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: 2
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: 3
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 its previous elements till X is not equal to 0.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to maximize the distance// between the two elements of the// array by at most X swapsvoid 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 codeint main(){ int a[] = { 100, 33, 10, 1 }; int x = 5, A = 100, B = 1; int n = sizeof(a) / sizeof(a[0]); max_distance(a, n, x, A, B); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to maximize the distance// between the two elements of the// array by at most X swapsstatic 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 codepublic 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
# Python3 program for the above approach# Function to maximize the distance # between the two elements of the # array by at most X swapsdef 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#
// C# program for the above approachusing System;class GFG{// Function to maximize the distance// between the two elements of the// array by at most X swapsstatic 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 codepublic 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 |
Javascript
<script>// JavaScript program for the above approach// Function to maximize the distance// between the two elements of the// array by at most X swapsfunction max_distance(a,n,x,A,B){ // Initialize the variables let 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) { document.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))) document.write(n - 1); else if ((g == n - 1) && (h == 0)) document.write(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--; } } document.write(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--; } } document.write(g - h); } }}// Driver codelet a=[ 100, 33, 10, 1];let x = 5, A = 100, B = 1;let n = a.length;max_distance(a, n, x, A, B);// This code is contributed by rag2127</script> |
Output:
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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