给定长度为N的数字字符串S和数字K ，任务是找到最大数目的不同字符串，其中最大的K出现在其中，如果S的任何两个相邻数字之和为K，则用K替换任意两个相邻的数字。
例子：
Input: S = “313”, K = 4
Output: 2
Explanation: Possible strings that can be generated are:
- Replacing S[0] and S[1] with K modifies S to “43”.
- Replacing S[1] and S[2] with K modifies S to “34”.
Input: S = “12352”, K = 7
Output: 1
Explanation: Only string possible is by replacing S[3] and S[4] with K i.e., S = “1237”.
方法：给定的问题可以通过观察得出，对于S的某个子串，有两种类型的结果可以起作用，即它可以是x和y数相等的“ xy”序列，即， “ xyxy…xyxy”或者它可以是具有一个额外x的xy序列，即“ xyxy…xyxyx” ，其中x + y = K。
- 情况1： “ xyxy…xyxy”形式的字符串可以转换为字符串“ KK…KK” ，其中出现K (子字符串的长度)/ 2次。因此，将形成的不同字符串的数量为1 。
- 情况2：形式为“ xyxy…xyxyx”的字符串有一个额外的“ x”，并且可以转换为字符串“ KK…KKx” ，其中出现K (子字符串-1的长度)/ 2次。令该值为M。通过观察，可以看出在转换后的字符串x将会有(M +1)个可能的位置。
请按照以下步骤解决问题：
- 用1初始化变量ans ，用0初始化标志，用-1初始化start_index ，其中ans将存储直到索引i的答案，并且start_index将用于存储每个子串的起始索引(从该序列开始)。
- 遍历字符串S并执行以下操作：
- 如果任何相邻的数字S [i]和S [i + 1]的总和为K且flag设置为false ，则将flag设置为true并将start_index设置为i 。
- 如果标志为true且S [i]和S [i + 1]不是K ，请将标志设置为false ，并且如果(start_index – i + 1)为奇数，则将ans乘以(start_index – i + 1 – 1)情况2中所述的/ 2 +1 。
- 遍历字符串S后，如果flag为true并且(N – start_index)为奇数，则将ans乘以(N – start_index – 1)/ 2 + 1 。
- 完成上述步骤后，将ans打印为所需答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the desired number
// of strings
void countStrings(string s, int k)
{
// Store the count of strings
int ans = 1;
// Store the length of the string
int len = s.size();
// Initialize variable to indicate
// the start of the substring
int flag = 0;
// Store the starting index of
// the substring
int start_ind;
// Traverse the string
for (int i = 0; i < len - 1; i++) {
// If sum of adjacent characters
// is K mark the starting index
// and set flag to 1
if (s[i] - '0' + s[i + 1] - '0'
== k
&& flag == 0) {
flag = 1;
start_ind = i;
}
// If sum of adjacent characters
// is not K and the flag variable
// is set, end the substring here
if (flag == 1
&& s[i] - '0'
+ s[i + 1] - '0'
!= k) {
// Set flag to 0 denoting
// the end of substring
flag = 0;
// Check if the length of the
// substring formed is odd
if ((i - start_ind + 1) % 2 != 0)
// Update the answer
ans *= (i - start_ind + 1 - 1)
/ 2
+ 1;
}
}
// If flag is set and end of string
// is reached, mark the end of substring
if (flag == 1
&& (len - start_ind) % 2 != 0)
// Update the answer
ans *= (len - start_ind) / 2 + 1;
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
string S = "313";
int K = 4;
// Function Call
countStrings(S, K);
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the desired number
// of strings
static void countStrings(String s, int k)
{
// Store the count of strings
int ans = 1;
// Store the length of the string
int len = s.length();
// Initialize variable to indicate
// the start of the substring
int flag = 0;
// Store the starting index of
// the substring
int start_ind = 0;
// Traverse the string
for(int i = 0; i < len - 1; i++)
{
// If sum of adjacent characters
// is K mark the starting index
// and set flag to 1
if (s.charAt(i) - '0' +
s.charAt(i + 1) - '0' == k && flag == 0)
{
flag = 1;
start_ind = i;
}
// If sum of adjacent characters
// is not K and the flag variable
// is set, end the substring here
if (flag == 1 && s.charAt(i) - '0' +
s.charAt(i + 1) - '0' != k)
{
