Find the minimum sum of distance to A and B from any integer point in a ring of size N

• Last Updated : 10 May, 2021

Given a circular ring which has marking from 1 to N. Given two numbers A and B, you can stand at any place(say X) and count the total sum of the distance(say Z i.e., distance from X to A + distance from X to B). The task is to choose X in such a way that Z is minimized. Print the value of Z thus obtained. Note that X cannot neither be equal to A nor be equal to B.
Examples:

Input: N = 6, A = 2, B = 4
Output:
Choose X as 3, so that distance from X to A is 1, and distance from X to B is 1.
Input: N = 4, A = 1, B = 2
Output:
Choose X as 3 or 4, both of them gives distance as 3.

Approach: There are two paths between positions A and B on the circle, one in clockwise direction and another in an anti-clockwise. An optimal value for Z is to choose X as any point on the minimum path between A and B then Z will be equal to the minimum distance between the positions except for the case when both the positions are adjacent to each other i.e. the minimum distance is 1. In that case, X cannot be chosen as the point between them as it must be different from both A and B and the result will be 3.
Below is the implementation of the above approach:

C++

 // C++ implementation of the approach#include using namespace std; // Function to return the minimum value of Zint findMinimumZ(int n, int a, int b){     // Change elements such that a < b    if (a > b)        swap(a, b);     // Distance from (a to b)    int distClock = b - a;     // Distance from (1 to a) + (b to n)    int distAntiClock = (a - 1) + (n - b + 1);     // Minimum distance between a and b    int minDist = min(distClock, distAntiClock);     // If both the positions are    // adjacent on the circle    if (minDist == 1)        return 3;     // Return the minimum Z possible    return minDist;} // Driver codeint main(){    int n = 4, a = 1, b = 2;    cout << findMinimumZ(n, a, b);     return 0;}

Java

 // Java implementation of the approachclass GFG{     // Function to return the minimum value of Z    static int findMinimumZ(int n, int a, int b)    {         // Change elements such that a < b        if (a > b)        {            swap(a, b);        }         // Distance from (a to b)        int distClock = b - a;         // Distance from (1 to a) + (b to n)        int distAntiClock = (a - 1) + (n - b + 1);         // Minimum distance between a and b        int minDist = Math.min(distClock, distAntiClock);         // If both the positions are        // adjacent on the circle        if (minDist == 1)        {            return 3;        }         // Return the minimum Z possible        return minDist;    }     private static void swap(int x, int y)    {        int temp = x;        x = y;        y = temp;    }     // Driver code    public static void main(String[] args)    {        int n = 4, a = 1, b = 2;        System.out.println(findMinimumZ(n, a, b));    }} /* This code contributed by PrinciRaj1992 */

Python3

 # Python 3 implementation of the approach# Function to return the minimum value of Zdef findMinimumZ(n, a, b):         # Change elements such that a < b    if (a > b):        temp = a        a = b        b = temp     # Distance from (a to b)    distClock = b - a     # Distance from (1 to a) + (b to n)    distAntiClock = (a - 1) + (n - b + 1)     # Minimum distance between a and b    minDist = min(distClock, distAntiClock)     # If both the positions are    # adjacent on the circle    if (minDist == 1):        return 3     # Return the minimum Z possible    return minDist # Driver codeif __name__ == '__main__':    n = 4    a = 1    b = 2    print(findMinimumZ(n, a, b))     # This code is contributed by# Surendra_Gangwar

C#

 // C# implementation of the approachusing System; class GFG{         // Function to return the minimum value of Z    static int findMinimumZ(int n, int a, int b)    {         // Change elements such that a < b        if (a > b)        {            swap(a, b);        }         // Distance from (a to b)        int distClock = b - a;         // Distance from (1 to a) + (b to n)        int distAntiClock = (a - 1) + (n - b + 1);         // Minimum distance between a and b        int minDist = Math.Min(distClock, distAntiClock);         // If both the positions are        // adjacent on the circle        if (minDist == 1)        {            return 3;        }         // Return the minimum Z possible        return minDist;    }     private static void swap(int x, int y)    {        int temp = x;        x = y;        y = temp;    }     // Driver code    static public void Main ()    {        int n = 4, a = 1, b = 2;        Console.WriteLine(findMinimumZ(n, a, b));    } } /* This code contributed by ajit*/

PHP

 \$b)                 \$a = \$a ^ \$b;        \$b = \$a ^ \$b;        \$a = \$a ^ \$b;     // Distance from (a to b)    \$distClock = \$b - \$a;     // Distance from (1 to a) + (b to n)    \$distAntiClock = (\$a - 1) + (\$n - \$b + 1);     // Minimum distance between a and b    \$minDist = min(\$distClock, \$distAntiClock);     // If both the positions are    // adjacent on the circle    if (\$minDist == 1)        return 3;     // Return the minimum Z possible    return \$minDist;} // Driver code \$n = 4;\$a = 1;\$b = 2;echo findMinimumZ(\$n, \$a, \$b);  // This code is contributed by akt_mit?>

Javascript


Output:
3

Time Complexity: O(1 )

Auxiliary Space: O(1)

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