给定两个数组A []和B [] ,它们都由前N个自然数组成,则任务是计算最后一个数组元素需要移到数组A []中任意位置的最小次数。 ]使数组A []和B []相等。
例子:
Input: A[] = {1, 2, 3, 4, 5}, B[] = {1, 5, 2, 3, 4}
Output:1
Explanation:
Initially, the array A[] is {1, 2, 3, 4, 5}. After moving the last array element, i.e. 5, and placing them between arr[0] (= 1) and arr[1](= 2) modifies the array to {1, 5, 2, 3, 4}, which is the same as the array B[].
Therefore, the minimum number of operations required to convert the array A[] to B[] is 1.
Input: A[] = {3, 2, 1}, B[] = {1, 2, 3}
Output: 2
Explanation:
Initially, the array A[] is {3, 2, 1}.
Operation 1: After moving the last array element, i.e. 1, to the beginning of the array, modifies the array to {1, 3, 2}.
Operation 2: After moving the last element of the array, i.e. 2 and placing them between the elements arr[0] (= 1) and arr[1] (= 3) modifies the array to {1, 2, 3}, which is the same as the array B[].
Therefore, the minimum number of operations required to convert the array A[] to B[] is 2.
方法:给定的问题可以通过找到第一个排列的前i个连续元素与第二个排列的子序列相同来解决,因此自最后一个运算起,操作数必须至少少(N – I )个(N – i)个元素可以最佳选择,并以所需的索引插入。因此, (N – i)是将数组A []转换为B []所需的最小步数。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
int minCount(int A[], int B[], int N)
{
// Stores the index in the first
// permutation A[] which is same
// as the subsequence in B[]
int i = 0;
// Find the first i elements in A[]
// which is a subsequence in B[]
for (int j = 0; j < N; j++) {
// If element A[i]
// is same as B[j]
if (A[i] == B[j]) {
i++;
}
}
// Return the count of
// operations required
return N - i;
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3, 4, 5 };
int B[] = { 1, 5, 2, 3, 4 };
int N = sizeof(A) / sizeof(A[0]);
cout << minCount(A, B, N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
static int minCount(int A[], int B[], int N)
{
// Stores the index in the first
// permutation A[] which is same
// as the subsequence in B[]
int i = 0;
// Find the first i elements in A[]
// which is a subsequence in B[]
for(int j = 0; j < N; j++)
{
// If element A[i]
// is same as B[j]
if (A[i] == B[j])
{
i++;
}
}
// Return the count of
// operations required
return N - i;
}
// Driver Code
public static void main (String[] args)
{
int A[] = { 1, 2, 3, 4, 5 };
int B[] = { 1, 5, 2, 3, 4 };
int N = A.length;
System.out.println(minCount(A, B, N));
}
}
// This code is contributed by AnkThon
Python3
# Python3 program for the above approach
# Function to count the minimum number
# of operations required to convert
# the array A[] into array B[]
def minCount(A, B, N):
# Stores the index in the first
# permutation A[] which is same
# as the subsequence in B[]
i = 0
# Find the first i elements in A[]
# which is a subsequence in B[]
for j in range(N):
# If element A[i]
# is same as B[j]
if (A[i] == B[j]):
i += 1
# Return the count of
# operations required
return N - i
# Driver Code
if __name__ == '__main__':
A = [ 1, 2, 3, 4, 5 ]
B = [ 1, 5, 2, 3, 4 ]
N = len(A)
print(minCount(A, B, N))
# This code is contributed by ipg2016107
C#
// C# program for the above approach
using System;
class GFG{
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
static int minCount(int[] A, int[] B, int N)
{
// Stores the index in the first
// permutation A[] which is same
// as the subsequence in B[]
int i = 0;
// Find the first i elements in A[]
// which is a subsequence in B[]
for(int j = 0; j < N; j++)
{
// If element A[i]
// is same as B[j]
if (A[i] == B[j])
{
i++;
}
}
// Return the count of
// operations required
return N - i;
}
// Driver Code
public static void Main(string[] args)
{
int[] A = { 1, 2, 3, 4, 5 };
int[] B = { 1, 5, 2, 3, 4 };
int N = A.Length;
Console.WriteLine(minCount(A, B, N));
}
}
// This code is contributed by ukasp
Javascript
1
时间复杂度: O(N)
辅助空间: O(1)