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# Largest lexicographic array with at-most K consecutive swaps

Given an array arr[], find the lexicographically largest array that can be obtained by performing at-most k consecutive swaps.

Examples :

```Input : arr[] = {3, 5, 4, 1, 2}
k = 3
Output : 5, 4, 3, 2, 1
Explanation :     Array given : 3 5 4 1 2
After swap 1 : 5 3 4 1 2
After swap 2 : 5 4 3 1 2
After swap 3 : 5 4 3 2 1
Input : arr[] = {3, 5, 1, 2, 1}
k = 3
Output : 5, 3, 2, 1, 1```

Brute Force Approach : Generate all permutation of the array and then pick the one which satisfies the condition of at most K swaps. The time complexity of this approach is O(n!).

Optimized Approach : In this greedy approach, first find the largest element present in the array which is greater than(if the 1st position element is not the greatest) the 1st position and which can be placed at the 1st position with at-most K swaps. After finding that element, note its index. Then, swap elements of the array and update K value. Apply this procedure for other positions till k is non-zero or array becomes lexicographically largest.

Below is the implementation of above approach :

## C++

 `// C++ program to find lexicographically``// maximum value after k swaps.``#include ``using` `namespace` `std;` `// Function which modifies the array``void` `KSwapMaximum(``int` `arr[], ``int` `n, ``int` `k)``{``    ``for` `(``int` `i = 0; i < n - 1 && k > 0; ++i) {` `        ``// Here, indexPosition is set where we``        ``// want to put the current largest integer``        ``int` `indexPosition = i;``        ``for` `(``int` `j = i + 1; j < n; ++j) {` `            ``// If we exceed the Max swaps``            ``// then break the loop``            ``if` `(k <= j - i)``                ``break``;` `            ``// Find the maximum value from i+1 to``            ``// max k or n which will replace``            ``// arr[indexPosition]``            ``if` `(arr[j] > arr[indexPosition])``                ``indexPosition = j;``        ``}` `        ``// Swap the elements from Maximum indexPosition``        ``// we found till now to the ith index``        ``for` `(``int` `j = indexPosition; j > i; --j)``            ``swap(arr[j], arr[j - 1]);` `        ``// Updates k after swapping indexPosition-i``        ``// elements``        ``k -= indexPosition - i;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 5, 4, 1, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `k = 3;` `    ``KSwapMaximum(arr, n, k);` `    ``// Print the final Array``    ``for` `(``int` `i = 0; i < n; ++i)``        ``cout << arr[i] << ``" "``;``}`

## Java

 `// Java program to find``// lexicographically``// maximum value after``// k swaps.``import` `java.io.*;` `class` `GFG``{``    ``static` `void` `SwapInts(``int` `array[],``                         ``int` `position1,``                         ``int` `position2)``    ``{``        ``// Swaps elements``        ``// in an array.``        ` `        ``// Copy the first``        ``// position's element``        ``int` `temp = array[position1];``        ` `        ``// Assign to the``        ``// second element``        ``array[position1] = array[position2];``        ` `        ``// Assign to the``        ``// first element``        ``array[position2] = temp;``    ``}``    ` `    ``// Function which``    ``// modifies the array``    ``static` `void` `KSwapMaximum(``int` `[]arr,``                             ``int` `n, ``int` `k)``    ``{``        ``for` `(``int` `i = ``0``;``                 ``i < n - ``1` `&& k > ``0``; ++i)``        ``{``    ` `            ``// Here, indexPosition``            ``// is set where we want to``            ``// put the current largest``            ``// integer``            ``int` `indexPosition = i;``            ``for` `(``int` `j = i + ``1``; j < n; ++j)``            ``{``    ` `                ``// If we exceed the``                ``// Max swaps then``                ``// break the loop``                ``if` `(k <= j - i)``                    ``break``;``    ` `                ``// Find the maximum value``                ``// from i+1 to max k or n``                ``// which will replace``                ``// arr[indexPosition]``                ``if` `(arr[j] > arr[indexPosition])``                    ``indexPosition = j;``            ``}``    ` `            ``// Swap the elements from``            ``// Maximum indexPosition``            ``// we found till now to``            ``// the ith index``            ``for` `(``int` `j = indexPosition; j > i; --j)``                ``SwapInts(arr, j, j - ``1``);``    ` `            ``// Updates k after swapping``            ``// indexPosition-i elements``            ``k -= indexPosition - i;``        ``}``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `[]arr = { ``3``, ``5``, ``4``, ``1``, ``2` `};``        ``int` `n = arr.length;``        ``int` `k = ``3``;``    ` `        ``KSwapMaximum(arr, n, k);``    ` `        ``// Print the final Array``        ``for` `(``int` `i = ``0``; i < n; ++i)``            ``System.out.print(arr[i] + ``" "``);``    ``}``}` `// This code is contributed by``// Manish Shaw(manishshaw1)`

