# Maximize distance between two elements of Array by at most X swaps

Given an array arr[] of unique elements and three integers X, A and B. The task is to print the maximum possible distance between elements A and B in atmost X swaps with the adjacent elements.
Examples:

Input: arr[] = {5, 1, 3, 2}, X = 1, A = 2, B = 3
Output:
Explanation:
3 is swapped with it’s adjacent element 1. So, the distance between element 3 and 2 is 2 and no more interchanges are possible since X = 1.

Input: arr = {100, 33, 10, 1}, X = 5, A = 100, B = 1
Output:
Explanation:
As the elements are already the beginning and ending element, the distance between them is already maximized.

Approach: The following steps can be followed to compute the result:

• In case the given X is 0, then |index(A)-index(B)| is returned as final answer.
• If the elements are already at index 0 and n-1, return N-1 as the distance is already maximized.
• Else, the greater index element is swapped X times with their adjacent element till it is less than the size of the array.
• After reaching the end of the array if X > 0; then the smaller index element is swapped with it’s previous elements till X is not equal to 0.

Below is the implementation of the above approach:

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to maximize the distance ` `// between the two elements of the ` `// array by at most X swaps ` `void` `max_distance(``int` `a[], ``int` `n, ``int` `x, ` `                  ``int` `A, ``int` `B) ` `{ ` `    ``// Initialize the variables ` `    ``int` `g = 0, h = 0; ` ` `  `    ``// Iterate till the length of the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// Check if current element is A ` `        ``if` `(a[i] == A) ` `            ``// Store index ` `            ``g = i; ` ` `  `        ``// Check if current element is B ` `        ``if` `(a[i] == B) ` `            ``// Store index ` `            ``h = i; ` `    ``} ` ` `  `    ``// If X = 0, swapping can not be done ` `    ``if` `(x == 0) { ` `        ``cout << ``abs``(g - h); ` `    ``} ` ` `  `    ``// If elements are at starting and ending ` `    ``// indices, then the distance ` `    ``// is already maximum ` `    ``else` `if` `((g == 0) && (h == (n - 1))) ` `        ``cout << n - 1 << endl; ` ` `  `    ``else` `if` `((g == n - 1) && (h == 0)) ` `        ``cout << n - 1 << endl; ` ` `  `    ``else` `{ ` ` `  `        ``// Greater index is incremented ` `        ``// till x > 0 and the ` `        ``// index of element < size of array ` `        ``if` `(h > g) { ` `            ``while` `((x > 0) && (h < n - 1)) { ` `                ``h++; ` `                ``x--; ` `            ``} ` ` `  `            ``// Check if reached the size of array ` `            ``// and x > 0, then the ` `            ``// smaller index is decremented ` `            ``if` `(x > 0) { ` `                ``while` `((x > 0) && (g > 0)) { ` `                    ``g--; ` `                    ``x--; ` `                ``} ` `            ``} ` `            ``cout << h - g << endl; ` `        ``} ` ` `  `        ``// Greater index is incremented till x>0 ` `        ``// and index of element < size of array ` `        ``else` `{ ` `            ``while` `((x > 0) && (g < n - 1)) { ` `                ``g++; ` `                ``x--; ` `            ``} ` ` `  `            ``// Check if reached the size of the array ` `            ``// and x > 0 the smaller index ` `            ``// is decremented ` `            ``if` `(x > 0) { ` `                ``while` `((x > 0) && (h > 0)) { ` `                    ``h--; ` `                    ``x--; ` `                ``} ` `            ``} ` `            ``cout << g - h << endl; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 100, 33, 10, 1 }; ` `    ``int` `x = 5, A = 100, B = 1; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``max_distance(a, n, x, A, B); ` `    ``return` `0; ` `} `

