Dada una array de enteros, encuentre la longitud de la subsecuencia más larga de modo que los elementos de la subsecuencia sean enteros consecutivos, los números consecutivos pueden estar en cualquier orden.
Ejemplos :
C++
// C++ program to find longest
// contiguous subsequence
#include <bits/stdc++.h>
using namespace std;
// Returns length of the longest
// contiguous subsequence
int findLongestConseqSubseq(int arr[], int n)
{
int ans = 0, count = 0;
// sort the array
sort(arr, arr + n);
vector<int> v;
v.push_back(arr[0]);
//insert repeated elements only once in the vector
for (int i = 1; i < n; i++)
{
if (arr[i] != arr[i - 1])
v.push_back(arr[i]);
}
// find the maximum length
// by traversing the array
for (int i = 0; i < v.size(); i++)
{
// Check if the current element is equal
// to previous element +1
if (i > 0 && v[i] == v[i - 1] + 1)
count++;
// reset the count
else
count = 1;
// update the maximum
ans = max(ans, count);
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 2, 3 };
int n = sizeof arr / sizeof arr[0];
cout << "Length of the Longest contiguous subsequence "
"is "
<< findLongestConseqSubseq(arr, n);
return 0;
}
Java
// Java program to find longest
// contiguous subsequence
import java.io.*;
import java.util.*;
class GFG
{
static int findLongestConseqSubseq(int arr[],
int n)
{
// Sort the array
Arrays.sort(arr);
int ans = 0, count = 0;
ArrayList<Integer> v = new ArrayList<Integer>();
v.add(10);
// Insert repeated elements
// only once in the vector
for (int i = 1; i < n; i++)
{
if (arr[i] != arr[i - 1])
v.add(arr[i]);
}
// Find the maximum length
// by traversing the array
for (int i = 0; i < v.size(); i++)
{
// Check if the current element is
// equal to previous element +1
if (i > 0 &&v.get(i) == v.get(i - 1) + 1)
count++;
else
count = 1;
// Update the maximum
ans = Math.max(ans, count);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
int n = arr.length;
System.out.println(
"Length of the Longest "
+ "contiguous subsequence is "
+ findLongestConseqSubseq(arr, n));
}
}
// This code is contributed by parascoding
Python3
# Python3 program to find longest
# contiguous subsequence
# Returns length of the longest
# contiguous subsequence
def findLongestConseqSubseq(arr, n):
ans = 0
count = 0
# Sort the array
arr.sort()
v = []
v.append(arr[0])
# Insert repeated elements only
# once in the vector
for i in range(1, n):
if (arr[i] != arr[i - 1]):
v.append(arr[i])
# Find the maximum length
# by traversing the array
for i in range(len(v)):
# Check if the current element is
# equal to previous element +1
if (i > 0 and v[i] == v[i - 1] + 1):
count += 1
# Reset the count
else:
count = 1
# Update the maximum
ans = max(ans, count)
return ans
# Driver code
arr = [ 1, 2, 2, 3 ]
n = len(arr)
print("Length of the Longest contiguous subsequence is",
findLongestConseqSubseq(arr, n))
# This code is contributed by avanitrachhadiya2155
C#
// C# program to find longest
// contiguous subsequence
using System;
using System.Collections.Generic;
class GFG{
static int findLongestConseqSubseq(int[] arr,
int n)
{
// Sort the array
Array.Sort(arr);
int ans = 0, count = 0;
List<int> v = new List<int>();
v.Add(10);
// Insert repeated elements
// only once in the vector
for(int i = 1; i < n; i++)
{
if (arr[i] != arr[i - 1])
v.Add(arr[i]);
}
// Find the maximum length
// by traversing the array
for(int i = 0; i < v.Count; i++)
{
// Check if the current element is
// equal to previous element +1
if (i > 0 && v[i] == v[i - 1] + 1)
count++;
else
count = 1;
// Update the maximum
ans = Math.Max(ans, count);
}
return ans;
}
// Driver code
static void Main()
{
int[] arr = { 1, 9, 3, 10, 4, 20, 2 };
int n = arr.Length;
Console.WriteLine("Length of the Longest " +
"contiguous subsequence is " +
findLongestConseqSubseq(arr, n));
}
}
// This code is contributed by divyeshrabadiya07
Javascript
<script>
// JavaScript program to find longest
// contiguous subsequence
// Returns length of the longest
// contiguous subsequence
function findLongestConseqSubseq(arr, n) {
let ans = 0, count = 0;
// sort the array
arr.sort(function (a, b) { return a - b; })
var v = [];
v.push(arr[0]);
//insert repeated elements only once in the vector
for (let i = 1; i < n; i++) {
if (arr[i] != arr[i - 1])
v.push(arr[i]);
}
// find the maximum length
// by traversing the array
for (let i = 0; i < v.length; i++) {
// Check if the current element is equal
// to previous element +1
if (i > 0 && v[i] == v[i - 1] + 1)
count++;
// reset the count
else
count = 1;
// update the maximum
ans = Math.max(ans, count);
}
return ans;
}
// Driver code
let arr = [1, 2, 2, 3];
let n = arr.length;
document.write(
"Length of the Longest contiguous subsequence is "
+findLongestConseqSubseq(arr, n)
);
// This code is contributed by Potta Lokesh
</script>
C++
// C++ program to find longest
// contiguous subsequence
#include <bits/stdc++.h>
using namespace std;
// Returns length of the longest
// contiguous subsequence
int findLongestConseqSubseq(int arr[], int n)
{
unordered_set<int> S;
int ans = 0;
// Hash all the array elements
for (int i = 0; i < n; i++)
S.insert(arr[i]);
// check each possible sequence from
// the start then update optimal length
for (int i = 0; i < n; i++)
{
// if current element is the starting
// element of a sequence
if (S.find(arr[i] - 1) == S.end())
{
// Then check for next elements
// in the sequence
int j = arr[i];
while (S.find(j) != S.end())
j++;
// update optimal length if
// this length is more
ans = max(ans, j - arr[i]);
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
int n = sizeof arr / sizeof arr[0];
cout << "Length of the Longest contiguous subsequence "
"is "
<< findLongestConseqSubseq(arr, n);
return 0;
}
Java
// Java program to find longest
// consecutive subsequence
import java.io.*;
import java.util.*;
class ArrayElements {
// Returns length of the longest
// consecutive subsequence
static int findLongestConseqSubseq(int arr[], int n)
{
HashSet<Integer> S = new HashSet<Integer>();
int ans = 0;
// Hash all the array elements
for (int i = 0; i < n; ++i)
S.add(arr[i]);
// check each possible sequence from the start
// then update optimal length
for (int i = 0; i < n; ++i)
{
// if current element is the starting
// element of a sequence
if (!S.contains(arr[i] - 1))
{
// Then check for next elements
// in the sequence
int j = arr[i];
while (S.contains(j))
j++;
// update optimal length if this
// length is more
if (ans < j - arr[i])
ans = j - arr[i];
}
}
return ans;
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
int n = arr.length;
System.out.println(
"Length of the Longest consecutive subsequence is "
+ findLongestConseqSubseq(arr, n));
}
}
// This code is contributed by Aakash Hasija
Python3
# Python program to find longest contiguous subsequence
def findLongestConseqSubseq(arr, n):
s = set()
ans = 0
# Hash all the array elements
for ele in arr:
s.add(ele)
# check each possible sequence from the start
# then update optimal length
for i in range(n):
# if current element is the starting
# element of a sequence
if (arr[i]-1) not in s:
# Then check for next elements in the
# sequence
j = arr[i]
while(j in s):
j += 1
# update optimal length if this length
# is more
ans = max(ans, j-arr[i])
return ans
# Driver code
if __name__ == '__main__':
n = 7
arr = [1, 9, 3, 10, 4, 20, 2]
print ("Length of the Longest contiguous subsequence is ",findLongestConseqSubseq(arr, n))
# Contributed by: Harshit Sidhwa
C#
using System;
using System.Collections.Generic;
// C# program to find longest consecutive subsequence
public class ArrayElements {
// Returns length of the
// longest consecutive subsequence
public static int findLongestConseqSubseq(int[] arr,
int n)
{
HashSet<int> S = new HashSet<int>();
int ans = 0;
// Hash all the array elements
for (int i = 0; i < n; ++i) {
S.Add(arr[i]);
}
// check each possible sequence from the start
// then update optimal length
for (int i = 0; i < n; ++i)
{
// if current element is the starting
// element of a sequence
if (!S.Contains(arr[i] - 1))
{
// Then check for next elements in the
// sequence
int j = arr[i];
while (S.Contains(j))
{
j++;
}
// update optimal length if this length
// is more
if (ans < j - arr[i])
{
ans = j - arr[i];
}
}
}
return ans;
}
// Driver code
public static void Main(string[] args)
