通过删除包含相同字符的子字符串来删除字符串的最少步骤

给定一个字符串str 。如果在一次操作中所有字符都相同，则只能删除一些连续的字符。任务是找到完全删除字符串所需的最少操作数。

例子：

Input: str = “abcddcba”
Output: 4
Delete dd, then the string is “abccba”
Delete cc, then the string is “abba”
Delete bb, then the string is “aa”
Delete aa, then the string is null.

Input: str = “abc”
Output: 3

方法：可以使用

令dp [l] [r]是子字符串s [l，l + 1，l + 2，…r]的答案。然后我们有两种情况：

• 子字符串的第一个字母与其余字母分开删除，然后dp [l] [r] = 1 + dp [l + 1] [r] 。
• 子字符串的第一个字母与其他一些字母(两个字母必须相等)一起删除，然后dp [l] [r] = dp [l + 1] [i-1] + dp [i] [r ]，考虑到升≤I≤r和s [I] = S [1]。

可以与备注一起递归调用以下两种情况，以避免重复的函数调用。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
const int N = 10;
  
// Function to return the minimum number of
// delete operations
int findMinimumDeletion(int l, int r, int dp[N][N], string s)
{
  
    if (l > r)
        return 0;
    if (l == r)
        return 1;
    if (dp[l][r] != -1)
        return dp[l][r];
  
    // When a single character is deleted
    int res = 1 + findMinimumDeletion(l + 1, r, dp, s);
  
    // When a group of consecutive characters
    // are deleted if any of them matches
    for (int i = l + 1; i <= r; ++i) {
  
        // When both the characters are same then
        // delete the letters in between them
        if (s[l] == s[i])
            res = min(res, findMinimumDeletion(l + 1, i - 1, dp, s)
                               + findMinimumDeletion(i, r, dp, s));
    }
  
    // Memoize
    return dp[l][r] = res;
}
  
// Driver code
int main()
{
    string s = "abcddcba";
    int n = s.length();
    int dp[N][N];
    memset(dp, -1, sizeof dp);
    cout << findMinimumDeletion(0, n - 1, dp, s);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
    static int N = 10;
  
    // Function to return the minimum number of
    // delete operations
    static int findMinimumDeletion(int l, int r, 
                            int dp[][], String s)
    {
  
        if (l > r) 
        {
            return 0;
        }
        if (l == r) 
        {
            return 1;
        }
        if (dp[l][r] != -1) 
        {
            return dp[l][r];
        }
  
        // When a single character is deleted
        int res = 1 + findMinimumDeletion(l + 1, r, dp, s);
  
        // When a group of consecutive characters
        // are deleted if any of them matches
        for (int i = l + 1; i <= r; ++i)
        {
  
            // When both the characters are same then
            // delete the letters in between them
            if (s.charAt(l) == s.charAt(i)) 
            {
                res = Math.min(res, findMinimumDeletion(l + 1, i - 1, dp, s)
                        + findMinimumDeletion(i, r, dp, s));
            }
        }
  
        // Memoize
        return dp[l][r] = res;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        String s = "abcddcba";
        int n = s.length();
        int dp[][] = new int[N][N];
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++) 
            {
                dp[i][j] = -1;
            }
        }
        System.out.println(findMinimumDeletion(0, n - 1, dp, s));
    }
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
  
# Function to return the minimum
# number of delete operations
def findMinimumDeletion(l, r, dp, s):
  
    if l > r:
        return 0
    if l == r:
        return 1
    if dp[l][r] != -1:
        return dp[l][r]
  
    # When a single character is deleted
    res = 1 + findMinimumDeletion(l + 1, r, 
                                     dp, s)
  
    # When a group of consecutive characters
    # are deleted if any of them matches
    for i in range(l + 1, r + 1): 
  
        # When both the characters are same then
        # delete the letters in between them
        if s[l] == s[i]:
            res = min(res, findMinimumDeletion(l + 1, i - 1, dp, s) +
                           findMinimumDeletion(i, r, dp, s))
      
    # Memoize
    dp[l][r] = res
    return res
  
# Driver code
if __name__ == "__main__":
  
    s = "abcddcba"
    n = len(s)
    N = 10
    dp = [[-1 for i in range(N)] 
              for j in range(N)]
    print(findMinimumDeletion(0, n - 1, dp, s))
  
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
    static int N = 10;
  
    // Function to return the minimum number of
    // delete operations
    static int findMinimumDeletion(int l, int r, 
                            int [,]dp, String s)
    {
  
        if (l > r) 
        {
            return 0;
        }
        if (l == r) 
        {
            return 1;
        }
        if (dp[l, r] != -1) 
        {
            return dp[l, r];
        }
  
        // When a single character is deleted
        int res = 1 + findMinimumDeletion(l + 1, r, dp, s);
  
        // When a group of consecutive characters
        // are deleted if any of them matches
        for (int i = l + 1; i <= r; ++i)
        {
  
            // When both the characters are same then
            // delete the letters in between them
            if (s[l] == s[i]) 
            {
                res = Math.Min(res, findMinimumDeletion(l + 1, i - 1, dp, s)
                        + findMinimumDeletion(i, r, dp, s));
            }
        }
  
        // Memoize
        return dp[l,r] = res;
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        String s = "abcddcba";
        int n = s.Length;
        int [,]dp = new int[N, N];
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++) 
            {
                dp[i, j] = -1;
            }
        }
        Console.WriteLine(findMinimumDeletion(0, n - 1, dp, s));
    }
}
  
// This code has been contributed by 29AjayKumar

PHP

 $r)
        return 0;
    if ($l == $r)
        return 1;
    if ($dp[$l][$r] != -1)
        return $dp[$l][$r];
  
    // When a single character is deleted
    $res = 1 + findMinimumDeletion($l + 1, $r, 
                                      $dp, $s);
  
    // When a group of consecutive characters
    // are deleted if any of them matches
    for ($i = $l + 1; $i <= $r; ++$i) 
    {
  
        // When both the characters are same then
        // delete the letters in between them
        if ($s[$l] == $s[$i])
            $res = min($res, findMinimumDeletion($l + 1, $i - 1, $dp, $s) + 
                             findMinimumDeletion($i, $r, $dp, $s));
    }
  
    // Memoize
    return $dp[$l][$r] = $res;
}
  
// Driver code
$s = "abcddcba";
$n = strlen($s);
$dp = array(array());
for($i = 0; $i < $GLOBALS['N']; $i++)
    for($j = 0; $j < $GLOBALS['N']; $j++)
        $dp[$i][$j] = -1 ;
          
echo findMinimumDeletion(0, $n - 1, $dp, $s);
  
// This code is contributed by Ryuga
?>

输出：

4