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

05 Aug, 2021
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 1st swap a[ ] becomes 1 9 2 8 1 4 9 9 9 After 2nd swap a[ ] becomes 9 1 2 8 1 4 9 9 9 After 3rd swap a[ ] becomes 9 1 8 2 1 4 9 9 9 After 4th swap a[ ] becomes 9 8 1 2 1 4 9 9 9Input : 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 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++
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; // Function to print the// elements of the arrayvoid 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 positionvoid 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 arrayvoid 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 codeint 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 implementation of the above approachclass GFG{ // Function to print the// elements of the arraystatic 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 positionstatic 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 arraystatic 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 codepublic 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 implementation of the above approachimport sys # Function to print the# elements of the arraydef 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 positiondef 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 arraydef 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 codeif __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# 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 arraystatic 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 positionstatic 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 arraystatic 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 codepublic 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 */
<script> // JavaScript implementation of the above approach  // Function to print the// elements of the arrayfunction 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 positionfunction 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 arrayfunction 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>
9 8 1 2 1 4 9 9 9
 
Time Complexity: O(N*N) where N is the length of given array 
princiraj1992
ankthon
gfgking
saurabh1990aror
number-digits
Arrays
Arrays
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