There are N distinct integers arranged on a circle. The distance between any two adjacent numbers is 1. The task is to travel on this circle starting with the smallest number, then moving to the second smallest, third smallest, and so on until the largest number and print the minimum travel distance.
Examples:
Input: arr[] = {3, 6, 5, 1, 2, 4}
Output: 8
1 -> 2 (Left to right)
2 -> 3 (Left to right)
3 -> 4 (Right to left)
4 -> 5 (Left to right or right to left)
5 -> 6 (Right to left)Input: arr[] = {14, 16, 8, 17, 12, 10, 4, 13, 11, 20}
Output: 27
Approach: If we want to travel between two elements with indices i and j, we have two possible paths, one of length |i – j| and the other of length n – |i – j| and we will take the one with the minimum distance.
Initially, we build an array of pairs of (element, index) and we sort it in increasing order of the elements. Then we can take every two consecutive pairs and find the shortest way to travel between these two elements.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>#define mod 1000000007using namespace std;// Function to return the minimum distanceint min_distance(int n, vector<int> arr){ // Declare a Pair[] array of size n vector<pair<int,int>> val(n); // Store all the (element, index) pairs for ( int i = 0; i < n; i++) val[i] = {arr[i], i}; // Sort the pairs in ascending order of the value sort(val.begin(), val.end()); int min_dist = 0; for (int i = 1; i < n; i++) { // Choose the minimum distance path // of the two available min_dist += min(abs(val[i].second - val[i - 1].second), n - abs(val[i].second - val[i - 1].second)); } return min_dist;}// Driver Codeint main(){ vector<int> arr = {3, 6, 5, 1, 2, 4}; int n = arr.size(); cout << (min_distance(n, arr));}// This code is contributed by mohit kumar 29 |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to return the minimum distance static int min_distance(int n, int[] arr) { // Declare a Pair[] array of size n Pair[] val = new Pair[n]; // Store all the (element, index) pairs for (int i = 0; i < n; i++) val[i] = new Pair(arr[i], i); // Sort the pairs in ascending order of the value Arrays.sort(val, com); int min_dist = 0; for (int i = 1; i < n; i++) { // Choose the minimum distance path // of the two available min_dist += Math.min(Math.abs(val[i].y - val[i - 1].y), n - Math.abs(val[i].y - val[i - 1].y)); } return min_dist; } // Comparator to be used in the // sorting of pairs based on the values static final Comparator<Pair> com = new Comparator<Pair>() { public int compare(Pair a, Pair b) { if (Integer.compare(a.x, b.x) != 0) return Integer.compare(a.x, b.x); else return Integer.compare(a.y, b.y); } }; // Class to represent a pair static class Pair { int x, y; Pair(int p, int q) { x = p; y = q; } } // Driver code public static void main(String[] args) { int[] arr = new int[] { 3, 6, 5, 1, 2, 4 }; int n = arr.length; System.out.println(min_distance(n, arr)); }} |
Python
# Python implementation of the approach # Function to return the minimum distance def min_distance(n, arr): # Declare a Pair[] array of size n val = [None] * n # Store all the (element, index) pairs for i in range(0, n): val[i] = (arr[i], i) # Sort the pairs in ascending order of the value val.sort() min_dist = 0 for i in range(1, n): # Choose the minimum distance path # of the two available min_dist += min(abs(val[i][1] - val[i - 1][1]), n - abs(val[i][1] - val[i - 1][1])) return min_dist # Driver code if __name__ == "__main__": arr = [3, 6, 5, 1, 2, 4] n = len(arr) print(min_distance(n, arr))# This code is contributed by Rituraj Jain |
Javascript
<script>// JavaScript implementation of the approach// Function to return the minimum distancefunction min_distance(n,arr){ // Declare a Pair[] array of size n let val = new Array(n); // Store all the (element, index) pairs for (let i = 0; i < n; i++) val[i] = [arr[i], i]; // Sort the pairs in ascending order of the value val.sort(function(a,b){return a[0]-b[0];}); let min_dist = 0; for (let i = 1; i < n; i++) { // Choose the minimum distance path // of the two available min_dist += Math.min(Math.abs(val[i][1] - val[i - 1][1]), n - Math.abs(val[i][1] - val[i - 1][1])); } return min_dist;}// Driver codelet arr=[3, 6, 5, 1, 2, 4];let n=arr.length;document.write(min_distance(n, arr))// This code is contributed by patel2127</script> |
C#
using System;using System.Collections.Generic;class GFG{ // Function to return the minimum distance public static int min_distance(int n, int[] arr) { // Declare a Pair[] array of size n Pair[] val = new Pair[n]; // Store all the (element, index) pairs for (int i = 0; i < n; i++) val[i] = new Pair(arr[i], i); // Sort the pairs in ascending order of the value Array.Sort(val, com); int min_dist = 0; for (int i = 1; i < n; i++) { // Choose the minimum distance path // of the two available min_dist += Math.Min(Math.Abs(val[i].y - val[i - 1].y), n - Math.Abs(val[i].y - val[i - 1].y)); } return min_dist; } // Comparator to be used in the // sorting of pairs based on the values static readonly IComparer<Pair> com = new PairComparer(); class PairComparer : IComparer<Pair> { public int Compare(Pair a, Pair b) { if (Comparer<int>.Default.Compare(a.x, b.x) != 0) return Comparer<int>.Default.Compare(a.x, b.x); else return Comparer<int>.Default.Compare(a.y, b.y); } } // Class to represent a pair class Pair { public int x, y; public Pair(int p, int q) { x = p; y = q; } } // Driver code public static void Main(string[] args) { int[] arr = new int[] { 3, 6, 5, 1, 2, 4 }; int n = arr.Length; Console.WriteLine(min_distance(n, arr)); }}// This code is contributed by Harshad |
Output
8
Complexity Analysis:
- Time Complexity: O(n * log(n))
- Auxiliary Space: O(n)
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