Maximum number formed from array with K number of adjacent swaps allowed

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Given an array a[ ] and the number of adjacent swap operations allowed are K. The task is to find the max number that can be formed using these swap operations.

Examples:

Input : a[]={ 1, 2, 9, 8, 1, 4, 9, 9, 9 }, K = 4
Output : 9 8 1 2 1 4 9 9 9
After 1^st swap a[ ] becomes 1 9 2 8 1 4 9 9 9
After 2^nd swap a[ ] becomes 9 1 2 8 1 4 9 9 9
After 3^rd swap a[ ] becomes 9 1 8 2 1 4 9 9 9
After 4^th swap a[ ] becomes 9 8 1 2 1 4 9 9 9

Input : a[]={2, 5, 8, 7, 9}, K = 2
Output : 8 2 5 7 9

Approach:

Starting from the first digit, check for the next K digits and store the index of the largest number.
Bring that greatest digit to the top by swapping the adjacent digits.
Reduce to the value of K by the number of adjacent swaps done.
Repeat the above steps until the number of swaps becomes zero.

Below is the implementation of the above approach

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the
// elements of the array
void print(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
    cout << endl;
}
 
// Exchange array elements one by
// one from  right to left side
// starting from the current position
// and ending at the target position
void swapMax(int* arr, int target_position,
                      int current_position)
{
    int aux = 0;
    for (int i = current_position;
         i > target_position; i--) {
        aux = arr[i - 1];
        arr[i - 1] = arr[i];
        arr[i] = aux;
    }
}
 
// Function to return the
// maximum number array
void maximizeArray(int* arr,
                   int length, int swaps)
{
    // Base condition
    if (swaps == 0)
        return;
 
    // Start from the first index
    for (int i = 0; i < length; i++) {
        int max_index = 0, max = INT_MIN;
 
        // Search for the next K elements
        int limit = (swaps + i) > length ? 
                        length : swaps + i;
 
        // Find index of the maximum
        // element in next K elements
        for (int j = i; j <= limit; j++) {
            if (arr[j] > max) {
                max = arr[j];
                max_index = j;
            }
        }
 
        // Update the value of
        // number of swaps
        swaps -= (max_index - i);
 
        // Update the array elements by
        // swapping adjacent elements
        swapMax(arr, i, max_index);
 
        if (swaps == 0)
            break;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 9, 8, 1, 4, 9, 9, 9 };
    int length = sizeof(arr) / sizeof(int);
    int swaps = 4;
    maximizeArray(arr, length, swaps);
 
    print(arr, length);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
 
// Function to print the
// elements of the array
static void print(int arr[], int n)
{
    for (int i = 0; i < n; i++)
    {
        System.out.print(arr[i] + " ");
    }
    System.out.println();
}
 
// Exchange array elements one by
// one from right to left side
// starting from the current position
// and ending at the target position
static void swapMax(int[] arr, int target_position,
                    int current_position)
{
    int aux = 0;
    for (int i = current_position;
        i > target_position; i--) 
    {
        aux = arr[i - 1];
        arr[i - 1] = arr[i];
        arr[i] = aux;
    }
}
 
// Function to return the
// maximum number array
static void maximizeArray(int[] arr,
                int length, int swaps)
{
    // Base condition
    if (swaps == 0)
        return;
 
    // Start from the first index
    for (int i = 0; i < length; i++) 
    {
        int max_index = 0, max = Integer.MIN_VALUE;
 
        // Search for the next K elements
        int limit = (swaps + i) > length ? 
                        length : swaps + i;
 
        // Find index of the maximum
        // element in next K elements
        for (int j = i; j <= limit; j++) 
        {
            if (arr[j] > max) 
            {
                max = arr[j];
                max_index = j;
            }
        }
 
        // Update the value of
        // number of swaps
        swaps -= (max_index - i);
 
        // Update the array elements by
        // swapping adjacent elements
        swapMax(arr, i, max_index);
 
        if (swaps == 0)
            break;
    }
}
 
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 1, 2, 9, 8, 1, 4, 9, 9, 9 };
    int length = arr.length;
    int swaps = 4;
    maximizeArray(arr, length, swaps);
 
    print(arr, length);
}
}
 
/* This code is contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the above approach 
import sys
 
# Function to print the 
# elements of the array 
def print_ele(arr, n) :
     
    for i in range(n) :
        print(arr[i],end=" "); 
         
    print(); 
 
# Exchange array elements one by 
# one from right to left side 
# starting from the current position 
# and ending at the target position 
def swapMax(arr, target_position, 
                    current_position) :
                         
    aux = 0; 
    for i in range(current_position, target_position,-1) :
        aux = arr[i - 1]; 
        arr[i - 1] = arr[i]; 
        arr[i] = aux; 
 
