Java程序用于具有相同左右旋转的数字的最长子序列

给定一个数字字符串S ，任务是找到左旋转等于右旋转的子序列的最大长度。

例子：

Input: S = “100210601”
Output: 4
Explanation:
The subsequence “0000” satisfies the necessary condition.
The subsequence “1010” generates the string “0101” on left rotation and string “0101” on right rotation. Since both the rotations are same. Therefore, “1010” satisfies the condition as well.
Therefore, the maximum length of such subsequence is 4.
Input: S = “252525”
Output: 6
Explanation:
The subsequence “252525” generates the string “525252” on left rotation and string “525252” on right rotation. Since both the rotations are same. Therefore, the “252525” satisfies the required condition.

编程需要懂一点英语

朴素方法：解决问题的最简单方法是生成给定字符串的所有可能子序列，并且对于每个子序列，检查其左旋转是否等于其右旋转。
时间复杂度： O(2 ^N * N)
辅助空间： O(N)

有效方法：为了优化上述方法，主要观察是子序列应该由单个字符组成，或者应该是由两个字符组成的偶数长度。

Illustration:
str = “2424”
Left rotation of the string = “4242”
Right rotation of the string = “4242”
As we can see, since the number is of even length having two characters appearing alternately, therefore, the left and right rotation of the given number is equal.
str = “24242”
Left rotation of the string = “42422”
Right rotation of the string = “22424”
As we can see, since the number is of odd length having two characters appearing alternately, therefore, the left and right rotation of the given number is not equal.

编程需要懂一点英语

请按照以下步骤解决问题：

生成所有可能的两位数。
对于生成的每个两位数，检查字符串中两个数字的交替出现。保持递增计数以存储特定组合的交替子序列的长度。
在字符串的整个遍历之后，检查两个数字是否相等。如果发现是真的，将计数更新为所需的答案。如果两个数字相等，则分别更新count或count – 1到答案，如果计数是偶数或奇数。
对所有可能的组合重复上述步骤并打印获得的最大计数。

下面是上述方法的实现：

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
  
// Function to find the longest subsequence
// having equal left and right rotation
static int findAltSubSeq(String s)
{
    // Length of the String
    int n = s.length(), ans = Integer.MIN_VALUE;
  
    // Iterate for all possible combinations
    // of a two-digit numbers
    for (int i = 0; i < 10; i++)
    {
        for (int j = 0; j < 10; j++)
        {
            int cur = 0, f = 0;
  
            // Check for alternate occurrence
            // of current combination
            for (int k = 0; k < n; k++) 
            {
                if (f == 0 && s.charAt(k) - '0' == i) 
                {
                    f = 1;
  
                    // Increment the current value
                    cur++;
                }
                else if (f == 1 && 
                         s.charAt(k) - '0' == j) 
                {
                    f = 0;
  
                    // Increment the current value
                    cur++;
                }
            }
  
            // If alternating sequence is
            // obtained of odd length
            if (i != j && cur % 2 == 1)
  
                // Reduce to even length
                cur--;
  
            // Update answer to store
            // the maximum
            ans = Math.max(cur, ans);
        }
    }
  
    // Return the answer
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    String s = "100210601";
    System.out.print(findAltSubSeq(s));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

时间复杂度： O(N)
辅助空间： O(1)

有关更多详细信息，请参阅关于具有相同左右旋转的数字的最长子序列的完整文章！