给定大小为N的二进制字符串和整数K ，任务是对该字符串执行K个操作并打印最终的字符串：
- 如果操作号为奇数，则将字符串反向，
- 如果操作数为偶数，则对字符串进行补充。
例子：
Input: str = “1011”, K = 2
Output: 0010
After the first step, string will be reversed and becomes “1101”.
After the second step, the string will be complemented and becomes “0010”.
Input: str = “1001”, K = 4
Output: 1001
After all operation the string will remain same.
天真的方法：
遍历所有K个步，如果当前步是奇数，则执行相反的操作，否则对字符串补码。
高效方法：观察给定的操作模式：
- 如果一个字符串被反转偶数次，则获得原始字符串。
- 类似地，如果一个字符串被补偶数次，则获得原始字符串。
- 因此，这些运算仅取决于K的奇偶校验。
- 因此，我们将计算要执行的反向操作的数量。如果奇偶校验是奇数，那么我们将其反转。否则字符串将保持不变。
- 同样，我们将计算要执行的补码运算的数量。如果奇偶校验是奇数，那么我们将对其进行补充。否则字符串将保持不变。
下面是上述方法的实现：
C++
// C++ program to perform K operations upon
// the string and find the modified string
#include
using namespace std;
// Function to perform K operations upon
// the string and find modified string
string ReverseComplement(
string s, int n, int k)
{
// Number of reverse operations
int rev = (k + 1) / 2;
// Number of complement operations
int complment = k - rev;
// If rev is odd parity
if (rev % 2)
reverse(s.begin(), s.end());
// If complment is odd parity
if (complment % 2) {
for (int i = 0; i < n; i++) {
// Complementing each position
if (s[i] == '0')
s[i] = '1';
else
s[i] = '0';
}
}
// Return the modified string
return s;
}
// Driver Code
int main()
{
string str = "10011";
int k = 5;
int n = str.size();
// Function call
cout << ReverseComplement(str, n, k);
return 0;
} Java
// Java program to perform K operations upon
// the String and find the modified String
class GFG{
// Function to perform K operations upon
// the String and find modified String
static String ReverseComplement(
char []s, int n, int k)
{
// Number of reverse operations
int rev = (k + 1) / 2;
// Number of complement operations
int complment = k - rev;
// If rev is odd parity
if (rev % 2 == 1)
s = reverse(s);
// If complment is odd parity
if (complment % 2 == 1) {
for (int i = 0; i < n; i++) {
// Complementing each position
if (s[i] == '0')
s[i] = '1';
else
s[i] = '0';
}
}
// Return the modified String
return String.valueOf(s);
}
static char[] reverse(char a[]) {
int i, n = a.length;
char t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// Driver Code
public static void main(String[] args)
{
String str = "10011";
int k = 5;
int n = str.length();
// Function call
System.out.print(ReverseComplement(str.toCharArray(), n, k));
}
}
// This code is contributed by Rajput-JiPython3
# Python3 program to perform K operations upon
# the string and find the modified string
# Function to perform K operations upon
# the string and find modified string
def ReverseComplement(s,n,k):
# Number of reverse operations
rev = (k + 1) // 2
# Number of complement operations
complment = k - rev
# If rev is odd parity
if (rev % 2):
s = s[::-1]
# If complment is odd parity
if (complment % 2):
for i in range(n):
# Complementing each position
if (s[i] == '0'):
s[i] = '1'
else:
s[i] = '0'
# Return the modified string
return s
# Driver Code
if __name__ == '__main__':
str1 = "10011"
k = 5
n = len(str1)
# Function call
print(ReverseComplement(str1, n, k))
# This code is contributed by Surendra_GangwarC#
// C# program to perform K operations upon
// the String and find the modified String
using System;
class GFG{
// Function to perform K operations upon
// the String and find modified String
static string ReverseComplement(char []s,
int n, int k)
{
// Number of reverse operations
int rev = (k + 1) / 2;
// Number of complement operations
int complment = k - rev;
// If rev is odd parity
if (rev % 2 == 1)
s = reverse(s);
// If complment is odd parity
if (complment % 2 == 1)
{
for (int i = 0; i < n; i++)
{
// Complementing each position
if (s[i] == '0')
s[i] = '1';
else
s[i] = '0';
}
}
// Return the modified String
return (new string(s));
}
static char[] reverse(char[] a)
{
int i, n = a.Length;
char t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// Driver Code
public static void Main()
{
string str = "10011";
int k = 5;
int n = str.Length;
// Function call
Console.Write(ReverseComplement(str.ToCharArray(), n, k));
}
}
// This code is contributed by rutvik_56输出：
11001
时间复杂度： O(N)