给定两个由不同元素组成的大小为N 的数组A[]和B[] ,任务是找到交换两个给定数组的最小成本。交换两个元素A[i]和B[j] 的成本是min(A[i], A[j]) 。总成本是所有掉期操作成本的累计总和。
注意:这里,交换后元素的顺序可能与原始数组不同。
例子:
Input: N = 3, A[] = {1, 4, 2}, B[] = {10, 6, 12}
Output: 5
Explanation:
Following swap operations will give the minimum cost:
swap(A[0], B[2]): cost = min(A[0], B[2]) = 1, A[ ] = {12, 4, 2}, B[ ] = {10, 6, 1}
swap(A[2], B[2]): cost = min(A[2], B[2]) = 1, A[ ] = {12, 4, 1}, B[ ] = {10, 6, 2}
swap(A[2], B[0]): cost = min(A[2], B[0]) = 1, A[ ] = {12, 4, 10}, B[ ] = {1, 6, 2}
swap(A[1], B[0]): cost = min(A[1], B[0]) = 1, A[ ] = {12, 1, 10}, B[ ] = {4, 6, 2}
swap(A[1], B[1]): cost = min(A[1], B[1]) = 1, A[ ] = {12, 6, 10}, B[ ] = {4, 1, 2}
Therefore, the minimum cost to swap two arrays = 1 + 1 + 1 + 1 + 1 = 5
Input: N = 2, A[] = {9, 12}, B[] = {3, 15}
Output: 9
方法:
请按照以下步骤解决问题:
- 同时遍历数组并从中找到最小元素,例如K 。
- 现在,在交换两个数组之前,每个元素都带有 K。因此,所需的交换次数为2*N – 1 。
- 打印K * (2 * N – 1)作为答案。
下面是上述方法的实现:
C++
// C++ program to implement
// the above appraoch
#include
using namespace std;
// Function to calculate and return the
// minimum cost required to swap two arrays
int getMinCost(vector A, vector B,
int N)
{
int mini = INT_MAX;
for (int i = 0; i < N; i++) {
mini = min(mini, min(A[i], B[i]));
}
// Return the total minimum cost
return mini * (2 * N - 1);
}
// Driver Code
int main()
{
int N = 3;
vector A = { 1, 4, 2 };
vector B = { 10, 6, 12 };
cout << getMinCost(A, B, N);
return 0;
}
Java
// Java program to implement
// the above appraoch
class GFG{
// Function to calculate and return the
// minimum cost required to swap two arrays
static int getMinCost(int [] A, int [] B,
int N)
{
int mini = Integer.MAX_VALUE;
for (int i = 0; i < N; i++)
{
mini = Math.min(mini,
Math.min(A[i], B[i]));
}
// Return the total minimum cost
return mini * (2 * N - 1);
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
int [] A = { 1, 4, 2 };
int [] B = { 10, 6, 12 };
System.out.print(getMinCost(A, B, N));
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 program to implement
# the above appraoch
import sys
# Function to calculate and return the
# minimum cost required to swap two arrays
def getMinCost(A, B, N):
mini = sys.maxsize
for i in range(N):
mini = min(mini, min(A[i], B[i]))
# Return the total minimum cost
return mini * (2 * N - 1)
# Driver Code
N = 3
A = [ 1, 4, 2 ]
B = [ 10, 6, 12 ]
print(getMinCost(A, B, N))
# This code is contributed by chitranayal
C#
// C# program to implement
// the above appraoch
using System;
class GFG{
// Function to calculate and return the
// minimum cost required to swap two arrays
static int getMinCost(int[] A, int[] B, int N)
{
int mini = int.MaxValue;
for (int i = 0; i < N; i++)
{
mini = Math.Min(mini, Math.Min(A[i], B[i]));
}
// Return the total minimum cost
return mini * (2 * N - 1);
}
// Driver Code
public static void Main(String[] args)
{
int N = 3;
int[] A = {1, 4, 2};
int[] B = {10, 6, 12};
Console.Write(getMinCost(A, B, N));
}
}
// This code is contributed by shikhasingrajput
Javascript
5
时间复杂度: O(N)
辅助空间: O(1)
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