Minimize elements to be added to a given array such that it contains another given array as its subsequence

Given an array A[] consisting of N distinct integers and another array B[] consisting of M integers, the task is to find the minimum number of elements to be added to the array B[] such that the array A[] becomes the subsequence of the array B[].

Examples:

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.

Naive Approach: The naive approach is to generate all the subsequences of the array B and then find that subsequence such that on adding a minimum number of elements from the array A to make it equal to the array A. Print the minimum count of element added.
Time Complexity: O(N*2^M)
Auxiliary Space: O(M+N) 

Efficient Approach: The above approach can be optimized using Dynamic Programming. The idea is to find the Longest Common Subsequence between the given two arrays A and B. The main observation is that the minimum number of elements to be added in B[] such that A[] becomes its subsequence can be found by subtracting the length of the longest common subsequence from the length of the array A[].

Therefore, the difference between the length of the array A[] and length of the Longest Common Subsequence is the required result.

Below is the implementation of the above approach:

3

Time Complexity: O(M*M), where N and M are the lengths of array A[] and B[] respectively.
Auxiliary Space: O(M*N)

Efficient approach : Space optimization

In previous approach the dp[i][j] is depend upon the current and previous row of 2D matrix. So to optimize space we use a 1D vectors dp to store previous value  and use prev to store the previous diagonal element and get the current computation. 

Implementation Steps:

• Define a vector dp of size m+1 and initialize its first element to 0.
• For each element j in B, iterate in reverse order from n to 1 and update dp[i] as follows:
  a. If A[i-1] == B[j-1], set dp[i] to the previous value of dp[i-1] (diagonal element).
  b. If A[i-1] != B[j-1], set dp[i] to the maximum value between dp[i] and dp[i-1]+1 (value on the left).
• Finally, return n - dp[m].

Implementation:

Output

3

Time Complexity: O(M*M), where N and M are the lengths of array A[] and B[] respectively.
Auxiliary Space: O(M)