有N 个不同的整数排列在一个圆上。任意两个相邻数字之间的距离为1 。任务是从最小的数字开始在这个圆上移动,然后移动到第二小的、第三小的,依此类推,直到最大的数字并打印最小旅行距离。
例子:
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
方法:如果我们想在索引为i和j 的两个元素之间穿行,我们有两条可能的路径,长度之一是|i – j|另一个长度为n – |i – j|我们将选择距离最小的那个。
最初,我们构建了一个(element, index)对的数组,并按元素的升序对其进行排序。然后我们可以每两个连续的对,找到在这两个元素之间旅行的最短路径。
下面是上述方法的实现:
CPP
#include
#define mod 1000000007
using namespace std;
// Function to return the minimum distance
int min_distance(int n, vector arr)
{
// Declare a Pair[] array of size n
vector> 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 Code
int main()
{
vector 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 approach
import 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 com = new Comparator() {
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
输出:
8
时间复杂度: O(n * log(n))
辅助空间: O(n)
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