要插入的最小字符数，以使三个连续字符都不相同
给定一个字符串str ，任务是修改字符串，使三个连续的字符都不相同。在单个操作中，可以在字符串的任何位置插入任何字符。找出所需的此类操作的最小数量。
例子：
Input : str = “aabbbcc”
Output: 1
“aabbdbcc” is the modified string.
Input: str = “geeksforgeeks”
Output: 0
方法：对于每三个相同的连续字符，必须在它们之间插入一个字符以使任何三个连续字符不同。但是需要尽量减少操作次数，因此必须在第二个字符之后插入字符。例如，如果字符串是“bbbb”，那么如果字符被插入到第二个位置，即“babbb”，那么仍然有三个连续的相同字符，需要另一个操作来解决这个问题，但是如果字符被插入到第三个位置即“bbabb”，那么只需要一个操作。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count of characters
// that are to be inserted in str such that no
// three consecutive characters are same
int getCount(string str, int n)
{
// To store the count of
// operations required
int cnt = 0;
int i = 0;
while (i < n - 2) {
// A character needs to be
// inserted after str[i + 1]
if (str[i] == str[i + 1]
&& str[i] == str[i + 2]) {
cnt++;
i = i + 2;
}
// Current three consecutive
// characters are not same
else
i++;
}
return cnt;
}
// Driver code
int main()
{
string str = "aabbbcc";
int n = str.length();
cout << getCount(str, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the count of characters
// that are to be inserted in str such that no
// three consecutive characters are same
static int getCount(char[] str, int n)
{
// To store the count of
// operations required
int cnt = 0;
int i = 0;
while (i < n - 2)
{
// A character needs to be
// inserted after str[i + 1]
if (str[i] == str[i + 1] &&
str[i] == str[i + 2])
{
cnt++;
i = i + 2;
}
// Current three consecutive
// characters are not same
else
{
i++;
}
}
return cnt;
}
// Driver code
static public void main(String[] arg)
{
String str = "aabbbcc";
int n = str.length();
System.out.println(getCount(str.toCharArray(), n));
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation of the approach
# Function to return the count of characters
# that are to be inserted in str1 such that no
# three consecutive characters are same
def getCount(str1, n):
# To store the count of
# operations required
cnt = 0;
i = 0;
while (i < n - 2):
# A character needs to be
# inserted after str1[i + 1]
if (str1[i] == str1[i + 1] and
str1[i] == str1[i + 2]):
cnt += 1
i = i + 2
# Current three consecutive
# characters are not same
else:
i += 1
return cnt
# Driver code
str1 = "aabbbcc"
n = len(str1)
print(getCount(str1, n))
# This code is contributed by Mohit KumarC#
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the count of characters
// that are to be inserted in str such that no
// three consecutive characters are same
static int getCount(string str, int n)
{
// To store the count of
// operations required
int cnt = 0;
int i = 0;
while (i < n - 2)
{
// A character needs to be
// inserted after str[i + 1]
if (str[i] == str[i + 1] &&
str[i] == str[i + 2])
{
cnt++;
i = i + 2;
}
// Current three consecutive
// characters are not same
else
{
i++;
}
}
return cnt;
}
// Driver code
static public void Main ()
{
string str = "aabbbcc";
int n = str.Length;
Console.WriteLine(getCount(str, n));
}
}
// This code is contributed by AnkitRai01Javascript
输出：
1