# Array element moved by k using single moves

Given a list of n integers containing numbers 1-n in a shuffled way and a integer K. N people are standing in a queue to play badminton. At first, the first two players in the queue play a game. Then the loser goes to the end of the queue, and the one who wins plays with the next person from the line, and so on. They play until someone wins k games consecutively. This player becomes the winner.

**Examples :**

Input: arr[] = {2, 1, 3, 4, 5} k = 2 Output: 5 Explanation: 2 plays with 1, 1 goes to end of queue. 2 plays with 3, 3 wins, 2 goes to end of queue. 3 plays with 4, so 3 goes to the end of the queue. 5 plays with everyone and wins as it is the largest of all elements. Input: arr[] = {3, 1, 2} k = 2 Output: 3 Explanation : 3 plays with 1. 3 wins. 1 goes to the end of the line. 3 plays with 2. 3 wins. 3 wins twice in a row.

A **naive approach** is to run two nested for loops and check for every element which one is more from i to n being the first loop and the second being from i+1 to n and then from 0 to n-1 and count the number of continuous smaller elements and get the answer.

This will not be efficient enough as it takes **O(n*n)** .

An **efficient approach** will be to run a loop from 1 to n and keep track of best (or maximum element) so far and number of smaller elements than this maximum. If current best loose, initialize the greater value to the best and the count to 1, as the winner won 1 time already. If at any step it has won k times, you get your answer. But if k >= n-1, then the maximum number will be the only answer as it will the most number of times being the greatest. If while iterating you don’t find any player that has won k times, then the maximum number which is in the list will always be our answer.

Below is the implementation to the above approach

## C++

`// CPP program to find winner of game ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `int` `winner(` `int` `a[], ` `int` `n, ` `int` `k) ` `{ ` ` ` `// if the number of steps is more then ` ` ` `// n-1, ` ` ` `if` `(k >= n - 1) ` ` ` `return` `n; ` ` ` ` ` `// initially the best is 0 and no of ` ` ` `// wins is 0. ` ` ` `int` `best = 0, times = 0; ` ` ` ` ` `// traverse through all the numbers ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// if the value of array is more ` ` ` `// then that of previous best ` ` ` `if` `(a[i] > best) { ` ` ` ` ` `// best is replaced by a[i] ` ` ` `best = a[i]; ` ` ` ` ` `// if not the first index ` ` ` `if` `(i) ` ` ` `times = 1; ` `// no of wins is 1 now ` ` ` `} ` ` ` ` ` `else` ` ` `times += 1; ` `// if it wins ` ` ` ` ` `// if any position has more then k wins ` ` ` `// then return ` ` ` `if` `(times >= k) ` ` ` `return` `best; ` ` ` `} ` ` ` ` ` `// Maximum element will be winner because ` ` ` `// we move smaller element at end and repeat ` ` ` `// the process. ` ` ` `return` `best; ` `} ` ` ` `// driver program to test the above function ` `int` `main() ` `{ ` ` ` `int` `a[] = { 2, 1, 3, 4, 5 }; ` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `int` `k = 2; ` ` ` `cout << winner(a, n, k); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java program to find winner of game ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `winner(` `int` `a[], ` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` `// if the number of steps is more then ` ` ` `// n-1, ` ` ` `if` `(k >= n - ` `1` `) ` ` ` `return` `n; ` ` ` ` ` `// initially the best is 0 and no of ` ` ` `// wins is 0. ` ` ` `int` `best = ` `0` `, times = ` `0` `; ` ` ` ` ` `// traverse through all the numbers ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` ` ` `// if the value of array is more ` ` ` `// then that of previous best ` ` ` `if` `(a[i] > best) { ` ` ` ` ` `// best is replaced by a[i] ` ` ` `best = a[i]; ` ` ` ` ` `// if not the first index ` ` ` `if` `(i == ` `1` `) ` ` ` ` ` `// no of wins is 1 now ` ` ` `times = ` `1` `; ` ` ` `} ` ` ` ` ` `else` ` ` ` ` `// if it wins ` ` ` `times += ` `1` `; ` ` ` ` ` `// if any position has more then ` ` ` `// k wins then return ` ` ` `if` `(times >= k) ` ` ` `return` `best; ` ` ` `} ` ` ` ` ` `// Maximum element will be winner ` ` ` `// because we move smaller element ` ` ` `// at end and repeat the process. ` ` ` `return` `best; ` ` ` `} ` ` ` ` ` `// driver program to test the above function ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `a[] = { ` `2` `, ` `1` `, ` `3` `, ` `4` `, ` `5` `}; ` ` ` `int` `n = a.length; ` ` ` `int` `k = ` `2` `; ` ` ` ` ` `System.out.println(winner(a, n, k)); ` ` ` `} ` `} ` ` ` `/*This code is contributed by Nikita Tiwari.*/` |