# Function to return the 
# maximum number array 
def maximizeArray(arr, length, swaps) : 
 
    # Base condition 
    if (swaps == 0) :
        return; 
 
    # Start from the first index 
    for i in range(length) :
        max_index = 0; max = -(sys.maxsize-1);
         
        # Search for the next K elements
        if (swaps + i) > length :
            limit = length
        else:
            limit = swaps + i
             
        # Find index of the maximum
        # element in next K elements
        for j in range(i, limit + 1) :
            if (arr[j] > max) :
                max = arr[j];
                max_index = j;
                 
        # Update the value of
        # number of swaps
        swaps -= (max_index - i); 
 
        # Update the array elements by 
        # swapping adjacent elements 
        swapMax(arr, i, max_index); 
 
        if (swaps == 0) :
            break; 
 
# Driver code 
if __name__ == "__main__" : 
 
    arr = [ 1, 2, 9, 8, 1, 4, 9, 9, 9 ]; 
    length = len(arr); 
    swaps = 4; 
    maximizeArray(arr, length, swaps); 
 
    print_ele(arr, length); 
 
# This code is contributed by AnkitRai01

C#

// C# program to find the sum 
// and product of k smallest and 
// k largest prime numbers in an array 
 
using System;
     
class GFG
{
 
// Function to print the
// elements of the array
static void print(int []arr, int n)
{
    for (int i = 0; i < n; i++)
    {
        Console.Write(arr[i] + " ");
    }
    Console.WriteLine();
}
 
// Exchange array elements one by
// one from right to left side
// starting from the current position
// and ending at the target position
static void swapMax(int[] arr, int target_position,
                    int current_position)
{
    int aux = 0;
    for (int i = current_position;
        i > target_position; i--) 
    {
        aux = arr[i - 1];
        arr[i - 1] = arr[i];
        arr[i] = aux;
    }
}
 
// Function to return the
// maximum number array
static void maximizeArray(int[] arr,
                int length, int swaps)
{
    // Base condition
    if (swaps == 0)
        return;
 
    // Start from the first index
    for (int i = 0; i < length; i++) 
    {
        int max_index = 0, max = int.MinValue;
 
        // Search for the next K elements
        int limit = (swaps + i) > length ? 
                        length : swaps + i;
 
        // Find index of the maximum
        // element in next K elements
        for (int j = i; j <= limit; j++) 
        {
            if (arr[j] > max) 
            {
                max = arr[j];
                max_index = j;
            }
        }
 
        // Update the value of
        // number of swaps
        swaps -= (max_index - i);
 
        // Update the array elements by
        // swapping adjacent elements
        swapMax(arr, i, max_index);
 
        if (swaps == 0)
            break;
    }
}
 
// Driver code
public static void Main(String[] args) 
{
    int []arr = { 1, 2, 9, 8, 1, 4, 9, 9, 9 };
    int length = arr.Length;
    int swaps = 4;
    maximizeArray(arr, length, swaps);
 
    print(arr, length);
}
}
 
/* This code is contributed by PrinciRaj1992 */

Javascript

<script>
 
// JavaScript implementation of the above approach
 
 
// Function to print the
// elements of the array
function print(arr, n) {
    for (let i = 0; i < n; i++) {
        document.write(arr[i] + " ");
    }
    document.write("<br>");
}
 
// Exchange array elements one by
// one from right to left side
// starting from the current position
// and ending at the target position
function swapMax(arr, target_position, current_position) {
    let aux = 0;
    for (let i = current_position; i > target_position; i--) {
        aux = arr[i - 1];
        arr[i - 1] = arr[i];
        arr[i] = aux;
    }
}
 
// Function to return the
// maximum number array
function maximizeArray(arr, length, swaps) {
    // Base condition
    if (swaps == 0)
        return;
 
    // Start from the first index
    for (let i = 0; i < length; i++) {
        let max_index = 0, max = Number.MIN_SAFE_INTEGER;
 
        // Search for the next K elements
        let limit = (swaps + i) > length ?
            length : swaps + i;
 
        // Find index of the maximum
        // element in next K elements
        for (let j = i; j <= limit; j++) {
            if (arr[j] > max) {
                max = arr[j];
                max_index = j;
            }
        }
 
        // Update the value of
        // number of swaps
        swaps -= (max_index - i);
 
        // Update the array elements by
        // swapping adjacent elements
        swapMax(arr, i, max_index);
 
        if (swaps == 0)
            break;
    }
}
 
// Driver code
 
let arr = [1, 2, 9, 8, 1, 4, 9, 9, 9];
let length = arr.length;
let swaps = 4;
maximizeArray(arr, length, swaps);
 
print(arr, length);
 
// This code is contributed by gfgking
 
</script>

Output:

9 8 1 2 1 4 9 9 9

Time Complexity: O(N*N) where N is the length of given array.
Auxiliary Space: O(1)

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