## Python3

 `# Python program to find``# lexicographically``# maximum value after``# k swaps.` `arr ``=` `[``3``, ``5``, ``4``, ``1``, ``2``]` `# Function which``# modifies the array``def` `KSwapMaximum(n, k) :``    ` `    ``global` `arr``    ``for` `i ``in` `range``(``0``, n ``-` `1``) :``        ``if` `(k > ``0``) :``        ` `            ``# Here, indexPosition``            ``# is set where we want to``            ``# put the current largest``            ``# integer``            ``indexPosition ``=` `i``            ``for` `j ``in` `range``(i ``+` `1``, n) :        ``    ` `                ``# If we exceed the Max swaps``                ``# then break the loop``                ``if` `(k <``=` `j ``-` `i) :``                    ``break``    ` `                ``# Find the maximum value``                ``# from i+1 to max k or n``                ``# which will replace``                ``# arr[indexPosition]``                ``if` `(arr[j] > arr[indexPosition]) :``                    ``indexPosition ``=` `j``            ` `            ``# Swap the elements from``            ``# Maximum indexPosition``            ``# we found till now to``            ``# the ith index``            ``for` `j ``in` `range``(indexPosition, i, ``-``1``) :``                ``t ``=` `arr[j]``                ``arr[j] ``=` `arr[j ``-` `1``]``                ``arr[j ``-` `1``] ``=` `t``    ` `            ``# Updates k after swapping``            ``# indexPosition-i elements``            ``k ``=` `k ``-` `indexPosition ``-` `i` `# Driver code``n ``=` `len``(arr)``k ``=` `3` `KSwapMaximum(n, k)` `# Print the final Array``for` `i ``in` `range``(``0``, n) :``    ``print` `(``"{} "` `.``            ``format``(arr[i]),``                 ``end ``=` `"")``    ` `# This code is contributed by``# Manish Shaw(manishshaw1)`

## C#

 `// C# program to find``// lexicographically``// maximum value after``// k swaps.``using` `System;` `class` `GFG``{``    ``static` `void` `SwapInts(``int``[] array,``                         ``int` `position1,``                         ``int` `position2)``    ``{``        ``// Swaps elements in an array.``        ` `        ``// Copy the first position's element``        ``int` `temp = array[position1];``        ` `        ``// Assign to the second element``        ``array[position1] = array[position2];``        ` `        ``// Assign to the first element``        ``array[position2] = temp;``    ``}``    ` `    ``// Function which``    ``// modifies the array``    ``static` `void` `KSwapMaximum(``int` `[]arr,``                             ``int` `n, ``int` `k)``    ``{``        ``for` `(``int` `i = 0;``                 ``i < n - 1 && k > 0; ++i)``        ``{``    ` `            ``// Here, indexPosition``            ``// is set where we want to``            ``// put the current largest``            ``// integer``            ``int` `indexPosition = i;``            ``for` `(``int` `j = i + 1; j < n; ++j)``            ``{``    ` `                ``// If we exceed the``                ``// Max swaps then``                ``// break the loop``                ``if` `(k <= j - i)``                    ``break``;``    ` `                ``// Find the maximum value``                ``// from i+1 to max k or n``                ``// which will replace``                ``// arr[indexPosition]``                ``if` `(arr[j] > arr[indexPosition])``                    ``indexPosition = j;``            ``}``    ` `            ``// Swap the elements from``            ``// Maximum indexPosition``            ``// we found till now to``            ``// the ith index``            ``for` `(``int` `j = indexPosition; j > i; --j)``                ``SwapInts(arr, j, j - 1);``    ` `            ``// Updates k after swapping``            ``// indexPosition-i elements``            ``k -= indexPosition - i;``        ``}``    ``}``    ` `    ``// Driver code``    ``static` `void` `Main()``    ``{``        ``int` `[]arr = ``new` `int``[]{ 3, 5, 4, 1, 2 };``        ``int` `n = arr.Length;``        ``int` `k = 3;``    ` `        ``KSwapMaximum(arr, n, k);``    ` `        ``// Print the final Array``        ``for` `(``int` `i = 0; i < n; ++i)``            ``Console.Write(arr[i] + ``" "``);``    ``}``}``// This code is contributed by``// Manish Shaw(manishshaw1)`

## PHP

 ` 0; ``\$i``++)``    ``{``        ` `        ``// Here, indexPosition``        ``// is set where we want to``        ``// put the current largest``        ``// integer``        ``\$indexPosition` `= ``\$i``;``        ``for` `(``\$j` `= ``\$i` `+ 1;``             ``\$j` `< ``\$n``; ``\$j``++)``        ``{` `            ``// If we exceed the Max swaps``            ``// then break the loop``            ``if` `(``\$k` `<= ``\$j` `- ``\$i``)``                ``break``;` `            ``// Find the maximum value``            ``// from i+1 to max k or n``            ``// which will replace``            ``// arr[indexPosition]``            ``if` `(``\$arr``[``\$j``] > ``\$arr``[``\$indexPosition``])``                ``\$indexPosition` `= ``\$j``;``        ``}``        ` `        ``// Swap the elements from``        ``// Maximum indexPosition``        ``// we found till now to``        ``// the ith index``        ``for` `(``\$j` `= ``\$indexPosition``;``             ``\$j` `> ``\$i``; ``\$j``--)``            ``swap(``\$arr``[``\$j``], ``\$arr``[``\$j` `- 1]);` `        ``// Updates k after swapping``        ``// indexPosition-i elements``        ``\$k` `-= ``\$indexPosition` `- ``\$i``;``    ``}``}` `// Driver code``\$arr` `= ``array``( 3, 5, 4, 1, 2 );``\$n` `= ``count``(``\$arr``);``\$k` `= 3;` `KSwapMaximum(``\$arr``, ``\$n``, ``\$k``);` `// Print the final Array``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``echo` `(``\$arr``[``\$i``].``" "``);``    ` `// This code is contributed by``// Manish Shaw(manishshaw1)``?>`

## Javascript

 ``

Output

`5 4 3 1 2 `

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

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