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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 Z``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 = 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``int` `main()``{``    ``int` `n = 4, a = 1, b = 2;``    ``cout << findMinimumZ(n, a, b);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``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``    ``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 Z``def` `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 code``if` `__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 approach``using` `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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