Dada una array que contiene n números. El problema es encontrar la longitud del subarreglo contiguo más largo tal que cada elemento del subarreglo sea estrictamente mayor que su elemento anterior en el mismo subarreglo. La complejidad del tiempo debe ser O(n).
Ejemplos:
Input : arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2}
Output : 5
The subarray is {3, 5, 7, 8, 9}
Input : arr[] = {12, 13, 1, 5, 4, 7, 8, 10, 10, 11}
Output : 4
The subarray is {4, 7, 8, 10}
Algoritmo:
lenOfLongIncSubArr(arr, n)
Declare max = 1, len = 1
for i = 1 to n-1
if arr[i] > arr[i-1]
len++
else
if max < len
max = len
len = 1
if max < len
max = len
return max
Implementación:
C++
// C++ implementation to find the length of
// longest increasing contiguous subarray
#include <bits/stdc++.h>
using namespace std;
// function to find the length of longest increasing
// contiguous subarray
int lenOfLongIncSubArr(int arr[], int n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
int max = 1, len = 1;
// traverse the array from the 2nd element
for (int i=1; i<n; i++)
{
// if current element if greater than previous
// element, then this element helps in building
// up the previous increasing subarray encountered
// so far
if (arr[i] > arr[i-1])
len++;
else
{
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if (max < len)
max = len;
// reset 'len' to 1 as from this element
// again the length of the new increasing
// subarray is being calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
max = len;
// required maximum length
return max;
}
// Driver program to test above
int main()
{
int arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Length = "
<< lenOfLongIncSubArr(arr, n);
return 0;
}
Java
// JAVA Code to find length of
// Longest increasing subarray
import java.util.*;
class GFG {
// function to find the length of longest
// increasing contiguous subarray
public static int lenOfLongIncSubArr(int arr[],
int n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
int max = 1, len = 1;
// traverse the array from the 2nd element
for (int i=1; i<n; i++)
{
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered
// so far
if (arr[i] > arr[i-1])
len++;
else
{
// check if 'max' length is less
// than the length of the current
// increasing subarray. If true,
// than update 'max'
if (max < len)
max = len;
// reset 'len' to 1 as from this
// element again the length of the
// new increasing subarray is being
// calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
max = len;
// required maximum length
return max;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2};
int n = arr.length;
System.out.println("Length = " +
lenOfLongIncSubArr(arr, n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python 3 implementation to find the length of
# longest increasing contiguous subarray
# function to find the length of longest
# increasing contiguous subarray
def lenOfLongIncSubArr(arr, n) :
# 'max' to store the length of longest
# increasing subarray
# 'len' to store the lengths of longest
# increasing subarray at different
# instants of time
m = 1
l = 1
# traverse the array from the 2nd element
for i in range(1, n) :
# if current element if greater than previous
# element, then this element helps in building
# up the previous increasing subarray encountered
# so far
if (arr[i] > arr[i-1]) :
l =l + 1
else :
# check if 'max' length is less than the length
# of the current increasing subarray. If true,
# then update 'max'
if (m < l) :
m = l
# reset 'len' to 1 as from this element
# again the length of the new increasing
# subarray is being calculated
l = 1
# comparing the length of the last
# increasing subarray with 'max'
if (m < l) :
m = l
# required maximum length
return m
# Driver program to test above
arr = [5, 6, 3, 5, 7, 8, 9, 1, 2]
n = len(arr)
print("Length = ", lenOfLongIncSubArr(arr, n))