{
int[] arr = new int[] { 1, 9, 3, 10, 4, 20, 2 };
int n = arr.Length;
Console.WriteLine(
"Length of the Longest consecutive subsequence is "
+ findLongestConseqSubseq(arr, n));
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Javascript program to find longest
// contiguous subsequence
// Returns length of the longest
// contiguous subsequence
function findLongestConseqSubseq(arr, n) {
let S = new Set();
let ans = 0;
// Hash all the array elements
for (let i = 0; i < n; i++)
S.add(arr[i]);
// check each possible sequence from
// the start then update optimal length
for (let i = 0; i < n; i++)
{
// if current element is the starting
// element of a sequence
if (!S.has(arr[i] - 1))
{
// Then check for next elements
// in the sequence
let j = arr[i];
while (S.has(j))
j++;
// update optimal length if
// this length is more
ans = Math.max(ans, j - arr[i]);
}
}
return ans;
}
// Driver code
let arr = [1, 9, 3, 10, 4, 20, 2];
let n = arr.length;
document.write("Length of the Longest contiguous subsequence is "
+ findLongestConseqSubseq(arr, n));
// This code is contributed by gfgking.
</script>
C++
// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
int findLongestConseqSubseq(int arr[], int N)
{
priority_queue<int, vector<int>, greater<int> > pq;
for (int i = 0; i < N; i++) {
// adding element from
// array to PriorityQueue
pq.push(arr[i]);
}
// Storing the first element
// of the Priority Queue
// This first element is also
// the smallest element
int prev = pq.top();
pq.pop();
// Taking a counter variable with value 1
int c = 1;
// Storing value of max as 1
// as there will always be
// one element
int max = 1;
while (!pq.empty()) {
// check if current peek
// element minus previous
// element is greater than
// 1 This is done because
// if it's greater than 1
// then the sequence
// doesn't start or is broken here
if (pq.top() - prev > 1) {
// Store the value of counter to 1
// As new sequence may begin
c = 1;
// Update the previous position with the
// current peek And remove it
prev = pq.top();
pq.pop();
}
// Check if the previous
// element and peek are same
else if (pq.top() - prev == 0) {
// Update the previous position with the
// current peek And remove it
prev = pq.top();
pq.pop();
}
// If the difference
// between previous element and peek is 1
else {
// Update the counter
// These are consecutive elements
c++;
// Update the previous position
// with the current peek And remove it
prev = pq.top();
pq.pop();
}
// Check if current longest
// subsequence is the greatest
if (max < c) {
// Store the current subsequence count as
// max
max = c;
}
}
return max;
}
// Driver Code
int main()
{
int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
int n = 7;
cout << "Length of the Longest consecutive subsequence "
"is "
<< findLongestConseqSubseq(arr, n);
return 0;
}
// this code is contributed by Manu Pathria
Java
// Java Program to find longest consecutive
// subsequence This Program uses Priority Queue
import java.io.*;
import java.util.PriorityQueue;
public class Longset_Sub
{
// return the length of the longest
// subsequence of consecutive integers
static int findLongestConseqSubseq(int arr[], int N)
{
PriorityQueue<Integer> pq
= new PriorityQueue<Integer>();
for (int i = 0; i < N; i++)
{
// adding element from
// array to PriorityQueue
pq.add(arr[i]);
}
// Storing the first element
// of the Priority Queue
// This first element is also
// the smallest element
int prev = pq.poll();
// Taking a counter variable with value 1
int c = 1;
// Storing value of max as 1
// as there will always be
// one element
int max = 1;
for (int i = 1; i < N; i++)
{
// check if current peek
// element minus previous
// element is greater than
// 1 This is done because
// if it's greater than 1
// then the sequence
// doesn't start or is broken here
if (pq.peek() - prev > 1)
{
// Store the value of counter to 1
// As new sequence may begin
c = 1;
// Update the previous position with the
// current peek And remove it
prev = pq.poll();
}
// Check if the previous
// element and peek are same
else if (pq.peek() - prev == 0)
{
// Update the previous position with the