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to maximize the distance ` `// between the two elements of the ` `// array by at most X swaps ` `static` `void` `max_distance(``int` `a[], ``int` `n, ``int` `x, ` `                         ``int` `A, ``int` `B) ` `{ ` `     `  `    ``// Initialize the variables ` `    ``int` `i, g = ``0``, h = ``0``; ` ` `  `    ``// Iterate till the length of the array ` `    ``for``(i = ``0``; i < n; i++)  ` `    ``{ ` `         `  `        ``// Check if current element is A ` `        ``if` `(a[i] == A) ` `         `  `            ``// Store index ` `            ``g = i; ` ` `  `        ``// Check if current element is B ` `        ``if` `(a[i] == B) ` `         `  `            ``// Store index ` `            ``h = i; ` `    ``} ` ` `  `    ``// If X = 0, swapping can not be done ` `    ``if` `(x == ``0``) ` `    ``{ ` `        ``System.out.print(Math.abs(g - h)); ` `    ``} ` ` `  `    ``// If elements are at starting and  ` `    ``// ending indices, then the distance ` `    ``// is already maximum ` `    ``else` `if` `((g == ``0``) && (h == (n - ``1``))) ` `        ``System.out.println(n - ``1``); ` ` `  `    ``else` `if` `((g == n - ``1``) && (h == ``0``)) ` `        ``System.out.println(n - ``1``); ` ` `  `    ``else` `    ``{ ` `         `  `        ``// Greater index is incremented ` `        ``// till x > 0 and theindex of  ` `        ``// element < size of array ` `        ``if` `(h > g)  ` `        ``{ ` `            ``while` `((x > ``0``) && (h < n - ``1``))  ` `            ``{ ` `                ``h++; ` `                ``x--; ` `            ``} ` ` `  `            ``// Check if reached the size of  ` `            ``// array and x > 0, then the ` `            ``// smaller index is decremented ` `            ``if` `(x > ``0``) ` `            ``{ ` `                ``while` `((x > ``0``) && (g > ``0``)) ` `                ``{ ` `                    ``g--; ` `                    ``x--; ` `                ``} ` `            ``} ` `            ``System.out.println(h - g); ` `        ``} ` ` `  `        ``// Greater index is incremented till x>0 ` `        ``// and index of element < size of array ` `        ``else` `        ``{ ` `            ``while` `((x > ``0``) && (g < n - ``1``)) ` `            ``{ ` `                ``g++; ` `                ``x--; ` `            ``} ` ` `  `            ``// Check if reached the size of the array ` `            ``// and x > 0 the smaller index ` `            ``// is decremented ` `            ``if` `(x > ``0``)  ` `            ``{ ` `                ``while` `((x > ``0``) && (h > ``0``)) ` `                ``{ ` `                    ``h--; ` `                    ``x--; ` `                ``} ` `            ``} ` `            ``System.out.println(g - h); ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String []args) ` `{ ` `    ``int` `a[] = { ``100``, ``33``, ``10``, ``1` `}; ` `    ``int` `x = ``5``, A = ``100``, B = ``1``; ` `    ``int` `n = a.length; ` `     `  `    ``max_distance(a, n, x, A, B); ` `} ` `} ` ` `  `// This code is contributed by chitranayal `

 `# Python3 program for the above approach ` ` `  `# Function to maximize the distance  ` `# between the two elements of the  ` `# array by at most X swaps ` `def` `max_distance(a, n, x, A, B): ` `     `  `    ``# Initialize the variables ` `    ``g ``=` `0` `    ``h ``=` `0` `     `  `    ``for` `i ``in` `range``(``0``, n): ` `         `  `        ``# Check if current element is A  ` `        ``if` `(a[i] ``=``=` `A): ` `             `  `            ``# Store index  ` `            ``g ``=` `i ` `             `  `        ``# Check if current element is B  ` `        ``if` `(a[i] ``=``=` `B): ` `             `  `            ``# Store index  ` `            ``h ``=` `i ` `             `  `    ``# If X = 0, swapping can not be done  ` `    ``if` `(x ``=``=` `0``): ` `        ``print``(``abs``(g ``-` `h)) ` `         `  `    ``# If elements are at starting and  ` `    ``# ending indices, then the distance  ` `    ``# is already maximum ` `    ``elif` `((g ``=``=` `0``) ``and` `(h ``=``=` `(n ``-` `1``))): ` `        ``print``(n ``-` `1``, end ``=` `'') ` `         `  `    ``elif` `((g ``=``=` `n ``-` `1``) ``and` `(h ``=``=` `0``)): ` `        ``print``(n ``-` `1``, end ``=` `'') ` `    ``else``: ` `         `  `        ``# Greater index is incremented  ` `        ``# till x > 0 and the  ` `        ``# index of element < size of array  ` `        ``if` `(h > g): ` `            ``while` `((x > ``0``) ``and` `(h < n ``-` `1``)): ` `                ``h ``+``=` `1` `                ``x ``-``=` `1` `                 `  `                ``# Check if reached the size  ` `                ``# of array and x > 0, then the  ` `                ``# smaller index is decremented  ` `                ``if` `(x > ``0``): ` `                    ``while` `((x > ``0``) ``and` `(g > ``0``)): ` `                        ``g ``-``=` `1` `                        ``x ``-``=` `1` `                         `  `                ``print``(h ``-` `g, end ``=` `'') ` `                 `  `        ``# Greater index is incremented till x>0  ` `        ``# and index of element < size of array  ` `        ``else``: ` `            ``while` `((x > ``0``) ``and` `(g < n ``-` `1``)): ` `                ``g ``+``=` `1` `                ``x ``-``=` `1` `                 `  `            ``# Check if reached the size of  ` `            ``# the array and x > 0 the smaller  ` `            ``# index is decremented  ` `            ``if` `(x > ``0``): ` `                ``while` `((x > ``0``) ``and` `(h > ``0``)): ` `                    ``h ``-``=` `1` `                    ``x ``-``=` `1` `                     `  `            ``print``(g ``-` `h, end ``=` `'') ` `             `  `# Driver code  ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``a ``=` `[ ``100``, ``33``, ``10``, ``1` `] ` `    ``x ``=` `5` `    ``A ``=` `100` `    ``B ``=` `1` `    ``n ``=` `len``(a) ` `     `  `    ``max_distance(a, n, x, A, B) ` `                     `  `# This code is contributed by virusbuddah_      `

