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# Steps required to visit M points in order on a circular ring of N points

Given an integer ‘n’, consider a circular ring containing ‘n’ points numbered from ‘1’ to ‘n’ such that you can move in the following way :

1 -> 2 -> 3 -> ….. -> n -> 1 -> 2 -> 3 -> ……

Also, given an array of integers (of size ‘m’), the task is to find the number of steps it’ll take to get to every point in the array in order starting at ‘1’

Examples :

```Input: n = 3, m = 3, arr[] = {2, 1, 2}
Output: 4
The sequence followed is 1->2->3->1->2

Input: n = 2, m = 1, arr[] = {2}
Output: 1
The sequence followed is 1->2```

Approach: Let’s denote the current position by cur and the next position by nxt

This gives us 2 cases:

1. If cur is smaller than nxt, you can move to it in nxt – cur steps.
2. Otherwise, you first need to reach the point n in n – cur steps, and then you can move to nxt in cur steps.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to count the steps required``int` `findSteps(``int` `n, ``int` `m, ``int` `a[])``{` `    ``// Start at 1``    ``int` `cur = 1;` `    ``// Initialize steps``    ``int` `steps = 0;``    ``for` `(``int` `i = 0; i < m; i++) {` `        ``// If nxt is greater than cur``        ``if` `(a[i] >= cur)``            ``steps += (a[i] - cur);``        ``else``            ``steps += (n - cur + a[i]);` `        ``// Now we are at a[i]``        ``cur = a[i];``    ``}``    ``return` `steps;``}` `// Driver code``int` `main()``{``    ``int` `n = 3, m = 3;``    ``int` `a[] = { 2, 1, 2 };``    ``cout << findSteps(n, m, a);``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `// Function to count the steps required``static` `int` `findSteps(``int` `n, ``int` `m,``                     ``int` `a[])``{` `    ``// Start at 1``    ``int` `cur = ``1``;` `    ``// Initialize steps``    ``int` `steps = ``0``;``    ``for` `(``int` `i = ``0``; i < m; i++)``    ``{` `        ``// If nxt is greater than cur``        ``if` `(a[i] >= cur)``            ``steps += (a[i] - cur);``        ``else``            ``steps += (n - cur + a[i]);` `        ``// Now we are at a[i]``        ``cur = a[i];``    ``}``    ``return` `steps;``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `n = ``3``, m = ``3``;``    ``int` `a[] = { ``2``, ``1``, ``2` `};``    ``System.out.println(findSteps(n, m, a));``}``}` `// This code is contributed by ihritik`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``// Function to count the``// steps required``static` `int` `findSteps(``int` `n,``                     ``int` `m, ``int` `[]a)``{` `    ``// Start at 1``    ``int` `cur = 1;` `    ``// Initialize steps``    ``int` `steps = 0;``    ``for` `(``int` `i = 0; i < m; i++)``    ``{` `        ``// If nxt is greater than cur``        ``if` `(a[i] >= cur)``            ``steps += (a[i] - cur);``        ``else``            ``steps += (n - cur + a[i]);` `        ``// Now we are at a[i]``        ``cur = a[i];``    ``}``    ``return` `steps;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 3, m = 3;``    ``int` `[]a = { 2, 1, 2 };``    ``Console.WriteLine(findSteps(n, m, a));``}``}` `// This code is contributed by ihritik`

## Python3

 `# Python3 implementation of the approach` `# Function to count the steps required``def` `findSteps(n, m, a):` `    ``# Start at 1``    ``cur ``=` `1` `    ``# Initialize steps``    ``steps ``=` `0``    ``for` `i ``in` `range``(``0``, m):` `        ``# If nxt is greater than cur``        ``if` `(a[i] >``=` `cur):``            ``steps ``+``=` `(a[i] ``-` `cur)``        ``else``:``            ``steps ``+``=` `(n ``-` `cur ``+` `a[i])` `        ``# Now we are at a[i]``        ``cur ``=` `a[i]``    ` `    ``return` `steps` `# Driver code``n ``=` `3``m ``=` `3``a ``=` `[``2``, ``1``, ``2` `]``print``(findSteps(n, m, a))` `# This code is contributed by ihritik`

## PHP

 `= ``\$cur``)``            ``\$steps` `+= (``\$a``[``\$i``] - ``\$cur``);``        ``else``            ``\$steps` `+= (``\$n` `- ``\$cur` `+ ``\$a``[``\$i``]);` `        ``// Now we are at a[i]``        ``\$cur` `= ``\$a``[``\$i``];``    ``}``    ``return` `\$steps``;``}` `// Driver code``\$n` `= 3;``\$m` `= 3;``\$a` `= ``array``(2, 1, 2 );``echo` `findSteps(``\$n``, ``\$m``, ``\$a``);` `// This code is contributed by ihritik``?>`

## Javascript

 ``

Output

`4`

Complexity Analysis:

• Time Complexity: O(M), since the loop runs for M times.
• Auxiliary Space: O(1), since no extra space has been taken.

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