// current peek And remove it
prev = pq.poll();
}
// if the difference
// between previous element and peek is 1
else
{
// Update the counter
// These are consecutive elements
c++;
// Update the previous position
// with the current peek And remove it
prev = pq.poll();
}
// Check if current longest
// subsequence is the greatest
if (max < c)
{
// Store the current subsequence count as
// max
max = c;
}
}
return max;
}
// Driver Code
public static void main(String args[])
throws IOException
{
int arr[] = { 1, 9, 3, 10, 4, 20, 2 };
int n = arr.length;
System.out.println(
"Length of the Longest consecutive subsequence is "
+ findLongestConseqSubseq(arr, n));
}
}
// This code is contributed by Sudipa Sarkar
Python3
# Python program for the above approach
import bisect
def findLongestConseqSubseq(arr,N):
pq = []
for i in range(N):
# adding element from
# array to PriorityQueue
bisect.insort(pq,arr[i])
# Storing the first element
# of the Priority Queue
# This first element is also
# the smallest element
prev = pq[0]
pq.pop(0)
# Taking a counter variable with value 1
c=1
# Storing value of max as 1
# as there will always be
# one element
max = 1
while(len(pq)):
# check if current peek
# element minus previous
# element is greater than
# 1 This is done because
# if it's greater than 1
# then the sequence
# doesn't start or is broken here
if(pq[0] - prev > 1):
# Store the value of counter to 1
# As new sequence may begin
c = 1
# Update the previous position with the
# current peek And remove it
prev = pq[0]
pq.pop(0)
# Check if the previous
# element and peek are same
elif(pq[0] - prev == 0):
# Update the previous position with the
# current peek And remove it
prev = pq[0]
pq.pop(0)
# If the difference
# between previous element and peek is 1
else:
# Update the counter
# These are consecutive elements
c = c +1
# Update the previous position
# with the current peek And remove it
prev = pq[0]
pq.pop(0)
# Check if current longest
# subsequence is the greatest
if(max < c):
# Store the current subsequence count as
# max
max = c
return max
# Driver Code
arr = [1, 9, 3, 10, 4, 20, 2]
n = 7
print("Length of the Longest consecutive subsequence is {}".format(findLongestConseqSubseq(arr, n)))
# This code is contributed by Pushpesh Raj
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// return the length of the longest
// subsequence of consecutive integers
static int findLongestConseqSubseq(int[] arr, int N)
{
List<int> pq = new List<int>();
for (int i = 0; i < N; i++)
{
// adding element from
// array to PriorityQueue
pq.Add(arr[i]);
pq.Sort();
}
// Storing the first element
// of the Priority Queue
// This first element is also
// the smallest element
int prev = pq[0];
// Taking a counter variable with value 1
int c = 1;
// Storing value of max as 1
// as there will always be
// one element
int max = 1;
for (int i = 1; i < N; i++)
{
// check if current peek
// element minus previous
// element is greater than
// 1 This is done because
// if it's greater than 1
// then the sequence
// doesn't start or is broken here
if (pq[0] - prev > 1)
{
// Store the value of counter to 1
// As new sequence may begin
c = 1;
// Update the previous position with the
// current peek And remove it
prev = pq[0];
pq.RemoveAt(0);
}
// Check if the previous
// element and peek are same
else if (pq[0] - prev == 0)
{
// Update the previous position with the
// current peek And remove it
prev = pq[0];
pq.RemoveAt(0);
}
// if the difference
// between previous element and peek is 1
else
{
// Update the counter
// These are consecutive elements
c++;
// Update the previous position
// with the current peek And remove it
prev = pq[0];
pq.RemoveAt(0);
}
// Check if current longest
// subsequence is the greatest
if (max < c)
{
// Store the current subsequence count as
// max
max = c;
}
}
return max;
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 9, 3, 10, 4, 20, 2 };
int n = arr.Length;
Console.WriteLine(
"Length of the Longest consecutive subsequence is "
+ findLongestConseqSubseq(arr, n));
}
}
// This code is contributed by code_hunt.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA