Given an arr[] consisting of N elements in the range [1, N], the task is to maximize the distance between smallest and largest array element by a single swap.
Examples:
Input: arr[] = {1, 4, 3, 2}
Output: 3
Explanation:
Swapping of arr[1] and arr[3] maximizes the distance.Input: arr[] = {1, 6, 5, 3, 4, 7, 2}
Output: 6
Explanation:
Swappinf of arr[5] and arr[6] maximizes the distance.
Approach
- Find the indices of 1 and N in the array.
- Let minIdx and maxIdx be the minimum and maximum of the two indices, respectively.
-
Now, maxIdx – minIdx is the current distance between the two elements. It can be maximized by the maximum possible from the following two swaps:
- Swapping a[minIdx] with a[0] increasing the distance by minIdx.
- Swapping a[maxIdx] with a[N – 1] increasing the distance by N – 1 – maxIdx.
Hence, maxIdx – minIdx + max(minIdx, N – 1 – maxIdx) gives us the required answer.
Illustration:
arr[] = {2, 4, 6, 1, 3, 5}
Smallest element = 1
Largest element = 6
minIdx = 2
maxIdx = 3
Required Answer = maxIdx – minIdx + max(minIdx, N – 1 – maxIdx)
= 3 – 2 + max(2, 5 – 3)
= 1 + 2 = 3
Below is the implementation of the above approach:
C++
// C++ program maximize the // distance between smallest // and largest array element // by a single swap #include <bits/stdc++.h> using namespace std; // Function to maximize the distance // between the smallest and largest // array element by a single swap int find_max_dist(int arr[], int N) { int minIdx = -1, maxIdx = -1; for (int i = 0; i < N; i++) { if (arr[i] == 1 || arr[i] == N) { if (minIdx == -1) minIdx = i; else { maxIdx = i; break; } } } return maxIdx - minIdx + max(minIdx, N - 1 - maxIdx); } // Driver Code int main() { int arr[] = { 1, 4, 3, 2 }; int N = sizeof(arr) / sizeof(arr[0]); cout << find_max_dist(arr, N) << endl; return 0; } |
Java
// Java program maximize the distance // between smallest and largest array // element by a single swap import java.util.*; class GFG{ // Function to maximize the distance // between the smallest and largest // array element by a single swap static int find_max_dist(int arr[], int N) { int minIdx = -1, maxIdx = -1; for(int i = 0; i < N; i++) { if (arr[i] == 1 || arr[i] == N) { if (minIdx == -1) minIdx = i; else { maxIdx = i; break; } } } return maxIdx - minIdx + Math.max(minIdx, N - 1 - maxIdx); } // Driver Code public static void main(String[] args) { int arr[] = { 1, 4, 3, 2 }; int N = arr.length; System.out.print(find_max_dist(arr, N) + "\n"); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program maximize the # distance between smallest # and largest array element # by a single swap # Function to maximize the distance # between the smallest and largest # array element by a single swap def find_max_dist(arr, N): minIdx, maxIdx = -1, -1 for i in range(N): if (arr[i] == 1 or arr[i] == N): if (minIdx == -1) : minIdx = i else : maxIdx = i break return (maxIdx - minIdx + max(minIdx, N - 1 - maxIdx)) # Driver code arr = [ 1, 4, 3, 2 ] N = len(arr) print(find_max_dist(arr, N)) # This code is contributed by divyeshrabadiya07 |
C#
// C# program maximize the distance // between smallest and largest array // element by a single swap using System; class GFG{ // Function to maximize the distance // between the smallest and largest // array element by a single swap static int find_max_dist(int []arr, int N) { int minIdx = -1, maxIdx = -1; for(int i = 0; i < N; i++) { if (arr[i] == 1 || arr[i] == N) { if (minIdx == -1) minIdx = i; else { maxIdx = i; break; } } } return maxIdx - minIdx + Math.Max(minIdx, N - 1 - maxIdx); } // Driver Code public static void Main(String[] args) { int []arr = { 1, 4, 3, 2 }; int N = arr.Length; Console.Write(find_max_dist(arr, N) + "\n"); } } // This code is contributed by Amit Katiyar |
3
Time Complexity: O(N)
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