# This code is contributed
# by Nikita Tiwari.
C#
// C# Code to find length of
// Longest increasing subarray
using System;
class GFG {
// function to find the length of longest
// increasing contiguous subarray
public static int lenOfLongIncSubArr(int[] arr,
int n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
int max = 1, len = 1;
// traverse the array from the 2nd element
for (int i = 1; i < n; i++) {
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered
// so far
if (arr[i] > arr[i - 1])
len++;
else {
// check if 'max' length is less
// than the length of the current
// increasing subarray. If true,
// than update 'max'
if (max < len)
max = len;
// reset 'len' to 1 as from this
// element again the length of the
// new increasing subarray is being
// calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
max = len;
// required maximum length
return max;
}
/* Driver program to test above function */
public static void Main()
{
int[] arr = { 5, 6, 3, 5, 7, 8, 9, 1, 2 };
int n = arr.Length;
Console.WriteLine("Length = " +
lenOfLongIncSubArr(arr, n));
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP implementation to find the length of
// longest increasing contiguous subarray
// function to find the length of
// longest increasing contiguous
// subarray
function lenOfLongIncSubArr($arr, $n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
$max = 1;
$len = 1;
// traverse the array from
// the 2nd element
for ($i = 1; $i < $n; $i++)
{
// if current element if
// greater than previous
// element, then this element
// helps in building up the
// previous increasing subarray
// encountered so far
if ($arr[$i] > $arr[$i-1])
$len++;
else
{
// check if 'max' length is
// less than the length
// of the current increasing
// subarray. If true,
// then update 'max'
if ($max < $len)
$max = $len;
// reset 'len' to 1 as
// from this element
// again the length of
// the new increasing
// subarray is being
// calculated
$len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if ($max < $len)
$max = $len;
// required maximum length
return $max;
}
// Driver Code
$arr = array(5, 6, 3, 5, 7, 8, 9, 1, 2);
$n = sizeof($arr);
echo "Length = ", lenOfLongIncSubArr($arr, $n);
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// Javascript implementation to find the length of
// longest increasing contiguous subarray
// function to find the length of longest increasing
// contiguous subarray
function lenOfLongIncSubArr(arr, n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
var max = 1, len = 1;
// traverse the array from the 2nd element
for (var i=1; i<n; i++)
{
// if current element if greater than previous
// element, then this element helps in building
// up the previous increasing subarray encountered
// so far
if (arr[i] > arr[i-1])
len++;
else
{
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if (max < len)
{
max = len;
}
// reset 'len' to 1 as from this element
// again the length of the new increasing
// subarray is being calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
{
max = len;
}
// required maximum length
return max;
}
// Driver program to test above
var arr = [5, 6, 3, 5, 7, 8, 9, 1, 2];
var n = arr.length;
document.write("Length = " + lenOfLongIncSubArr(arr, n));
// This code is contributed by shivani.
</script>
Producción
Length = 5
Complejidad de tiempo: O(n)
¿Cómo imprimir el subarreglo?
Podemos imprimir el subarreglo haciendo un seguimiento del índice con la mayor longitud.
Implementación:
C++
// C++ implementation to find the length of
// longest increasing contiguous subarray
#include <bits/stdc++.h>
using namespace std;
// function to find the length of longest increasing
// contiguous subarray
void printLogestIncSubArr(int arr[], int n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
int max = 1, len = 1, maxIndex = 0;
// traverse the array from the 2nd element
for (int i=1; i<n; i++)
{
// if current element if greater than previous
// element, then this element helps in building
// up the previous increasing subarray encountered
// so far
if (arr[i] > arr[i-1])
len++;
else
{
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if (max < len)
{
max = len;
// index assign the starting index of
// longest increasing contiguous subarray.
maxIndex = i - max;
}
// reset 'len' to 1 as from this element
// again the length of the new increasing
// subarray is being calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
{
max = len;
maxIndex = n - max;
}
// Print the elements of longest increasing
// contiguous subarray.
for (int i=maxIndex; i<max+maxIndex; i++)
cout << arr[i] << " ";
}
// Driver program to test above
int main()
{
int arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2};
int n = sizeof(arr) / sizeof(arr[0]);
printLogestIncSubArr(arr, n);
return 0;
}
// This code is contributed by Dharmendra kumar
Java
// JAVA Code For Longest increasing subarray
import java.util.*;
class GFG {
// function to find the length of longest
// increasing contiguous subarray
public static void printLogestIncSubArr(int arr[],
int n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
int max = 1, len = 1, maxIndex = 0;
// traverse the array from the 2nd element
for (int i = 1; i < n; i++)
{
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered
// so far
if (arr[i] > arr[i-1])
len++;
else
{
// check if 'max' length is less
// than the length of the current
// increasing subarray. If true,
// then update 'max'
if (max < len)
{
max = len;
// index assign the starting
// index of longest increasing
// contiguous subarray.
maxIndex = i - max;
}
// reset 'len' to 1 as from this
// element again the length of the
// new increasing subarray is
// being calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
{
max = len;
maxIndex = n - max;
}
// Print the elements of longest
// increasing contiguous subarray.
