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 elements that need 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 need 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: Refer to the previous post of this article for the simplest approach to solve the problem.
Time Complexity: O(N * 2M)
Auxiliary Space: O(M + N)
Dynamic Programming Approach: Refer to the previous post of this article for the Longest Common Subsequence based approach.
Time Complexity: O(N * M)
Auxiliary Space: O(N * M)
Efficient Approach: The idea is similar to finding the Longest Increasing Subsequence(LIS) from the array B[]. Follow the steps below to solve the problem:
- Consider elements of array B[] which are present in the array A[], and store the indices of each element of the array A[] in a Map
- Then, find the LIS array subseq[] using Binary Search which consists of the indices in increasing order.
- Finally, the minimum number of elements to be inserted into array B[] is equal to N – len(LIS), where len(LIS) is calculated using Binary Search in the above steps.
Below is the implementation of the above approach:
// C++ program for the // above approach #include <bits/stdc++.h> using namespace std;
// Function to return minimum // element to be added in array // B so that array A become // subsequence of array B int minElements( int A[], int B[],
int N, int M)
{ // Stores indices of the
// array elements
map< int , int > map;
// Iterate over the array
for ( int i = 0; i < N; i++)
{
// Store the indices of
// the array elements
map[A[i]] = i;
}
// Stores the LIS
vector< int > subseq;
int l = 0, r = -1;
for ( int i = 0; i < M; i++)
{
// Check if element B[i]
// is in array A[]
if (map.find(B[i]) !=
map.end())
{
int e = map[B[i]];
// Perform Binary Search
while (l <= r)
{
// Find the value of
// mid m
int m = l + (r - l) / 2;
// Update l and r
if (subseq[m] < e)
l = m + 1;
else
r = m - 1;
}
// If found better element
// 'e' for pos r + 1
if (r + 1 < subseq.size())
{
subseq[r + 1] = e;
}
// Otherwise, extend the
// current subsequence
else
{
subseq.push_back(e);
}
l = 0;
r = subseq.size() - 1;
}
}
// Return the answer
return N - subseq.size();
} // Driver code int main()
{ // Given arrays
int A[] = {1, 2, 3, 4, 5};
int B[] = {2, 5, 6, 4, 9, 12};
int M = sizeof (A) /
sizeof (A[0]);
int N = sizeof (B) /
sizeof (B[0]);
// Function Call
cout << minElements(A, B,
M, N);
return 0;
} // This code is contributed by divyeshrabadiya07 |
// Java program for the above approach import java.util.*;
import java.lang.*;
class GFG {
// Function to return minimum element
// to be added in array B so that array
// A become subsequence of array B
static int minElements(
int [] A, int [] B, int N, int M)
{
// Stores indices of the
// array elements
Map<Integer, Integer> map
= new HashMap<>();
// Iterate over the array
for ( int i = 0 ;
i < A.length; i++) {
// Store the indices of
// the array elements
map.put(A[i], i);
}
// Stores the LIS
ArrayList<Integer> subseq
= new ArrayList<>();
int l = 0 , r = - 1 ;
for ( int i = 0 ; i < M; i++) {
// Check if element B[i]
// is in array A[]
if (map.containsKey(B[i])) {
int e = map.get(B[i]);
// Perform Binary Search
while (l <= r) {
// Find the value of
// mid m
int m = l + (r - l) / 2 ;
// Update l and r
if (subseq.get(m) < e)
l = m + 1 ;
else
r = m - 1 ;
}
// If found better element
// 'e' for pos r + 1
if (r + 1 < subseq.size()) {
subseq.set(r + 1 , e);
}
// Otherwise, extend the
// current subsequence
else {
subseq.add(e);
}
l = 0 ;
r = subseq.size() - 1 ;
}
}
// Return the answer
return N - subseq.size();
}
// Driver Code
public static void main(String[] args)
{
// Given arrays
int [] A = { 1 , 2 , 3 , 4 , 5 };
int [] B = { 2 , 5 , 6 , 4 , 9 , 12 };
int M = A.length;
int N = B.length;
