1, n):
        if (ord(str[i]) == ord('0')):
            zeros += 1
        else:
            ones += 1
 
        # If condition fails
        # changes need to be made
        if (zeros > ones):
            zeros -= 1
            ones += 1
            count += 1
 
    # Return the required count
    return count
 
# Driver code
if __name__ == '__main__':
    str = "0000"
    n = len(str)
    print(minChanges(str, n))
 
# This code contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the minimum
// changes required
static int minChanges(char[] str, int n)
{
 
    // To store the count of minimum changes,
    // number of ones and the number of zeroes
    int count = 0, zeros = 0, ones = 0;
 
    // First character has to be '1'
    if (str[0] != '1')
    {
        count++;
        ones++;
    }
 
    for (int i = 1; i < n; i++)
    {
        if (str[i] == '0')
            zeros++;
        else
            ones++;
 
        // If condition fails
        // changes need to be made
        if (zeros > ones)
        {
            zeros--;
            ones++;
            count++;
        }
    }
 
    // Return the required count
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    char []str = "0000".ToCharArray();
    int n = str.Length;
    Console.Write(minChanges(str, n));
}
}
 
// This code contributed by Rajput-Ji

PHP

 $ones)
        {
            $zeros--;
            $ones++;
            $count++;
        }
    }
 
    // Return the required count
    return $count;
}
 
// Driver code
$str = "0000";
$n = strlen($str);
echo minChanges($str, $n);
 
// This code is contributed by mits
?>

Javascript

 
// Javascript implementation of the approach
 
// Function to return the minimum
// changes required
function minChanges(str, n)
{
   
    // To store the count of minimum changes,
    // number of ones and the number of zeroes
    let count = 0, zeros = 0, ones = 0;
   
    // First character has to be '1'
    if (str[0] != '1')
    {
        count++;
        ones++;
    }
   
    for (let i = 1; i < n; i++)
    {
        if (str[i] == '0')
            zeros++;
        else
            ones++;
   
        // If condition fails
        // changes need to be made
        if (zeros > ones)
        {
            zeros--;
            ones++;
            count++;
        }
    }
   
    // Return the required count
    return count;
} 
     
// Driver Code
 
     let str = "0000".split('');
    let n = str.length;
     document.write(minChanges(str, n));
             

输出：

2

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