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to maximize the distance ` `// between the two elements of the ` `// array by at most X swaps ` `static` `void` `max_distance(``int` `[]a, ``int` `n, ``int` `x, ` `                         ``int` `A, ``int` `B) ` `{ ` `     `  `    ``// Initialize the variables ` `    ``int` `i, g = 0, h = 0; ` ` `  `    ``// Iterate till the length of the array ` `    ``for``(i = 0; i < n; i++)  ` `    ``{ ` `         `  `        ``// Check if current element is A ` `        ``if` `(a[i] == A) ` `         `  `            ``// Store index ` `            ``g = i; ` ` `  `        ``// Check if current element is B ` `        ``if` `(a[i] == B) ` `         `  `            ``// Store index ` `            ``h = i; ` `    ``} ` ` `  `    ``// If X = 0, swapping can not be done ` `    ``if` `(x == 0) ` `    ``{ ` `        ``Console.Write(Math.Abs(g - h)); ` `    ``} ` ` `  `    ``// If elements are at starting and  ` `    ``// ending indices, then the distance ` `    ``// is already maximum ` `    ``else` `if` `((g == 0) && (h == (n - 1))) ` `        ``Console.WriteLine(n - 1); ` ` `  `    ``else` `if` `((g == n - 1) && (h == 0)) ` `        ``Console.WriteLine(n - 1); ` ` `  `    ``else` `    ``{ ` `         `  `        ``// Greater index is incremented ` `        ``// till x > 0 and theindex of  ` `        ``// element < size of array ` `        ``if` `(h > g)  ` `        ``{ ` `            ``while` `((x > 0) && (h < n - 1))  ` `            ``{ ` `                ``h++; ` `                ``x--; ` `            ``} ` ` `  `            ``// Check if reached the size of  ` `            ``// array and x > 0, then the ` `            ``// smaller index is decremented ` `            ``if` `(x > 0) ` `            ``{ ` `                ``while` `((x > 0) && (g > 0)) ` `                ``{ ` `                    ``g--; ` `                    ``x--; ` `                ``} ` `            ``} ` `            ``Console.WriteLine(h - g); ` `        ``} ` ` `  `        ``// Greater index is incremented till x>0 ` `        ``// and index of element < size of array ` `        ``else` `        ``{ ` `            ``while` `((x > 0) && (g < n - 1)) ` `            ``{ ` `                ``g++; ` `                ``x--; ` `            ``} ` ` `  `            ``// Check if reached the size of the  ` `            ``// array and x > 0 the smaller index ` `            ``// is decremented ` `            ``if` `(x > 0)  ` `            ``{ ` `                ``while` `((x > 0) && (h > 0)) ` `                ``{ ` `                    ``h--; ` `                    ``x--; ` `                ``} ` `            ``} ` `            ``Console.WriteLine(g - h); ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `[]a = { 100, 33, 10, 1 }; ` `    ``int` `x = 5, A = 100, B = 1; ` `    ``int` `n = a.Length; ` `     `  `    ``max_distance(a, n, x, A, B); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:
```3
```

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