*chevron_right*

*filter_none*

## Python3

`# Python code to find ` `# winner of game ` ` ` `# function to find the winner ` `def` `winner( a, n, k): ` ` ` ` ` `# if the number of steps is ` ` ` `# more then n-1 ` ` ` `if` `k >` `=` `n ` `-` `1` `: ` ` ` `return` `n ` ` ` ` ` `# initially the best is 0 ` ` ` `# and no of wins is 0. ` ` ` `best ` `=` `0` ` ` `times ` `=` `0` ` ` ` ` `# traverse through all the numbers ` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# if the value of array is more ` ` ` `# then that of previous best ` ` ` `if` `a[i] > best: ` ` ` ` ` `# best is replaced by a[i] ` ` ` `best ` `=` `a[i] ` ` ` ` ` `# if not the first index ` ` ` `if` `i ` `=` `=` `True` `: ` ` ` ` ` `# no of wins is 1 now ` ` ` `times ` `=` `1` ` ` `else` `: ` ` ` `times ` `+` `=` `1` `# if it wins ` ` ` ` ` `# if any position has more ` ` ` `# then k wins then return ` ` ` `if` `times >` `=` `k: ` ` ` `return` `best ` ` ` ` ` `# Maximum element will be winner ` ` ` `# because we move smaller element ` ` ` `# at end and repeat the process. ` ` ` `return` `best ` ` ` ` ` `# driver code ` `a ` `=` `[ ` `2` `, ` `1` `, ` `3` `, ` `4` `, ` `5` `] ` `n ` `=` `len` `(a) ` `k ` `=` `2` `print` `(winner(a, n, k)) ` ` ` `# This code is contributed by "Abhishek Sharma 44" ` |

*chevron_right*

*filter_none*

## C#

`// C# program to find winner of game ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `winner(` `int` `[] a, ` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` ` ` `// if the number of steps is more ` ` ` `// then n-1, ` ` ` `if` `(k >= n - 1) ` ` ` `return` `n; ` ` ` ` ` `// initially the best is 0 and no of ` ` ` `// wins is 0. ` ` ` `int` `best = 0, times = 0; ` ` ` ` ` `// traverse through all the numbers ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// if the value of array is more ` ` ` `// then that of previous best ` ` ` `if` `(a[i] > best) { ` ` ` ` ` `// best is replaced by a[i] ` ` ` `best = a[i]; ` ` ` ` ` `// if not the first index ` ` ` `if` `(i == 1) ` ` ` ` ` `// no of wins is 1 now ` ` ` `times = 1; ` ` ` `} ` ` ` ` ` `else` ` ` ` ` `// if it wins ` ` ` `times += 1; ` ` ` ` ` `// if any position has more ` ` ` `// then k wins then return ` ` ` `if` `(times >= k) ` ` ` `return` `best; ` ` ` `} ` ` ` ` ` `// Maximum element will be winner ` ` ` `// because we move smaller element ` ` ` `// at end and repeat the process. ` ` ` `return` `best; ` ` ` `} ` ` ` ` ` `// driver program to test the above ` ` ` `// function ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[] a = { 2, 1, 3, 4, 5 }; ` ` ` `int` `n = a.Length; ` ` ` `int` `k = 2; ` ` ` ` ` `Console.WriteLine(winner(a, n, k)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to find winner of game ` ` ` `function` `winner(` `$a` `, ` `$n` `, ` `$k` `) ` `{ ` ` ` `// if the number of steps ` ` ` `// is more then n-1, ` ` ` `if` `(` `$k` `>= ` `$n` `- 1) ` ` ` `return` `$n` `; ` ` ` ` ` `// initially the best is 0 ` ` ` `// and no. of wins is 0. ` ` ` `$best` `= 0; ` `$times` `= 0; ` ` ` ` ` `// traverse through all the numbers ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` ` ` `// if the value of array is more ` ` ` `// then that of previous best ` ` ` `if` `(` `$a` `[` `$i` `] > ` `$best` `) ` ` ` `{ ` ` ` ` ` `// best is replaced by a[i] ` ` ` `$best` `= ` `$a` `[` `$i` `]; ` ` ` ` ` `// if not the first index ` ` ` `if` `(` `$i` `) ` ` ` ` ` `// no of wins is 1 now ` ` ` `$times` `= 1; ` ` ` `} ` ` ` ` ` `else` ` ` ` ` `// if it wins ` ` ` `$times` `+= 1; ` ` ` ` ` `// if any position has more then ` ` ` `// k wins then return ` ` ` `if` `(` `$times` `>= ` `$k` `) ` ` ` `return` `$best` `; ` ` ` `} ` ` ` ` ` `// Maximum element will be winner ` ` ` `// because we move smaller element ` ` ` `// at end and repeat the process. ` ` ` `return` `$best` `; ` `} ` ` ` `// Driver Code ` `$a` `= ` `array` `( 2, 1, 3, 4, 5 ); ` `$n` `= sizeof(` `$a` `); ` `$k` `= 2; ` `echo` `(winner(` `$a` `, ` `$n` `, ` `$k` `)); ` ` ` `// This code is contributed by Ajit. ` `?> ` |

*chevron_right*

*filter_none*

**Output :**

5

**Time complexity** : O(n)

## Recommended Posts:

- Check whether it is possible to make both arrays equal by modifying a single element
- Collect maximum points in an array with k moves
- Find the index of the left pointer after possible moves in the array
- Find the minimum number of moves to reach end of the array
- Minimum number of moves to make a binary array K periodic
- Count minimum number of "move-to-front" moves to sort an array
- Find single in an array of 2n+1 integer elements
- Reduce the array to a single integer with the given operation
- Move all zeroes to end of array | Set-2 (Using single traversal)
- Closest greater element for every array element from another array
- Replace each element by the difference of the total size of the array and frequency of that element
- Range Query on array whose each element is XOR of index value and previous element
- Sum of (maximum element - minimum element) for all the subsets of an array.
- Find last element after deleting every second element in array of n integers
- Replace every array element by Bitwise Xor of previous and next element

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.