Longest increasing subarray
Last Updated :
07 Feb, 2023
Given an array containing n numbers. The problem is to find the length of the longest contiguous subarray such that every element in the subarray is strictly greater than its previous element in the same subarray. Time Complexity should be O(n).
Examples:
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}
Algorithm:
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
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int lenOfLongIncSubArr( int arr[], int n)
{
int max = 1, len = 1;
for ( int i=1; i<n; i++)
{
if (arr[i] > arr[i-1])
len++;
else
{
if (max < len)
max = len;
len = 1;
}
}
if (max < len)
max = len;
return max;
}
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
import java.util.*;
class GFG {
public static int lenOfLongIncSubArr( int arr[],
int n)
{
int max = 1 , len = 1 ;
for ( int i= 1 ; i<n; i++)
{
if (arr[i] > arr[i- 1 ])
len++;
else
{
if (max < len)
max = len;
len = 1 ;
}
}
if (max < len)
max = len;
return max;
}
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));
}
}
|
Python3
def lenOfLongIncSubArr(arr, n) :
m = 1
l = 1
for i in range ( 1 , n) :
if (arr[i] > arr[i - 1 ]) :
l = l + 1
else :
if (m < l) :
m = l
l = 1
if (m < l) :
m = l
return m
arr = [ 5 , 6 , 3 , 5 , 7 , 8 , 9 , 1 , 2 ]
n = len (arr)
print ( "Length = " , lenOfLongIncSubArr(arr, n))
|
C#
using System;
class GFG {
public static int lenOfLongIncSubArr( int [] arr,
int n)
{
int max = 1, len = 1;
for ( int i = 1; i < n; i++) {
if (arr[i] > arr[i - 1])
len++;
else {
if (max < len)
max = len;
len = 1;
}
}
if (max < len)
max = len;
return max;
}
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));
}
}
|
PHP
<?php
function lenOfLongIncSubArr( $arr , $n )
{
$max = 1;
$len = 1;
for ( $i = 1; $i < $n ; $i ++)
{
if ( $arr [ $i ] > $arr [ $i -1])
$len ++;
else
{
if ( $max < $len )
$max = $len ;
$len = 1;
}
}
if ( $max < $len )
$max = $len ;
return $max ;
}
$arr = array (5, 6, 3, 5, 7, 8, 9, 1, 2);
$n = sizeof( $arr );
echo "Length = " , lenOfLongIncSubArr( $arr , $n );
?>
|
Javascript
<script>
function lenOfLongIncSubArr(arr, n)
{
var max = 1, len = 1;
for ( var i=1; i<n; i++)
{
if (arr[i] > arr[i-1])
len++;
else
{
if (max < len)
{
max = len;
}
len = 1;
}
}
if (max < len)
{
max = len;
}
return max;
}
var arr = [5, 6, 3, 5, 7, 8, 9, 1, 2];
var n = arr.length;
document.write( "Length = " + lenOfLongIncSubArr(arr, n));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
How to print the subarray?
We can print the subarray by keeping track of the index with the largest length.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void printLongestIncSubArr( int arr[], int n)
{
int max = 1, len = 1, maxIndex = 0;
for ( int i=1; i<n; i++)
{
if (arr[i] > arr[i-1])
len++;
else
{
if (max < len)
{
max = len;
maxIndex = i - max;
}
len = 1;
}
}
if (max < len)
{
max = len;
maxIndex = n - max;
}
for ( int i=maxIndex; i<max+maxIndex; i++)
cout << arr[i] << " " ;
}
int main()
{
int arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2};
int n = sizeof (arr) / sizeof (arr[0]);
printLongestIncSubArr(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG {
public static void printLongestIncSubArr( int arr[],
int n)
{
int max = 1 , len = 1 , maxIndex = 0 ;
for ( int i = 1 ; i < n; i++)
{
if (arr[i] > arr[i- 1 ])
len++;
else
{
if (max < len)
{
max = len;
maxIndex = i - max;
}
len = 1 ;
}
}
if (max < len)
{
max = len;
maxIndex = n - max;
}
for ( int i = maxIndex; i < max+maxIndex; i++)
System.out.print(arr[i] + " " );
}
public static void main(String[] args)
{
int arr[] = { 5 , 6 , 3 , 5 , 7 , 8 , 9 , 1 , 2 };
int n = arr.length;
printLongestIncSubArr(arr, n);
}
}
|
Python3
def printLongestIncSubArr( arr, n) :
m = 1
l = 1
maxIndex = 0
for i in range ( 1 , n) :
if (arr[i] > arr[i - 1 ]) :
l = l + 1
else :
if (m < l) :
m = l
maxIndex = i - m
l = 1
if (m < l) :
m = l
maxIndex = n - m
for i in range (maxIndex, (m + maxIndex)) :
print (arr[i] , end = " " )
arr = [ 5 , 6 , 3 , 5 , 7 , 8 , 9 , 1 , 2 ]
n = len (arr)
printLongestIncSubArr(arr, n)
|
C#
using System;
class GFG {
public static void printLongestIncSubArr( int [] arr,
int n)
{
int max = 1, len = 1, maxIndex = 0;
for ( int i = 1; i < n; i++) {
if (arr[i] > arr[i - 1])
len++;
else
{
if (max < len) {
max = len;
maxIndex = i - max;
}
len = 1;
}
}
if (max < len) {
max = len;
maxIndex = n - max;
}
for ( int i = maxIndex; i < max + maxIndex; i++)
Console.Write(arr[i] + " " );
}
public static void Main()
{
int [] arr = { 5, 6, 3, 5, 7, 8, 9, 1, 2 };
int n = arr.Length;
printLongestIncSubArr(arr, n);
}
}
|
PHP
<?php
function printLongestIncSubArr(& $arr , $n )
{
$max = 1;
$len = 1;
$maxIndex = 0;
for ( $i = 1; $i < $n ; $i ++)
{
if ( $arr [ $i ] > $arr [ $i - 1])
$len ++;
else
{
if ( $max < $len )
{
$max = $len ;
$maxIndex = $i - $max ;
}
$len = 1;
}
}
if ( $max < $len )
{
$max = $len ;
$maxIndex = $n - $max ;
}
for ( $i = $maxIndex ;
$i < ( $max + $maxIndex ); $i ++)
echo ( $arr [ $i ] . " " ) ;
}
$arr = array (5, 6, 3, 5, 7,
8, 9, 1, 2);
$n = sizeof( $arr );
printLongestIncSubArr( $arr , $n );
?>
|
Javascript
<script>
function printLongestIncSubArr(arr, n)
{
var max = 1, len = 1, maxIndex = 0;
for ( var i=1; i<n; i++)
{
if (arr[i] > arr[i-1])
len++;
else
{
if (max < len)
{
max = len;
maxIndex = i - max;
}
len = 1;
}
}
if (max < len)
{
max = len;
maxIndex = n - max;
}
for ( var i=maxIndex; i<max+maxIndex; i++)
document.write( arr[i] + " " );
}
var arr = [5, 6, 3, 5, 7, 8, 9, 1, 2];
var n = arr.length;
printLongestIncSubArr(arr, n);
</script>
|
Time Complexity: O(N) where N is the number of elements in the array.
Auxiliary Space: O(1)
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