使字符串满足给定条件所需的最少操作
给定一个字符串str ，任务是用最少的给定操作使字符串以相同的字符开始和结束。在单个操作中，可以删除字符串的任何字符。请注意，结果字符串的长度必须大于1 ，然后打印-1是不可能的。
例子：
Input: str = “geeksforgeeks”
Output: 3
Remove the first and the last two characters
and the string becomes “eeksforgee”
Input: str = “abcda”
Output: 0
方法：如果字符串需要以字符ch开始和结束，那么最佳方法是删除第一次出现ch之前的所有字符和最后一次出现ch之后的所有字符。找出从'a'到'z'的每个可能的ch值需要删除的字符数，并选择删除操作次数最少的字符数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int MAX = 26;
// Function to return the minimum
// operations required
int minOperation(string str, int len)
{
// To store the first and the last
// occurrence of all the characters
int first[MAX], last[MAX];
// Set the first and the last occurrence
// of all the characters to -1
for (int i = 0; i < MAX; i++) {
first[i] = -1;
last[i] = -1;
}
// Update the occurrences of the characters
for (int i = 0; i < len; i++) {
int index = (str[i] - 'a');
// Only set the first occurrence if
// it hasn't already been set
if (first[index] == -1)
first[index] = i;
last[index] = i;
}
// To store the minimum operations
int minOp = -1;
for (int i = 0; i < MAX; i++) {
// If the frequency of the current
// character in the string
// is less than 2
if (first[i] == -1 || first[i] == last[i])
continue;
// Count of characters to be
// removed so that the string
// starts and ends at the
// current character
int cnt = len - (last[i] - first[i] + 1);
if (minOp == -1 || cnt < minOp)
minOp = cnt;
}
return minOp;
}
// Driver code
int main()
{
string str = "abcda";
int len = str.length();
cout << minOperation(str, len);
return 0;
} Java
// Java implementation of the approach
class GFG
{
final static int MAX = 26;
// Function to return the minimum
// operations required
static int minOperation(String str, int len)
{
// To store the first and the last
// occurrence of all the characters
int first[] = new int[MAX];
int last[] = new int[MAX];
// Set the first and the last occurrence
// of all the characters to -1
for (int i = 0; i < MAX; i++)
{
first[i] = -1;
last[i] = -1;
}
// Update the occurrences of the characters
for (int i = 0; i < len; i++)
{
int index = (str.charAt(i) - 'a');
// Only set the first occurrence if
// it hasn't already been set
if (first[index] == -1)
first[index] = i;
last[index] = i;
}
// To store the minimum operations
int minOp = -1;
for (int i = 0; i < MAX; i++)
{
// If the frequency of the current
// character in the string
// is less than 2
if (first[i] == -1 ||
first[i] == last[i])
continue;
// Count of characters to be
// removed so that the string
// starts and ends at the
// current character
int cnt = len - (last[i] - first[i] + 1);
if (minOp == -1 || cnt < minOp)
minOp = cnt;
}
return minOp;
}
// Driver code
public static void main (String[] args)
{
String str = "abcda";
int len = str.length();
System.out.println(minOperation(str, len));
}
}
// This code is contributed by AnkitRai01Python3
# Python implementation of the approach
MAX = 26;
# Function to return the minimum
# operations required
def minOperation(str, len):
# To store the first and the last
# occurrence of all the characters
first, last = [0] * MAX, [0] * MAX;
# Set the first and the last occurrence
# of all the characters to -1
for i in range(MAX):
first[i] = -1;
last[i] = -1;
# Update the occurrences of the characters
for i in range(len):
index = (ord(str[i]) - ord('a'));
# Only set the first occurrence if
# it hasn't already been set
if (first[index] == -1):
first[index] = i;
last[index] = i;
# To store the minimum operations
minOp = -1;
for i in range(MAX):
# If the frequency of the current
# character in the string
# is less than 2
if (first[i] == -1 or first[i] == last[i]):
continue;
# Count of characters to be
# removed so that the string
# starts and ends at the
# current character
cnt = len - (last[i] - first[i] + 1);
if (minOp == -1 or cnt < minOp):
minOp = cnt;
return minOp;
# Driver code
str = "abcda";
len = len(str);
print( minOperation(str, len));
# This code is contributed by 29AjayKumarC#
// C# implementation of the approach
using System;
class GFG
{
readonly static int MAX = 26;
// Function to return the minimum
// operations required
static int minOperation(String str, int len)
{
// To store the first and the last
// occurrence of all the characters
int []first = new int[MAX];
int []last = new int[MAX];
// Set the first and the last occurrence
// of all the characters to -1
for (int i = 0; i < MAX; i++)
{
first[i] = -1;
last[i] = -1;
}
// Update the occurrences of the characters
for (int i = 0; i < len; i++)
{
int index = (str[i] - 'a');
// Only set the first occurrence if
// it hasn't already been set
if (first[index] == -1)
first[index] = i;
last[index] = i;
}
// To store the minimum operations
int minOp = -1;
for (int i = 0; i < MAX; i++)
{
// If the frequency of the current
// character in the string
// is less than 2
if (first[i] == -1 ||
first[i] == last[i])
continue;
// Count of characters to be
// removed so that the string
// starts and ends at the
// current character
int cnt = len - (last[i] - first[i] + 1);
if (minOp == -1 || cnt < minOp)
minOp = cnt;
}
return minOp;
}
// Driver code
public static void Main (String[] args)
{
String str = "abcda";
int len = str.Length;
Console.WriteLine(minOperation(str, len));
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
0