// Set flag to 0 denoting
// the end of substring
flag = 0;
// Check if the length of the
// substring formed is odd
if ((i - start_ind + 1) % 2 != 0)
// Update the answer
ans *= (i - start_ind + 1 - 1) / 2 + 1;
}
}
// If flag is set and end of string
// is reached, mark the end of substring
if (flag == 1 && (len - start_ind) % 2 != 0)
// Update the answer
ans *= (len - start_ind) / 2 + 1;
// Print the answer
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
String S = "313";
int K = 4;
// Function Call
countStrings(S, K);
}
}
// This code is contributed by jana_sayantanPython3
# Python program to implement
# the above approach
# Function to find the desired number
# of strings
def countStrings(s, k) :
# Store the count of strings
ans = 1
# Store the length of the string
lenn = len(s)
# Initialize variable to indicate
# the start of the substring
flag = 0
# Traverse the string
for i in range(lenn - 1):
# If sum of adjacent characters
# is K mark the starting index
# and set flag to 1
if (ord(s[i]) - ord('0') + ord(s[i + 1]) - ord('0')
== k
and flag == 0) :
flag = 1
start_ind = i
# If sum of adjacent characters
# is not K and the flag variable
# is set, end the substring here
if (flag == 1
and ord(s[i]) - ord('0')
+ ord(s[i + 1]) - ord('0')
!= k) :
# Set flag to 0 denoting
# the end of substring
flag = 0
# Check if the length of the
# substring formed is odd
if ((i - start_ind + 1) % 2 != 0) :
# Update the answer
ans *= (i - start_ind + 1 - 1) // 2 + 1
# If flag is set and end of string
# is reached, mark the end of substring
if (flag == 1
and (lenn - start_ind) % 2 != 0):
# Update the answer
ans *= (lenn - start_ind) // 2 + 1
# Prthe answer
print(ans)
# Driver Code
S = "313"
K = 4
# Function Call
countStrings(S, K)
# This code is contributed by susmitakundugoaldangaC#
// C# program for the above approach
using System;
class GFG{
// Function to find the desired number
// of strings
static void countStrings(String s, int k)
{
// Store the count of strings
int ans = 1;
// Store the length of the string
int len = s.Length;
// Initialize variable to indicate
// the start of the substring
int flag = 0;
// Store the starting index of
// the substring
int start_ind = 0;
// Traverse the string
for(int i = 0; i < len - 1; i++)
{
// If sum of adjacent characters
// is K mark the starting index
// and set flag to 1
if (s[i] - '0' +
s[i + 1] - '0' == k && flag == 0)
{
flag = 1;
start_ind = i;
}
// If sum of adjacent characters
// is not K and the flag variable
// is set, end the substring here
if (flag == 1 && s[i] - '0' +
s[i + 1] - '0' != k)
{
// Set flag to 0 denoting
// the end of substring
flag = 0;
// Check if the length of the
// substring formed is odd
if ((i - start_ind + 1) % 2 != 0)
// Update the answer
ans *= (i - start_ind + 1 - 1) / 2 + 1;
}
}
// If flag is set and end of string
// is reached, mark the end of substring
if (flag == 1 && (len - start_ind) % 2 != 0)
// Update the answer
ans *= (len - start_ind) / 2 + 1;
// Print the answer
Console.WriteLine(ans);
}
// Driver Code
public static void Main(String[] args)
{
String S = "313";
int K = 4;
// Function Call
countStrings(S, K);
}
}
// This code is contributed by Princi Singh输出：
2时间复杂度： O(N)
辅助空间： O(N)