给定一个大小为N个元素的二进制圆数组和两个表示圆数组中索引的正整数x和y 。任务是检查从索引x到索引y的顺时针或逆时针路径，我们将遇到最小数目的位翻转。如果计数相等，则输出“顺时针”或“逆时针”以及最小位翻转的值。
例子：
Input : arr[] = { 0, 0, 0, 1, 1, 0 }
x = 0, y = 5
Output : Anti-clockwise 0
The path 0 -> 1 -> 2 -> 3 -> 4 -> 5, we have only 1 value change i.e from index 2 to 3.
The path 0 -> 5 have 0 value change.
So, the answer is Anti-clockwise 0.
Input : s = { 1, 1, 0, 1, 1 }
x = 2, y = 0
Output : Clockwise 1
想法是通过顺时针旋转一次并存储count1，然后逆时针旋转并存储count2进行检查。然后通过比较count1和count2输出。
如何顺时针或逆时针行驶?
在x> y的数组中将很难沿顺时针方向移动，而在y> x的逆时针方向上将很难相同。因此，我们将给定的二进制数组存储在字符串“ S”中。为了使其呈圆形，我们将S附加到S上，即S = S +S。我们将在x和y中进行调整以顺时针或逆时针移动。
现在，如果y> x并顺时针旋转，则很容易从x迭代到y并计算翻转位的数量。
如果y> x并逆时针旋转，我们将添加| S |。到x，然后从y迭代到x，并计算翻转位的数量。
现在，如果x> y，我们将交换x和y并使用上述方法计算答案。然后输出与结果相反的结果。
要计算翻转位的数量，只需存储索引的当前位，然后检查下一个索引是否与当前位相同。如果是，则不执行其他任何操作，将当前位更改为下一个索引的位，并将最小位增加1。
以下是此方法的实现：
C++
// CPP program to find direction with minimum flips
#include
using namespace std;
// finding which path have minimum flip bit and
// the minimum flip bits
void minimumFlip(string s, int x, int y)
{
// concatenating given strin to itself,
// to make it circular
s = s + s;
// check x is greater than y.
// marking if output need to
// be opposite.
bool isOpposite = false;
if (x > y) {
swap(x, y);
isOpposite = true;
}
// iterate Clockwise
int valClockwise = 0;
char cur = s[x];
for (int i = x; i <= y; i++) {
// if current bit is not equal
// to next index bit.
if (s[i] != cur) {
cur = s[i];
valClockwise++;
}
}
// iterate Anti-Clockwise
int valAnticlockwise = 0;
cur = s[y];
x += s.length();
for (int i = y; i <= x; i++) {
// if current bit is not equal
// to next index bit.
if (s[i] != cur) {
cur = s[i];
valAnticlockwise++;
}
}
// Finding whether Clockwise or Anti-clockwise
// path take minimum flip.
if (valClockwise <= valAnticlockwise) {
if (!isOpposite)
cout << "Clockwise "
<< valClockwise << endl;
else
cout << "Anti-clockwise "
<< valAnticlockwise << endl;
}
else {
if (!isOpposite)
cout << "Anti-clockwise "
<< valAnticlockwise << endl;
else
cout << "Clockwise "
<< valClockwise << endl;
}
}
// Driven Program
int main()
{
int x = 0, y = 8;
string s = "000110";
minimumFlip(s, x, y);
return 0;
} Java
// Java program to find direction
// with minimum flips
class GFG
{
// finding which path have
// minimum flip bit and
// the minimum flip bits
static void minimumFlip(String s,
int x, int y)
{
// concatenating given strin to
// itself, to make it circular
s = s + s;
// check x is greater than y.
// marking if output need to
// be opposite.
boolean isOpposite = false;
if (x > y)
{
swap(x, y);
isOpposite = true;
}
// iterate Clockwise
int valClockwise = 0;
char cur = s.charAt(x);
for (int i = x; i <= y; i++)
{
// if current bit is not equal
// to next index bit.
if (s.charAt(i) != cur)
{
cur = s.charAt(i);
valClockwise++;
}
}
// iterate Anti-Clockwise
int valAnticlockwise = 0;
cur = s.charAt(y);
x += s.length();
for (int i = y; i < x; i++)
{
// if current bit is not equal
// to next index bit.
if (s.charAt(i) != cur)
{
cur = s.charAt(i);
valAnticlockwise++;
}
}
// Finding whether Clockwise
// or Anti-clockwise path
// take minimum flip.
if (valClockwise <= valAnticlockwise)
{
if (!isOpposite)
{
System.out.println("Clockwise " +
valClockwise);
}
else
{
System.out.println("Anti-clockwise " +
valAnticlockwise);
}
}
else if (!isOpposite)
{
System.out.println("Anti-clockwise " +
valAnticlockwise);
}
else
{
System.out.println("Clockwise " +
valClockwise);
}
}
static void swap(int a, int b)
{
int c = a;
a = b;
b = c;
}
// Driver code
public static void main(String[] args)
{
int x = 0, y = 8;
String s = "000110";
minimumFlip(s, x, y);
}
}
// This code is contributed by 29AjayKumarPython3
# Python 3 program to find direction
# with minimum flips
# finding which path have minimum flip bit
# and the minimum flip bits
def minimumFlip(s, x, y):
# concatenating given string to itself,
# to make it circular
s = s + s
# check x is greater than y.
# marking if output need to
# be opposite.
isOpposite = False
if (x > y):
temp = y
y = x;
x = temp
isOpposite = True
# iterate Clockwise
valClockwise = 0
cur = s[x]
for i in range(x, y + 1, 1):
# if current bit is not equal
# to next index bit.
if (s[i] != cur):
cur = s[i]
valClockwise += 1
# iterate Anti-Clockwise
valAnticlockwise = 0
cur = s[y]
x += len(s) - 1
for i in range(y, x + 1, 1):
# if current bit is not equal
# to next index bit.
if (s[i] != cur):
cur = s[i]
valAnticlockwise += 1
# Finding whether Clockwise or Anti-clockwise
# path take minimum flip.
if (valClockwise <= valAnticlockwise):
if (isOpposite == False):
print("Clockwise", valClockwise)
else:
print("Anti-clockwise",
valAnticlockwise)
else:
if (isOpposite == False):
print("Anti-clockwise",
valAnticlockwise)
else:
print("Clockwise", valClockwise)
# Driver Code
if __name__ == '__main__':
x = 0
y = 8
s = "000110"
minimumFlip(s, x, y)
# This code is contributed by
# Surendra_GangwarC#
// C# program to find direction
// with minimum flips
using System;
class GFG
{
// finding which path have
// minimum flip bit and
// the minimum flip bits
static void minimumFlip(String s,
int x, int y)
{
// concatenating given strin to
// itself, to make it circular
s = s + s;
// check x is greater than y.
// marking if output need to
// be opposite.
bool isOpposite = false;
if (x > y)
{
swap(x, y);
isOpposite = true;
}
// iterate Clockwise
int valClockwise = 0;
char cur = s[x];
for (int i = x; i <= y; i++)
{
// if current bit is not equal
// to next index bit.
if (s[i] != cur)
{
cur = s[i];
valClockwise++;
}
}
// iterate Anti-Clockwise
int valAnticlockwise = 0;
cur = s[y];
x += s.Length;
for (int i = y; i < x; i++)
{
// if current bit is not equal
// to next index bit.
if (s[i] != cur)
{
cur = s[i];
valAnticlockwise++;
}
}
// Finding whether Clockwise
// or Anti-clockwise path
// take minimum flip.
if (valClockwise <= valAnticlockwise)
{
if (!isOpposite)
{
Console.WriteLine("Clockwise " +
valClockwise);
}
else
{
Console.WriteLine("Anti-clockwise " +
valAnticlockwise);
}
}
else if (!isOpposite)
{
Console.WriteLine("Anti-clockwise " +
valAnticlockwise);
}
else
{
Console.WriteLine("Clockwise " +
valClockwise);
}
}
static void swap(int a, int b)
{
int c = a;
a = b;
b = c;
}
// Driver code
public static void Main(String[] args)
{
int x = 0, y = 8;
String s = "000110";
minimumFlip(s, x, y);
}
}
// This code contributed by Rajput-Ji输出：
Clockwise 2