给定一个由N 个整数组成的数组arr[] ,如果从索引i移动到索引j的成本是绝对差,则任务是从0开始,按升序查找访问所有数组元素的总成本在i和j之间。
例子:
Input: arr[ ] = { 4, 3, 2, 5, 1 }
Output: 11
Explanation:
Jump from index 0 to index 4. Cost = abs(4 – 0) = 4.
Jump from index 4 to index 2. Cost = abs(4 – 2) = 2.
Jump from index 2 to index1. Cost = abs(2 – 1) = 1.
Jump from index 1 to index 0. Cost = abs(1 – 0) = 1.
Jump from index 0 to index 3. Cost = abs(0 – 3) = 3.
Therefore, the total cost of visiting all array elements in ascending order = (4 + 2 + 1 + 1 + 3 = 11).
Input: arr[ ] = { 1, 2, 3 }
Output: 2
方法:这个想法是使用对向量进行排序的概念。请按照以下步骤解决问题:
- 初始化一对vector
> ,比如v,来存储元素对和它们各自的位置。 - 遍历数组arr[]并将{arr[i], i}对推入向量v 中。
- 初始化两个变量,比如ans = 0和last = 0,以存储所需的总成本和最后访问元素的索引。
- 按升序对成对的向量进行排序。
- 遍历向量v并通过abs(v[i].second – last)增加ans 。最后更新作为最后=常用3 [I]。第二。
- 完成上述步骤后,将得到的答案打印为ans。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to calculate total
// cost of visiting array
// elements in increasing order
int calculateDistance(int arr[], int N)
{
// Stores the pair of element
// and their positions
vector > v;
// Traverse the array arr[]
for (int i = 0; i < N; i++)
// Push the pair {arr[i], i} in v
v.push_back({ arr[i], i });
// Sort the vector in ascending order.
sort(v.begin(), v.end());
// Stores the total cost
int ans = 0;
// Stores the index of last element visited
int last = 0;
// Traverse the vector v
for (auto j : v) {
// Increment ans
ans += abs(j.second - last);
// Assign
last = j.second;
}
// Return ans
return ans;
}
// Driver Code
int main()
{
int arr[] = { 4, 3, 2, 5, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << calculateDistance(arr, N);
}
Java
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Pair class
static class Pair {
int first;
int second;
Pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to calculate total
// cost of visiting array
// elements in increasing order
static int calculateDistance(int arr[], int N)
{
// Stores the pair of element
// and their positions
Pair v[] = new Pair[N];
// Traverse the array arr[]
for (int i = 0; i < N; i++)
// Push the pair {arr[i], i} in v
v[i] = new Pair(arr[i], i);
// Sort the vector in ascending order.
Arrays.sort(v, (p1, p2) -> {
if (p1.first != p2.first)
return p1.first - p2.first;
return p1.second - p2.second;
});
// Stores the total cost
int ans = 0;
// Stores the index of last element visited
int last = 0;
// Traverse the vector v
for (Pair j : v) {
// Increment ans
ans += Math.abs(j.second - last);
// Assign
last = j.second;
}
// Return ans
return ans;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 4, 3, 2, 5, 1 };
int N = arr.length;
// Function call
System.out.println(calculateDistance(arr, N));
}
}
// This code is contributed by Kingash.
Python3
# Python3 implementation of the above approach
# Function to calculate total
# cost of visiting array
# elements in increasing order
def calculateDistance(arr, N):
# Stores the pair of element
# and their positions
v = []
# Traverse the array arr[]
for i in range(N):
# Push the pair {arr[i], i} in v
v.append([arr[i], i])
# Sort the vector in ascending order.
v.sort()
# Stores the total cost
ans = 0
# Stores the index of last element visited
last = 0
# Traverse the vector v
for j in v:
# Increment ans
ans += abs(j[1] - last)
# Assign
last = j[1]
# Return ans
return ans
# Driver Code
if __name__ == "__main__" :
arr = [ 4, 3, 2, 5, 1 ]
N = len(arr)
print(calculateDistance(arr, N))
# This code is contributed by AnkThon
11
时间复杂度: O(N)
辅助空间: O(N)
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