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# Distributing M items in a circle of size N starting from K-th position

M items are to be delivered in a circle of size N. Find the position where the M-th item will be delivered if we start from a given position K. Note that items are distributed at adjacent positions starting from K.

Examples :

```Input : N = 5 // Size of circle
M = 2 // Number of items
K = 1 // Starting position
Output : 2
The first item will be given to 1st
position. Second (or last) item will
be delivered to 2nd position

Input : N = 5
M = 8
K = 2
Output : 4
The last item will be delivered to
4th position```

We check if the number of items to be distributed is greater than our remaining positions in current cycle of circle or not. If yes, then we simply return m + k – 1 (We distribute items in same cycle starting from k). Else we compute number of remaining items after completing current cycle and return mod of remaining items.

Below is the implementation of the above idea:

## C++

 `// C++ program to find the position where``// last item is delivered.``#include ``using` `namespace` `std;` `// n ==> Size of circle``// m ==> Number of items``// k ==> Initial position``int` `lastPosition(``int` `n, ``int` `m, ``int` `k)``{``    ``// n - k + 1 is number of positions``    ``// before we reach beginning of circle``    ``// If m is less than this value, then``    ``// we can simply return (m-1)th position``    ``if` `(m <= n - k + 1)``        ``return` `m + k - 1;` `    ``// Let us compute remaining items before``    ``// we reach beginning.``    ``m = m - (n - k + 1);` `    ``// We compute m % n to skip all complete``    ``// rounds. If we reach end, we return n``    ``// else we return m % n``    ``return` `(m % n == 0) ? n : (m % n);``}` `// Driver code``int` `main()``{``    ``int` `n = 5;``    ``int` `m = 8;``    ``int` `k = 2;``    ``cout << lastPosition(n, m, k);``    ``return` `0;``}`

## Java

 `// Java program to find the position where``// last item is delivered.``class` `GFG {` `    ``// n ==> Size of circle``    ``// m ==> Number of items``    ``// k ==> Initial position``    ``static` `int` `lastPosition(``int` `n, ``int` `m, ``int` `k)``    ``{` `        ``// n - k + 1 is number of positions``        ``// before we reach beginning of circle``        ``// If m is less than this value, then``        ``// we can simply return (m-1)th position``        ``if` `(m <= n - k + ``1``)``            ``return` `m + k - ``1``;` `        ``// Let us compute remaining items before``        ``// we reach beginning.``        ``m = m - (n - k + ``1``);` `        ``// We compute m % n to skip all complete``        ``// rounds. If we reach end, we return n``        ``// else we return m % n``        ``return` `(m % n == ``0``) ? n : (m % n);``    ``}` `    ``// Driver Program to test above function``    ``public` `static` `void` `main(String arg[])``    ``{``        ``int` `n = ``5``;``        ``int` `m = ``8``;``        ``int` `k = ``2``;``        ``System.out.print(lastPosition(n, m, k));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to find the position where``# last item is delivered.` `# n ==> Size of circle``# m ==> Number of items``# k ==> Initial position``def` `lastPosition(n, m, k):``    ` `    ``# n - k + 1 is number of positions``    ``# before we reach beginning of circle``    ``# If m is less than this value, then``    ``# we can simply return (m-1)th position``    ``if` `(m <``=` `n ``-` `k ``+` `1``):``       ``return` `m ``+` `k ``-` `1`` ` `    ``# Let us compute remaining items before``    ``# we reach beginning.``    ``m ``=` `m ``-` `(n ``-` `k ``+` `1``)`` ` `    ``# We compute m % n to skip all complete``    ``# rounds. If we reach end, we return n``    ``# else we return m % n``    ``if``(m ``%` `n ``=``=` `0``):``        ``return` `n``    ``else``:``        ``return` `m ``%` `n`` ` `# Driver code``n ``=` `5``m ``=` `8``k ``=` `2``print``(lastPosition(n, m, k))` `# This code is contributed by Sachin Bisht`

## C#

 `// C# program to find the position where``// last item is delivered.``using` `System;` `class` `GFG {` `    ``// n ==> Size of circle``    ``// m ==> Number of items``    ``// k ==> Initial position``    ``static` `int` `lastPosition(``int` `n, ``int` `m, ``int` `k)``    ``{` `        ``// n - k + 1 is number of positions``        ``// before we reach beginning of circle``        ``// If m is less than this value, then``        ``// we can simply return (m-1)th position``        ``if` `(m <= n - k + 1)``            ``return` `m + k - 1;` `        ``// Let us compute remaining items before``        ``// we reach beginning.``        ``m = m - (n - k + 1);` `        ``// We compute m % n to skip all complete``        ``// rounds. If we reach end, we return n``        ``// else we return m % n``        ``return` `(m % n == 0) ? n : (m % n);``    ``}` `    ``// Driver Program to test above function``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 5;``        ``int` `m = 8;``        ``int` `k = 2;` `        ``Console.WriteLine(lastPosition(n, m, k));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ` Size of circle``// m ==> Number of items``// k ==> Initial position` `function` `lastPosition(``\$n``, ``\$m``, ``\$k``)``{``    ``// n - k + 1 is number of``    ``// positions before we reach``    ``// beginning of circle.``    ``// If m is less than this value,``    ``// then we can simply return``    ``// (m-1)th position``    ``if` `(``\$m` `<= ``\$n` `- ``\$k` `+ 1)``    ``return` `\$m` `+ ``\$k` `- 1;` `    ``// Let us compute remaining items``    ``// before we reach beginning.``    ``\$m` `= ``\$m` `- (``\$n` `- ``\$k` `+ 1);` `    ``// We compute m % n to skip``    ``// all complete rounds. If we``    ``// reach end, we return n``    ``// else we return m % n``    ``return` `(``\$m` `% ``\$n` `== 0) ? ``\$n` `: (``\$m` `% ``\$n``);``}` `// Driver code``\$n` `= 5;``\$m` `= 8;``\$k` `= 2;``echo` `lastPosition(``\$n``, ``\$m``, ``\$k``);` `// This code is contributed by ajit``?>`

## Javascript

 ``

Output :

`4`

Time Complexity: O(1)
Auxiliary Space: O(1)

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