Dadas dos arrays firstArr[] , que consisten solo en elementos distintos, y secondArr[] , la tarea es encontrar la longitud de LCS entre estas 2 arrays.
Ejemplos:
Entrada: firstArr[] = {3, 5, 1, 8}, secondArr[] = {3, 3, 5, 3, 8}
Salida: 3.
Explicación: LCS entre estas dos arrays es {3, 5, 8} .Entrada : primeraArr[] = {1, 2, 1}, segundaArr[] = {3}
Salida: 0
Enfoque ingenuo: siga los pasos a continuación para resolver el problema utilizando el enfoque más simple posible:
- Inicialice una array dp[][] tal que dp[i][j] almacene la subsecuencia común más larga de firstArr[ :i] y secondArr[ :j] .
- Recorra la array firstArr[] y para cada elemento de array de la array firstArr[] , recorra la array secondArr[] .
- Si firstArr[i] = secondArr[j]: Establecer dp[i][j] = dp[i – 1][j – 1] + 1 .
- De lo contrario: Establezca dp[i][j] = max(dp[ i – 1][j], dp[i][j – 1]) .
Complejidad temporal: O(N * M), donde N y M son los tamaños de las arrays firstArr[] y secondArr[] respectivamente.
Espacio Auxiliar: O(N * M)
Enfoque eficiente: para optimizar el enfoque anterior, siga los pasos a continuación:
- Inicialice un Map , digamos mp , para almacenar las asignaciones map[firstArr[i]] = i , es decir, asigne elementos de la primera array a sus respectivos índices.
- Dado que los elementos que están presentes en secondArr[] pero no en firstArr[] no son útiles en absoluto, ya que nunca pueden ser parte de una subsecuencia común, recorra la array secondArr[] y para cada elemento de la array, verifique si es presentes en el Mapa o no .
- Si se determina que es cierto, inserte map[ secondArr [i]] en una array temporal. De lo contrario, ignóralo.
- Encuentre la subsecuencia creciente más larga (LIS) de la array temporal obtenida e imprima su longitud como la respuesta requerida.
Ilustración:
firstArr[] = {3, 5, 1, 8}
secondArr={3, 3, 4, 5, 3, 8}
Asignación : 3->0, 5->1, 1->2, 8->3 ( De firstArr)
tempArr[] = {0, 0, 1, 0, 3}
Por lo tanto, la longitud de LIS de tempArr[] = 3 ({0, 1, 3})
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to 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
// Java program to 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
# Python3 program to 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#
// C# program to 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.
Javascript
<script>
// Javascript program to 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>
Producción:
3
Complejidad temporal: O(NlogN)
Espacio auxiliar: O(N)