Maximum length of Strictly Increasing Sub | Java implementation of the approach ; Function to return the maximum length of strictly increasing subarray after removing atmost one element ; Create two arrays pre and pos ; Find out the contribution of the current element in array [ 0 , i ] and update pre [ i ] ; Find out the contribution of the current element in array [ N - 1 , i ] and update pos [ i ] ; Calculate the maximum length of the stricly increasing subarray without removing any element ; Calculate the maximum length of the strictly increasing subarray after removing the current element ; Driver code
class GFG { static int maxIncSubarr ( int a [ ] , int n ) { int pre [ ] = new int [ n ] ; int pos [ ] = new int [ n ] ; pre [ 0 ] = 1 ; pos [ n - 1 ] = 1 ; int l = 0 ; for ( int i = 1 ; i < n ; i ++ ) { if ( a [ i ] > a [ i - 1 ] ) pre [ i ] = pre [ i - 1 ] + 1 ; else pre [ i ] = 1 ; } l = 1 ; for ( int i = n - 2 ; i >= 0 ; i -- ) { if ( a [ i ] < a [ i + 1 ] ) pos [ i ] = pos [ i + 1 ] + 1 ; else pos [ i ] = 1 ; } int ans = 0 ; l = 1 ; for ( int i = 1 ; i < n ; i ++ ) { if ( a [ i ] > a [ i - 1 ] ) l ++ ; else l = 1 ; ans = Math . max ( ans , l ) ; } for ( int i = 1 ; i <= n - 2 ; i ++ ) { if ( a [ i - 1 ] < a [ i + 1 ] ) ans = Math . max ( pre [ i - 1 ] + pos [ i + 1 ] , ans ) ; } return ans ; } public static void main ( String [ ] args ) { int arr [ ] = { 1 , 2 } ; int n = arr . length ; System . out . println ( maxIncSubarr ( arr , n ) ) ; } }