查找最大长度奇校验子串

给定一个二进制字符串str ，任务是找到str具有奇校验的子字符串的最大长度。如果二进制字符串包含奇数个1 ，则称其为奇校验。

例子：

Input: str = “1001110” 
Output: 6 
“001110” is the valid sub-string.

Input: str = “101101” 
Output: 5 
 

方法：

1. 计算给定字符串中1的数量并将其存储在变量cnt中。
2. 如果cnt = 0则不可能有奇校验的子字符串，因此结果将为0 。
3. 如果cnt是奇数，那么结果将是完整的字符串。
4. 现在对于cnt为偶数且> 0的情况，所需的子字符串将从索引0开始并在最后一次出现1之前结束，或者在第一次出现1之后开始并在给定字符串的末尾结束.
5. 在上一步的两个子串中选择长度较大的一个。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
    // finds the index of character of string
    int indexOf(string s, char c, int i)
    {
        for(; i < s.length(); i++)
        if(s[i] == c)
        return i;
         
        return -1;
    }
     
    // finds the last index of character of string
    int lastIndexOf(string s,char c,int i)
    {
        for(; i >= 0; i--)
        if(s[i] == c)
            return i;
         
        return -1;
    }
 
    // Function to return the maximum
    // length of the sub-string which
    // contains odd number of 1s
    int maxOddParity(string str, int n)
    {
 
        // Find the count of 1s in
        // the given string
        int cnt = 0;
        for (int i = 0; i < n; i++)
            if (str[i] == '1')
                cnt++;
 
        // If there are only 0s in the string
        if (cnt == 0)
            return 0;
 
        // If the count of 1s is odd then
        // the complete string has odd parity
        if (cnt % 2 == 1)
            return n;
 
        // Index of the first and the second
        // occurrences of '1' in the string
        int firstOcc = indexOf(str,'1',0);
        int secondOcc = indexOf(str,'1', firstOcc + 1);
 
        // Index of the last and the second last
        // occurrences of '1' in the string
        int lastOcc = lastIndexOf(str,'1',str.length()-1);
        int secondLastOcc = lastIndexOf(str,'1', lastOcc - 1);
 
        // Result will the sub-string ending just
        // before the last occurrence of '1' or the
        // sub-string starting just after the first
        // occurrence of '1'
        // choose the one with the maximum length
        return max(lastOcc, n - firstOcc - 1);
    }
 
    // Driver code
    int main()
    {
        string str = "101101";
        int n = str.length();
        cout<<(maxOddParity(str, n));
    }
 
// This code is contributed by Arnab Kundu

Java

// Java implementation of the approach
public class GFG {
 
    // Function to return the maximum
    // length of the sub-string which
    // contains odd number of 1s
    static int maxOddParity(String str, int n)
    {
 
        // Find the count of 1s in
        // the given string
        int cnt = 0;
        for (int i = 0; i < n; i++)
            if (str.charAt(i) == '1')
                cnt++;
 
        // If there are only 0s in the string
        if (cnt == 0)
            return 0;
 
        // If the count of 1s is odd then
        // the complete string has odd parity
        if (cnt % 2 == 1)
            return n;
 
        // Index of the first and the second
        // occurrences of '1' in the string
        int firstOcc = str.indexOf('1');
        int secondOcc = str.indexOf('1', firstOcc + 1);
 
        // Index of the last and the second last
        // occurrences of '1' in the string
        int lastOcc = str.lastIndexOf('1');
        int secondLastOcc = str.lastIndexOf('1', lastOcc - 1);
 
        // Result will the sub-string ending just
        // before the last occurrence of '1' or the
        // sub-string starting just after the first
        // occurrence of '1'
        // choose the one with the maximum length
        return Math.max(lastOcc, n - firstOcc - 1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "101101";
        int n = str.length();
        System.out.print(maxOddParity(str, n));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the maximum
# length of the sub-string which
# contains odd number of 1s
def maxOddParity(string, n):
 
    # Find the count of 1s in
    # the given string
    cnt = 0
    for i in range(n):
        if string[i] != '1':
            cnt += 1
 
    # If there are only 0s in the string
    if cnt == 0:
        return 0
 
    # If the count of 1s is odd then
    # the complete string has odd parity
    if cnt % 2 == 1:
        return n
 
    # Index of the first and the second
    # occurrences of '1' in the string
    firstOcc = string.index('1')
    secondOcc = string.index('1', firstOcc + 1)
 
    # Index of the last and the second last
    # occurrences of '1' in the string
    lastOcc = string.rindex('1')
    secondLastOcc = string.rindex('1', 0, lastOcc)
 
    # Result will the sub-string ending just
    # before the last occurrence of '1' or the
    # sub-string starting just after the first
    # occurrence of '1'
    # choose the one with the maximum length
    return max(lastOcc, n - firstOcc - 1)
 
# Driver Code
if __name__ == "__main__":
    string = "101101"
    n = len(string)
    print(maxOddParity(string, n))
 
# This code is contributed by
# sanjeev2552

C#

// C# implementation of the above approach
using System;
 
class GFG
{
 
    // Function to return the maximum
    // length of the sub-string which
    // contains odd number of 1s
    static int maxOddParity(String str, int n)
    {
 
        // Find the count of 1s in
        // the given string
        int cnt = 0;
        for (int i = 0; i < n; i++)
            if (str[i] == '1')
                cnt++;
 
        // If there are only 0s in the string
        if (cnt == 0)
            return 0;
 
        // If the count of 1s is odd then
        // the complete string has odd parity
        if (cnt % 2 == 1)
            return n;
 
        // Index of the first and the second
        // occurrences of '1' in the string
        int firstOcc = str.IndexOf('1');
        int secondOcc = str.IndexOf('1', firstOcc + 1);
 
        // Index of the last and the second last
        // occurrences of '1' in the string
        int lastOcc = str.LastIndexOf('1');
        int secondLastOcc = str.LastIndexOf('1', lastOcc - 1);
 
        // Result will the sub-string ending just
        // before the last occurrence of '1' or the
        // sub-string starting just after the first
        // occurrence of '1'
        // choose the one with the maximum length
        return Math.Max(lastOcc, n - firstOcc - 1);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String str = "101101";
        int n = str.Length;
        Console.WriteLine(maxOddParity(str, n));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// Javascript implementation of the above approach
 
// Function to return the maximum
// length of the sub-string which
// contains odd number of 1s
function maxOddParity(str, n)
{
     
    // Find the count of 1s in
    // the given string
    var cnt = 0;
    for(var i = 0; i < n; i++)
        if (str[i] == '1')
            cnt++;
 
    // If there are only 0s in the string
    if (cnt == 0)
        return 0;
 
    // If the count of 1s is odd then
    // the complete string has odd parity
    if (cnt % 2 == 1)
        return n;
 
    // Index of the first and the second
    // occurrences of '1' in the string
    var firstOcc = str.indexOf('1');
    var secondOcc = str.indexOf(
        '1', firstOcc + 1);
 
    // Index of the last and the second last
    // occurrences of '1' in the string
    var lastOcc = str.lastIndexOf('1');
    var secondLastOcc = str.lastIndexOf(
        '1', lastOcc - 1);
 
    // Result will the sub-string ending just
    // before the last occurrence of '1' or the
    // sub-string starting just after the first
    // occurrence of '1'
    // choose the one with the maximum length
    return Math.max(lastOcc, n - firstOcc - 1);
}
 
// Driver code
var str = "101101";
var n = str.length;
 
document.write(maxOddParity(str, n));
 
// This code is contributed by bunnyram19
 

输出：

5