// Function Call
System.out.println(
minElements(A, B, M, N));
}
} |
# Python3 program for the above approach # Function to return minimum element # to be added in array B so that array # A become subsequence of array B def minElements(A, B, N, M):
# Stores indices of the
# array elements
map = {}
# Iterate over the array
for i in range ( len (A)):
# Store the indices of
# the array elements
map [A[i]] = i
# Stores the LIS
subseq = []
l = 0
r = - 1
for i in range (M):
# Check if element B[i]
# is in array A[]
if B[i] in map :
e = map [B[i]]
# Perform Binary Search
while (l < = r):
# Find the value of
# mid m
m = l + (r - l) / / 2
# Update l and r
if (subseq[m] < e):
l = m + 1
else :
r = m - 1
# If found better element
# 'e' for pos r + 1
if (r + 1 < len (subseq)):
subseq[r + 1 ] = e
# Otherwise, extend the
# current subsequence
else :
subseq.append(e)
l = 0
r = len (subseq) - 1
# Return the answer
return N - len (subseq)
# Driver Code if __name__ = = '__main__' :
# Given arrays
A = [ 1 , 2 , 3 , 4 , 5 ]
B = [ 2 , 5 , 6 , 4 , 9 , 12 ]
M = len (A)
N = len (B)
# Function call
print (minElements(A, B, M, N))
# This code is contributed by mohit kumar 29 |
// C# program for // the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to return minimum element // to be added in array B so that array // A become subsequence of array B static int minElements( int [] A, int [] B,
int N, int M)
{ // Stores indices of the
// array elements
Dictionary< int ,
int > map = new Dictionary< int ,
int >();
// Iterate over the array
for ( int i = 0;
i < A.Length; i++)
{
// Store the indices of
// the array elements
map.Add(A[i], i);
}
// Stores the LIS
List< int > subseq = new List< int >();
int l = 0, r = -1;
for ( int i = 0; i < M; i++)
{
// Check if element B[i]
// is in array []A
if (map.ContainsKey(B[i]))
{
int e = map[B[i]];
// Perform Binary Search
while (l <= r)
{
// Find the value of
// mid m
int m = l + (r - l) / 2;
// Update l and r
if (subseq[m] < e)
l = m + 1;
else
r = m - 1;
}
// If found better element
// 'e' for pos r + 1
if (r + 1 < subseq.Count)
{
subseq[r + 1] = e;
}
// Otherwise, extend the
// current subsequence
else
{
subseq.Add(e);
}
l = 0;
r = subseq.Count - 1;
}
}
// Return the answer
return N - subseq.Count;
} // Driver Code public static void Main(String[] args)
{ // Given arrays
int [] A = {1, 2, 3, 4, 5};
int [] B = {2, 5, 6, 4, 9, 12};
int M = A.Length;
int N = B.Length;
// Function Call
Console.WriteLine(minElements(A, B,
M, N));
} } // This code is contributed by Princi Singh |
<script> // Javascript program for the // above approach // Function to return minimum // element to be added in array // B so that array A become // subsequence of array B function minElements(A, B, N, M)
{ // Stores indices of the
// array elements
var map = new Map();
// Iterate over the array
for ( var i = 0; i < N; i++)
{
// Store the indices of
// the array elements
map.set(A[i], i);
}
// Stores the LIS
var subseq = [];
var l = 0, r = -1;
for ( var i = 0; i < M; i++)
{
// Check if element B[i]
// is in array A[]
if (map.has(B[i]))
{
var e = map.get(B[i]);
// Perform Binary Search
while (l <= r)
{
// Find the value of
// mid m
var m = l + parseInt((r - l) / 2);
// Update l and r
if (subseq[m] < e)
l = m + 1;
else
r = m - 1;
}
// If found better element
// 'e' for pos r + 1
if (r + 1 < subseq.length)
{
subseq[r + 1] = e;
}
// Otherwise, extend the
// current subsequence
else
{
subseq.push(e);
}
l = 0;
r = subseq.length - 1;
}
}
// Return the answer
return N - subseq.length;
} // Driver code // Given arrays var A = [1, 2, 3, 4, 5];
var B = [2, 5, 6, 4, 9, 12];
var M = A.length;
var N = B.length;
// Function Call document.write( minElements(A, B, M, N));
</script> |
3
Time Complexity: O(N logN)
Auxiliary Space: O(N)