Given two arrays firstArr[], consisting of distinct elements only, and secondArr[], the task is to find the length of LCS between these 2 arrays.
Examples:
Input: firstArr[] = {3, 5, 1, 8}, secondArr[] = {3, 3, 5, 3, 8}
Output: 3.
Explanation: LCS between these two arrays is {3, 5, 8}.Input : firstArr[] = {1, 2, 1}, secondArr[] = {3}
Output: 0
Naive Approach: Follow the steps below to solve the problem using the simplest possible approach:
- Initialize an array dp[][] such that dp[i][j] stores longest common subsequence of firstArr[ :i] and secondArr[ :j].
- Traverse the array firstArr[] and for every array element of the array firstArr[], traverse the array secondArr[].
- If firstArr[i] = secondArr[j]: Set dp[i][j] = dp[i - 1][j - 1] + 1.
- Otherwise: Set dp[i][j] = max(dp[ i - 1][j], dp[i][j - 1]).
Time Complexity: O(N * M), where N and M are the sizes of the arrays firstArr[] and secondArr[] respectively.
Auxiliary Space: O(N * M)
Efficient Approach: To optimize the above approach, follow the steps below:
- Initialize a Map, say mp, to store the mappings map[firstArr[i]] = i, i.e. map elements of the first array to their respective indices.
- Since the elements which are present in secondArr[] but not in the firstArr[] are not useful at all, as they can never be a part of a common subsequence, traverse the array secondArr[] andfor each array element, check if it is present in the Map or not.
- If found to be true, push map[secondArr[i]] into a temporary Array. Otherwise ignore it.
- Find the Longest Increasing Subsequence (LIS) of the obtained temporary array and print its length as the required answer.
Illustration:
firstArr[] = {3, 5, 1, 8}
secondArr={3, 3, 4, 5, 3, 8}
Mapping: 3->0, 5->1, 1->2, 8->3 (From firstArr)
tempArr[] = {0, 0, 1, 0, 3}
Therefore, length of LIS of tempArr[] = 3 ({0, 1, 3})
Below is the implementation of the above approach:
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Longest Common
// Subsequence between the two arrays
int LCS(vector<int>& firstArr,
vector<int>& secondArr)
{
// Maps elements of firstArr[]
// to their respective indices
unordered_map<int, int> mp;
// Traverse the array firstArr[]
for (int i = 0; i < firstArr.size(); i++) {
mp[firstArr[i]] = i + 1;
}
// Stores the indices of common elements
// between firstArr[] and secondArr[]
vector<int> tempArr;
// Traverse the array secondArr[]
for (int i = 0; i < secondArr.size(); i++) {
// If current element exists in the Map
if (mp.find(secondArr[i]) != mp.end()) {
tempArr.push_back(mp[secondArr[i]]);
}
}
// Stores lIS from tempArr[]
vector<int> tail;
tail.push_back(tempArr[0]);
for (int i = 1; i < tempArr.size(); i++) {
if (tempArr[i] > tail.back())
tail.push_back(tempArr[i]);
else if (tempArr[i] < tail[0])
tail[0] = tempArr[i];
else {
auto it = lower_bound(tail.begin(),
tail.end(),
tempArr[i]);
*it = tempArr[i];
}
}
return (int)tail.size();
}
// Driver Code
int main()
{
vector<int> firstArr = { 3, 5, 1, 8 };
vector<int> secondArr = { 3, 3, 5, 3, 8 };
cout << LCS(firstArr, secondArr);
return 0;
}
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find the Longest Common
// Subsequence between the two arrays
static int LCS(int[] firstArr,int[] secondArr)
{
// Maps elements of firstArr[]
// to their respective indices
HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
// Traverse the array firstArr[]
for (int i = 0; i < firstArr.length; i++)
{
mp.put(firstArr[i], i + 1);
}
// Stores the indices of common elements
// between firstArr[] and secondArr[]
Vector<Integer> tempArr = new Vector<>();
// Traverse the array secondArr[]
for (int i = 0; i < secondArr.length; i++)
{
// If current element exists in the Map
if (mp.containsKey(secondArr[i]))
{
tempArr.add(mp.get(secondArr[i]));
}
}
// Stores lIS from tempArr[]
Vector<Integer> tail = new Vector<>();
