text: Count of subarrays of size K which is a permutation of numbers from 1 to K | Java program to implement the above approach ; Save index of numbers of the array ; Update min and max index with the current index and check if it 's a valid permutation ; Driver Code
code: import java . util . * ; class GFG { public static int find_permutations ( Vector < Integer > arr ) { int cnt = 0 ; int max_ind = - 1 , min_ind = 10000000 ; int n = arr . size ( ) ; HashMap < Integer , Integer > index_of = new HashMap < > ( ) ; for ( int i = 0 ; i < n ; i ++ ) { index_of . put ( arr . get ( i ) , i + 1 ) ; } for ( int i = 1 ; i <= n ; i ++ ) { max_ind = Math . max ( max_ind , index_of . get ( i ) ) ; min_ind = Math . min ( min_ind , index_of . get ( i ) ) ; if ( max_ind - min_ind + 1 == i ) cnt ++ ; } return cnt ; } public static void main ( String [ ] args ) { Vector < Integer > nums = new Vector < Integer > ( ) ; nums . add ( 2 ) ; nums . add ( 3 ) ; nums . add ( 1 ) ; nums . add ( 5 ) ; nums . add ( 4 ) ; System . out . print ( find_permutations ( nums ) ) ; } }