给定长度为N且整数K的二进制字符串S ，任务是检查是否有可能翻转K 0 ，以使所得的字符串不包含任何一对相邻的1 。如果可能的话，请打印“是” 。否则，打印“否” 。
Input: S = “01001001000”, K = 1
Output: Yes
Explanation:
Modifying S = “01001001000” to “01001001001” or modifying S = “01001001000″ to “01001001010″ satisfies the given condition.
Input: S = “000001000”, K = 4
Output: No
方法：遍历字符串并用两个相邻字符均为“ 0”的“ 1”替换那些“ 0” ，然后翻转“ 0”之一以计算翻转的所有可能位置，这样就可以了。 t影响原始字符串。请按照以下步骤解决问题：
- 初始化s变量，例如cnt为0，以存储可以翻转的位置数。
- 使用变量i遍历字符串并执行以下步骤：
- 如果当前字符为“ 1” ，则将i加2 。
- 除此以外：
- 如果i = 0，并且s [i + 1] =’0’：将cnt加1并将i加2 。否则，将i增加1 。
- 如果i =(N – 1)，并且s [i – 1] =’0’：将cnt递增1 ， i递增2 。否则，将i增加1 。
- 否则，如果s [i + 1] =’0’且s [i – 1] =’0’：将cnt递增1 ， i递增2。否则，将i递增1 。
- 完成上述步骤后，如果cnt的值至少为K，则打印“是” ，否则，打印“否” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if k '0's can
// be flipped such that the string
// does not contain any pair of adjacent '1's
void canPlace(string s, int n, int k)
{
// Store the count of flips
int cnt = 0;
// Variable to iterate the string
int i = 0;
// Iterate over characters
// of the string
while (i < n) {
// If the current character
// is '1', increment i by 2
if (s[i] == '1') {
i += 2;
}
// Otherwise, 3 cases arises
else {
// If the current index
// is the starting index
if (i == 0) {
// If next character is '0'
if (s[i + 1] == '0') {
cnt++;
i += 2;
}
// Increment i by 1
else
i++;
}
// If the current index
// is the last index
else if (i == n - 1) {
// If previous character is '0'
if (s[i - 1] == '0') {
cnt++;
i += 2;
}
else
i++;
}
// For remaining characters
else {
// If both the adjacent
// characters are '0'
if (s[i + 1] == '0' && s[i - 1] == '0') {
cnt++;
i += 2;
}
else
i++;
}
}
}
// If cnt is at least K, print "Yes"
if (cnt >= k) {
cout << "Yes";
}
// Otherwise, print "No"
else {
cout << "No";
}
}
// Driver Code
int main()
{
string S = "10001";
int K = 1;
int N = S.size();
canPlace(S, N, K);
return 0;
} Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG
{
// Function to check if k '0's can
// be flipped such that the string
// does not contain any pair of adjacent '1's
public static void canPlace(String s, int n, int k)
{
// Store the count of flips
int cnt = 0;
// Variable to iterate the string
int i = 0;
// Iterate over characters
// of the string
while (i < n)
{
// If the current character
// is '1', increment i by 2
if (s.charAt(i) == '1')
{
i += 2;
}
// Otherwise, 3 cases arises
else
{
// If the current index
// is the starting index
if (i == 0)
{
// If next character is '0'
if (s.charAt(i + 1) == '0')
{
cnt++;
i += 2;
}
// Increment i by 1
else
i++;
}
// If the current index
// is the last index
else if (i == n - 1)
{
// If previous character is '0'
if (s.charAt(i - 1) == '0')
{
cnt++;
i += 2;
}
else
i++;
}
// For remaining characters
else
{
// If both the adjacent
// characters are '0'
if (s.charAt(i + 1) == '0' && s.charAt(i - 1) == '0')
{
cnt++;
i += 2;
}
else
i++;
}
}
}
// If cnt is at least K, print "Yes"
if (cnt >= k)
{
System.out.println("Yes");
}
// Otherwise, print "No"
else
{
System.out.println("No");
}
}
// Driver code
public static void main (String[] args)
{
String S = "10001";
int K = 1;
int N = 5;
canPlace(S, N, K);
}
}
// This code is contributed by manupatharia.Python3
# Python3 program for the above approach
# Function to check if k '0's can
# be flipped such that the string
# does not contain any pair of adjacent '1's
def canPlace(s, n, k) :
# Store the count of flips
cnt = 0
# Variable to iterate the string
i = 0
# Iterate over characters
# of the string
while (i < n) :
# If the current character
# is '1', increment i by 2
if (s[i] == '1') :
i += 2
# Otherwise, 3 cases arises
else :
# If the current index
# is the starting index
if (i == 0) :
# If next character is '0'
if (s[i + 1] == '0') :
cnt += 1
i += 2
# Increment i by 1
else :
i += 1
# If the current index
# is the last index
elif (i == n - 1) :
# If previous character is '0'
if (s[i - 1] == '0') :
cnt += 1
i += 2
else :
i += 1
# For remaining characters
else :
# If both the adjacent
# characters are '0'
if (s[i + 1] == '0' and s[i - 1] == '0') :
cnt += 1
i += 2
else :
i += 1
# If cnt is at least K, print "Yes"
if (cnt >= k) :
print("Yes")
# Otherwise, print "No"
else :
print("No")
# Driver code
S = "10001"
K = 1
N = len(S)
canPlace(S, N, K)
# This code is contributed by divyeshrabadiya07.C#
// C# program for the above approach
using System;
class GFG
{
// Function to check if k '0's can
// be flipped such that the string
// does not contain any pair of adjacent '1's
static void canPlace(string s, int n, int k)
{
// Store the count of flips
int cnt = 0;
// Variable to iterate the string
int i = 0;
// Iterate over characters
// of the string
while (i < n) {
// If the current character
// is '1', increment i by 2
if (s[i] == '1') {
i += 2;
}
// Otherwise, 3 cases arises
else {
// If the current index
// is the starting index
if (i == 0) {
// If next character is '0'
if (s[i + 1] == '0') {
cnt++;
i += 2;
}
// Increment i by 1
else
i++;
}
// If the current index
// is the last index
else if (i == n - 1) {
// If previous character is '0'
if (s[i - 1] == '0') {
cnt++;
i += 2;
}
else
i++;
}
// For remaining characters
else {
// If both the adjacent
// characters are '0'
if (s[i + 1] == '0' && s[i - 1] == '0') {
cnt++;
i += 2;
}
else
i++;
}
}
}
// If cnt is at least K, print "Yes"
if (cnt >= k)
{
Console.WriteLine("Yes");
}
// Otherwise, print "No"
else
{
Console.WriteLine("No");
}
}
// Driver Code
public static void Main(String[] args)
{
string S = "10001";
int K = 1;
int N = S.Length;
canPlace(S, N, K);
}
}
// This code is contributed by susmitakundugoaldanga.输出：
Yes时间复杂度： O(n)
辅助空间： O(1)