tail.add(tempArr.get(0));
for (int i = 1; i < tempArr.size(); i++)
{
if (tempArr.get(i) > tail.lastElement())
tail.add(tempArr.get(i));
else if (tempArr.get(i) < tail.get(0))
tail.add(0, tempArr.get(i));
}
return (int)tail.size();
}
// Driver Code
public static void main(String[] args)
{
int[] firstArr = { 3, 5, 1, 8 };
int[] secondArr = { 3, 3, 5, 3, 8 };
System.out.print(LCS(firstArr, secondArr));
}
}
// This code is contributed by gauravrajput1
# Python3 program to implement
# the above approach
from bisect import bisect_left
# Function to find the Longest Common
# Subsequence between the two arrays
def LCS(firstArr, secondArr):
# Maps elements of firstArr[]
# to their respective indices
mp = {}
# Traverse the array firstArr[]
for i in range(len(firstArr)):
mp[firstArr[i]] = i + 1
# Stores the indices of common elements
# between firstArr[] and secondArr[]
tempArr = []
# Traverse the array secondArr[]
for i in range(len(secondArr)):
# If current element exists in the Map
if (secondArr[i] in mp):
tempArr.append(mp[secondArr[i]])
# Stores lIS from tempArr[]
tail = []
tail.append(tempArr[0])
for i in range(1, len(tempArr)):
if (tempArr[i] > tail[-1]):
tail.append(tempArr[i])
elif (tempArr[i] < tail[0]):
tail[0] = tempArr[i]
else :
it = bisect_left(tail, tempArr[i])
it = tempArr[i]
return len(tail)
# Driver Code
if __name__ == '__main__':
firstArr = [3, 5, 1, 8 ]
secondArr = [3, 3, 5, 3, 8 ]
print (LCS(firstArr, secondArr))
# This code is contributed by mohit kumar 29
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to find the longest Common
// Subsequence between the two arrays
static int LCS(int[] firstArr,int[] secondArr)
{
// Maps elements of firstArr[]
// to their respective indices
Dictionary<int,int> mp = new Dictionary<int,int>();
// Traverse the array firstArr[]
for (int i = 0; i < firstArr.Length; i++)
{
mp.Add(firstArr[i], i + 1);
}
// Stores the indices of common elements
// between firstArr[] and secondArr[]
List<int> tempArr = new List<int>();
// Traverse the array secondArr[]
for (int i = 0; i < secondArr.Length; i++)
{
// If current element exists in the Map
if (mp.ContainsKey(secondArr[i]))
{
tempArr.Add(mp[secondArr[i]]);
}
}
// Stores lIS from tempArr[]
List<int> tail = new List<int>();
tail.Add(tempArr[0]);
for (int i = 1; i < tempArr.Count; i++)
{
if (tempArr[i] > tail[tail.Count-1])
tail.Add(tempArr[i]);
else if (tempArr[i] < tail[0])
tail.Insert(0, tempArr[i]);
}
return (int)tail.Count;
}
// Driver Code
public static void Main(String[] args)
{
int[] firstArr = { 3, 5, 1, 8 };
int[] secondArr = { 3, 3, 5, 3, 8 };
Console.Write(LCS(firstArr, secondArr));
}
}
// This code is contributed by Rajput-Ji.
<script>
// Javascript program to implement
// the above approach
// Function to find the longest Common
// Subsequence between the two arrays
function LCS(firstArr, secondArr)
{
// Maps elements of firstArr[]
// to their respective indices
let mp = new Map()
// Traverse the array firstArr[]
for(let i = 0; i < firstArr.length; i++)
{
mp.set(firstArr[i], i + 1);
}
// Stores the indices of common elements
// between firstArr[] and secondArr[]
let tempArr = [];
// Traverse the array secondArr[]
for(let i = 0; i < secondArr.length; i++)
{
// If current element exists in the Map
if (mp.has(secondArr[i]))
{
tempArr.push(mp.get(secondArr[i]));
}
}
// Stores lIS from tempArr[]
let tail = [];
tail.push(tempArr[0]);
for(let i = 1; i < tempArr.length; i++)
{
if (tempArr[i] > tail[tail.length - 1])
tail.push(tempArr[i]);
else if (tempArr[i] < tail[0])
tail.unshift(0, tempArr[i]);
}
return tail.length;
}
// Driver Code
let firstArr = [ 3, 5, 1, 8 ];
let secondArr = [ 3, 3, 5, 3, 8 ];
document.write(LCS(firstArr, secondArr));
// This code is contributed by gfgking
</script>
Output:
3
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
