为了使所有数字或交替数字相同而要删除的最小数字

给定一个数字字符串str ，任务是找到要从字符串删除的最小位数，使其满足以下任一条件：

• 字符串的所有元素都是相同的。
• 偶数位置的所有元素都相同，奇数位置的所有元素都相同，这意味着字符串与每个数字的出现均相等。

例子：

Input: s = “95831” 
Output: 3 
Explanation: 
In this examples, remove any three elements form the string to make it alternating, i.e. “95” has 9 at even index and 5 at odd index and hence it satisfies second condition. 
Input: s = “100120013” 
Output: 5 
Explanation: 
In this case, either make the string 0000 or make the string 1010. In both the cases the minimum element must be removed from the string will be 5. 
 

方法：想法是使用贪婪方法。步骤如下：

1. 由于结果字符串中的所有字符都是交替且相同的，因此不同数字的最小子字符串的长度为2。
2. 因为，只有10种不同类型的数字从0到9 。这个想法是迭代长度为2的每个可能的字符串，并发现由它们形成的子序列的出现。
3. 因此，找到上述两位数字字符串的字符串的第一个和第二个字符的所有可能组合，并贪婪地构建以这些字符开头的s的最长可能子序列。
4. 在上述步骤中，字符串长度和具有交替数字的子序列的最大长度之间的差是所需的结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find longest possible
// subsequence of s beginning with x and y
int solve(string s, int x, int y)
{
    int res = 0;
 
    // Iterate over the string
    for (auto c : s) {
        if (c - '0' == x) {
 
            // Increment count
            res++;
 
            // Swap the positions
            swap(x, y);
        }
    }
 
    if (x != y && res % 2 == 1)
        --res;
 
    // Return the result
    return res;
}
 
// Function that finds all the
// possible pairs
int find_min(string s)
{
    int count = 0;
    for (int i = 0; i < 10; i++) {
 
        for (int j = 0; j < 10; j++) {
 
            // Update count
            count = max(count,
                        solve(s, i, j));
        }
    }
 
    // Return the answer
    return count;
}
 
// Driver Code
int main()
{
    // Given string s
    string s = "100120013";
 
    // Find the size of the string
    int n = s.size();
 
    // Function Call
    int answer = find_min(s);
 
    // This value is the count of
    // minimum element to be removed
    cout << (n - answer);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find longest possible
// subsequence of s beginning with x and y
static int solve(String s, int x, int y)
{
    int res = 0;
 
    // Iterate over the String
    for (char c : s.toCharArray())
    {
        if (c - '0' == x)
        {
 
            // Increment count
            res++;
 
            // Swap the positions
            x = x+y;
            y = x-y;
            x = x-y;
        }
    }
 
    if (x != y && res % 2 == 1)
        --res;
 
    // Return the result
    return res;
}
 
// Function that finds all the
// possible pairs
static int find_min(String s)
{
    int count = 0;
    for (int i = 0; i < 10; i++)
    {
        for (int j = 0; j < 10; j++)
        {
 
            // Update count
            count = Math.max(count,
                             solve(s, i, j));
        }
    }
 
    // Return the answer
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given String s
    String s = "100120013";
 
    // Find the size of the String
    int n = s.length();
 
    // Function Call
    int answer = find_min(s);
 
    // This value is the count of
    // minimum element to be removed
    System.out.print((n - answer));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
 
# Function to find longest possible
# subsequence of s beginning with x and y
def solve(s, x, y):
 
    res = 0
 
    # Iterate over the string
    for c in s:
        if(ord(c) - ord('0') == x):
 
            # Increment count
            res += 1
 
            # Swap the positions
            x, y = y, x
 
    if(x != y and res % 2 == 1):
        res -= 1
 
    # Return the result
    return res
 
# Function that finds all the
# possible pairs
def find_min(s):
 
    count = 0
    for i in range(10):
        for j in range(10):
 
            # Update count
            count = max(count, solve(s, i, j))
 
    # Return the answer
    return count
 
# Driver Code
 
# Given string s
s = "100120013"
 
# Find the size of the string
n = len(s)
 
# Function call
answer = find_min(s)
 
# This value is the count of
# minimum element to be removed
print(n - answer)
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find longest possible
// subsequence of s beginning with x and y
static int solve(String s, int x, int y)
{
    int res = 0;
 
    // Iterate over the String
    foreach (char c in s.ToCharArray())
    {
        if (c - '0' == x)
        {
 
            // Increment count
            res++;
 
            // Swap the positions
            x = x + y;
            y = x - y;
            x = x - y;
        }
    }
 
    if (x != y && res % 2 == 1)
        --res;
 
    // Return the result
    return res;
}
 
// Function that finds all the
// possible pairs
static int find_min(String s)
{
    int count = 0;
    for (int i = 0; i < 10; i++)
    {
        for (int j = 0; j < 10; j++)
        {
 
            // Update count
            count = Math.Max(count,
                             solve(s, i, j));
        }
    }
 
    // Return the answer
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given String s
    String s = "100120013";
 
    // Find the size of the String
    int n = s.Length;
 
    // Function Call
    int answer = find_min(s);
 
    // This value is the count of
    // minimum element to be removed
    Console.Write((n - answer));
}
}
 
// This code is contributed by gauravrajput1

输出：

5

时间复杂度： O(N)
辅助空间： O(N)