给定一个由N 个不同整数组成的数组A[]和另一个由M 个整数组成的数组B[] ，任务是找到要添加到数组B[] 中的最小元素数，使得数组A[]成为数组B[] 的子序列。
例子：
Input: N = 5, M = 6, A[] = {1, 2, 3, 4, 5}, B[] = {2, 5, 6, 4, 9, 12}
Output: 3
Explanation:
Below are the element that are needed to be added:
1) Add 1 before element 2 of B[]
2) Add 3 after element 6 of B[]
3) Add 5 in the last position of B[].
Therefore, the resulting array B[] is {1, 2, 5, 6, 3, 4, 9, 12, 5}.
Hence, A[] is the subsequence of B[] after adding 3 elements.
Input: N = 5, M = 5, A[] = {3, 4, 5, 2, 7}, B[] = {3, 4, 7, 9, 2}
Output: 2
Explanation:
Below are the elements that are needed to be added:
1) Add 5 after element 4.
2) Add 2 after element 5.
Therefore, the resulting array B[] is {3, 4, 5, 2, 7, 9, 2}.
Hence 2 elements are required to be added.
朴素的方法：朴素的方法是生成数组B 的所有子序列，然后找到该子序列，以便从数组A 中添加最少数量的元素使其等于数组A 。打印添加的元素的最小计数。
时间复杂度： O(N*2 M )
辅助空间： O(M+N)
高效方法：上述方法可以使用动态规划进行优化。这个想法是在给定的两个数组A和B之间找到最长公共子序列。主要观察结果是，可以通过从数组A[]的长度中减去最长公共子序列的长度来找到要在B[] 中添加以使A[]成为其子序列的最小元素数。
因此，数组A[]的长度与最长公共子序列的长度之差就是需要的结果。
下面是上述方法的实现：
C++14
// C++14 program for the above approach
#include
using namespace std;
// Function that finds the minimum number
// of the element must be added to make A
// as a subsequence in B
int transformSubsequence(int n, int m,
vector A,
vector B)
{
// Base Case
if (B.size() == 0)
return n;
// dp[i][j] indicates the length of
// LCS of A of length i & B of length j
vector> dp(n + 1,
vector(m + 1, 0));
for(int i = 0; i < n + 1; i++)
{
for(int j = 0; j < m + 1; j++)
{
// If there are no elements
// either in A or B then the
// length of lcs is 0
if (i == 0 or j == 0)
dp[i][j] = 0;
// If the element present at
// ith and jth index of A and B
// are equal then include in LCS
else if (A[i - 1] == B[j - 1])
dp[i][j] = 1 + dp[i - 1][j - 1];
// If they are not equal then
// take the max
else
dp[i][j] = max(dp[i - 1][j],
dp[i][j - 1]);
}
}
// Return difference of length
// of A and lcs of A and B
return n - dp[n][m];
}
// Driver Code
int main()
{
int N = 5;
int M = 6;
// Given sequence A and B
vector A = { 1, 2, 3, 4, 5 };
vector B = { 2, 5, 6, 4, 9, 12 };
// Function call
cout << transformSubsequence(N, M, A, B);
return 0;
}
// This code is contributed by mohit kumar 29 Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function that finds the minimum number
// of the element must be added to make A
// as a subsequence in B
static int transformSubsequence(int n, int m,
int []A, int []B)
{
// Base Case
if (B.length == 0)
return n;
// dp[i][j] indicates the length of
// LCS of A of length i & B of length j
int [][]dp = new int[n + 1][m + 1];
for(int i = 0; i < n + 1; i++)
{
for(int j = 0; j < m + 1; j++)
{
// If there are no elements
// either in A or B then the
// length of lcs is 0
if (i == 0 || j == 0)
dp[i][j] = 0;
// If the element present at
// ith and jth index of A and B
// are equal then include in LCS
else if (A[i - 1] == B[j - 1])
dp[i][j] = 1 + dp[i - 1][j - 1];
// If they are not equal then
// take the max
else
dp[i][j] = Math.max(dp[i - 1][j],
dp[i][j - 1]);
}
}
// Return difference of length
// of A and lcs of A and B
return n - dp[n][m];
}
// Driver Code
public static void main(String[] args)
{
int N = 5;
int M = 6;
// Given sequence A and B
int []A = {1, 2, 3, 4, 5};
int []B = {2, 5, 6, 4, 9, 12};
// Function call
System.out.print(transformSubsequence(N, M, A, B));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program for the above approach
# Function that finds the minimum number
# of the element must be added to make A
# as a subsequence in B
def transformSubsequence(n, m, A, B):
# Base Case
if B is None or len(B) == 0:
return n
# dp[i][j] indicates the length of
# LCS of A of length i & B of length j
dp = [[0 for col in range(m + 1)]
for row in range(n + 1)]
for i in range(n + 1):
for j in range(m + 1):
# If there are no elements
# either in A or B then the
# length of lcs is 0
if i == 0 or j == 0:
dp[i][j] = 0
# If the element present at
# ith and jth index of A and B
# are equal then include in LCS
elif A[i-1] == B[j-1]:
dp[i][j] = 1 + dp[i-1][j-1]
# If they are not equal then
# take the max
else:
dp[i][j] = max(dp[i-1][j], dp[i][j-1])
# Return difference of length
# of A and lcs of A and B
return n - dp[n][m]
# Driver Code
if __name__ == "__main__":
N = 5
M = 6
# Given Sequence A and B
A = [1, 2, 3, 4, 5]
B = [2, 5, 6, 4, 9, 12]
# Function Call
print(transformSubsequence(N, M, A, B))C#
// C# program for
// the above approach
using System;
class GFG{
// Function that finds the minimum number
// of the element must be added to make A
// as a subsequence in B
static int transformSubsequence(int n, int m,
int []A, int []B)
{
// Base Case
if (B.Length == 0)
return n;
// dp[i,j] indicates the length of
// LCS of A of length i & B of length j
int [,]dp = new int[n + 1, m + 1];
for(int i = 0; i < n + 1; i++)
{
for(int j = 0; j < m + 1; j++)
{
// If there are no elements
// either in A or B then the
// length of lcs is 0
if (i == 0 || j == 0)
dp[i, j] = 0;
// If the element present at
// ith and jth index of A and B
// are equal then include in LCS
else if (A[i - 1] == B[j - 1])
dp[i, j] = 1 + dp[i - 1, j - 1];
// If they are not equal then
// take the max
else
dp[i, j] = Math.Max(dp[i - 1, j],
dp[i, j - 1]);
}
}
// Return difference of length
// of A and lcs of A and B
return n - dp[n, m];
}
// Driver Code
public static void Main(String[] args)
{
int N = 5;
int M = 6;
// Given sequence A and B
int []A = {1, 2, 3, 4, 5};
int []B = {2, 5, 6, 4, 9, 12};
// Function call
Console.Write(transformSubsequence(N, M,
A, B));
}
}
// This code is contributed by Rajput-JiJavascript
输出：
3时间复杂度： O(M*M)，其中 N 和 M 分别是数组 A[] 和 B[] 的长度。
辅助空间： O(M*N)
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