for (int i = maxIndex; i < max+maxIndex; i++)
System.out.print(arr[i] + " ");
}
/* Driver program to test above function */
public static void main(String[] args)
{
int arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2};
int n = arr.length;
printLogestIncSubArr(arr, n);
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python 3 implementation to find the length of # longest increasing contiguous subarray # function to find the length of longest increasing # contiguous subarray def printLogestIncSubArr( arr, n) : # 'max' to store the length of longest # increasing subarray # 'len' to store the lengths of longest # increasing subarray at different # instants of time m = 1 l = 1 maxIndex = 0 # traverse the array from the 2nd element for i in range(1, n) : # if current element if greater than previous # element, then this element helps in building # up the previous increasing subarray # encountered so far if (arr[i] > arr[i-1]) : l =l + 1 else : # check if 'max' length is less than the length # of the current increasing subarray. If true, # then update 'max' if (m < l) : m = l # index assign the starting index of # longest increasing contiguous subarray. maxIndex = i - m # reset 'len' to 1 as from this element # again the length of the new increasing # subarray is being calculated l = 1 # comparing the length of the last # increasing subarray with 'max' if (m < l) : m = l maxIndex = n - m # Print the elements of longest # increasing contiguous subarray. for i in range(maxIndex, (m+maxIndex)) : print(arr[i] , end=" ") # Driver program to test above arr = [5, 6, 3, 5, 7, 8, 9, 1, 2] n = len(arr) printLogestIncSubArr(arr, n) # This code is contributed # by Nikita Tiwari
C#
// C# Code to print
// Longest increasing subarray
using System;
class GFG {
// function to find the length of longest
// increasing contiguous subarray
public static void printLogestIncSubArr(int[] arr,
int n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
int max = 1, len = 1, maxIndex = 0;
// traverse the array from the 2nd element
for (int i = 1; i < n; i++) {
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered
// so far
if (arr[i] > arr[i - 1])
len++;
else
{
// check if 'max' length is less
// than the length of the current
// increasing subarray. If true,
// then update 'max'
if (max < len) {
max = len;
// index assign the starting
// index of longest increasing
// contiguous subarray.
maxIndex = i - max;
}
// reset 'len' to 1 as from this
// element again the length of the
// new increasing subarray is
// being calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len) {
max = len;
maxIndex = n - max;
}
// Print the elements of longest
// increasing contiguous subarray.
for (int i = maxIndex; i < max + maxIndex; i++)
Console.Write(arr[i] + " ");
}
/* Driver program to test above function */
public static void Main()
{
int[] arr = { 5, 6, 3, 5, 7, 8, 9, 1, 2 };
int n = arr.Length;
printLogestIncSubArr(arr, n);
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP implementation to find
// the length of longest increasing
// contiguous subarray
// function to find the length of
// longest increasing contiguous subarray
function printLogestIncSubArr(&$arr, $n)
{
// 'max' to store the length of
// longest increasing subarray
// 'len' to store the lengths of
// longest increasing subarray at
// different instants of time
$max = 1;
$len = 1;
$maxIndex = 0;
// traverse the array from
// the 2nd element
for ($i = 1; $i < $n; $i++)
{
// if current element if greater
// than previous element, then
// this element helps in building
// up the previous increasing
// subarray encountered so far
if ($arr[$i] > $arr[$i - 1])
$len++;
else
{
// check if 'max' length is less
// than the length of the current
// increasing subarray. If true,
// then update 'max'
if ($max < $len)
{
$max = $len;
// index assign the starting
// index of longest increasing
// contiguous subarray.
$maxIndex = $i - $max;
}
// reset 'len' to 1 as from this
// element again the length of
// the new increasing subarray
// is being calculated
$len = 1;
}
}
// comparing the length of
// the last increasing
// subarray with 'max'
if ($max < $len)
{
$max = $len;
$maxIndex = $n - $max;
}
// Print the elements of
// longest increasing
// contiguous subarray.
for ($i = $maxIndex;
$i < ($max + $maxIndex); $i++)
echo($arr[$i] . " ") ;
}
// Driver Code
$arr = array(5, 6, 3, 5, 7,
8, 9, 1, 2);
$n = sizeof($arr);
printLogestIncSubArr($arr, $n);
// This code is contributed
// by Shivi_Aggarwal
?>
Javascript
<script>
// Javascript implementation to find the length of
// longest increasing contiguous subarray
// function to find the length of longest increasing
// contiguous subarray
function printLogestIncSubArr(arr, n)
{
// 'max' to store the length of longest
// increasing subarray
// 'len' to store the lengths of longest
// increasing subarray at different
// instants of time
var max = 1, len = 1, maxIndex = 0;
// traverse the array from the 2nd element
for (var i=1; i<n; i++)
{
// if current element if greater than previous
// element, then this element helps in building
// up the previous increasing subarray encountered
// so far
if (arr[i] > arr[i-1])
len++;
else
{
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if (max < len)
{
max = len;
// index assign the starting index of
// longest increasing contiguous subarray.
maxIndex = i - max;
}
// reset 'len' to 1 as from this element
// again the length of the new increasing
// subarray is being calculated
len = 1;
}
}
// comparing the length of the last
// increasing subarray with 'max'
if (max < len)
{
max = len;
maxIndex = n - max;
}
// Print the elements of longest increasing
// contiguous subarray.
for (var i=maxIndex; i<max+maxIndex; i++)
document.write( arr[i] + " ");
}
// Driver program to test above
var arr = [5, 6, 3, 5, 7, 8, 9, 1, 2];
var n = arr.length;
printLogestIncSubArr(arr, n);
// This code is contributed by rrrtnx.
</script>
Producción
3 5 7 8 9
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA