Minimum bit changes in Binary Circular array to reach a index

Given a Binary Circular Array of size N elements and two positive integers x and y indicating the indices in the circular array. The task is check which path, clockwise or anti-clockwise, from index x to index y, we face the minimum number bit flips. Output “Clockwise” or “Anti-clockwise” and the value of minimum bit flip, in case of equal count output “Clockwise”.

Examples:

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

The idea is to check by going once Clockwise and store the count1 and then going anti-clockwise and store the count2. Then output by comparing count1 and count2.

How to travel clockwise or anticlockwise?
It will be hard to travel clockwise in the array where x > y and same in case of anticlockwise where y > x. So, we will store the given binary array in the string “S”. And to make it circular, we will append S to S i.e S = S + S. We will make the adjustment in x and y to travel clockwise or anticlockwise.
Now, if y > x and to go clockwise, it will be easy to iterate from x to y and calculate the number of flip bits.
If y > x and to go anti-clockwise, we will add |S| to x then iterate from y to x and calculate the number of flip bits.

Now, if x > y, we will swap x and y and calculate the answer using above approach. Then output the opposite of the result .



To calculate the number of flip bits, just store the current bit of index and check if next index have the same bit as current. If yes then do nothing else change the current bit to the bit of the next index and increment minimum bit by 1.

Below is the implementation of this approach:

C++

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// CPP program to find direction with minimum flips
#include <bits/stdc++.h>
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;
}

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Java

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// 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 29AjayKumar

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Output:

Clockwise 2


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