Alternate XOR operations on sorted array | Java implementation of the approach ; Function to find the maximum and the minimum elements from the array after performing the given operation k times ; To store the current sequence of elements ; To store the next sequence of elements after xoring with current elements ; Store the frequency of elements of arr [ ] in arr1 [ ] ; Storing all precomputed XOR values so that we don 't have to do it again and again  as XOR is a costly operation ; Perform the operations k times ; The value of count decide on how many elements we have to apply XOR operation ; If current element is present in the array to be modified ; Suppose i = m and arr1 [ i ] = num , it means ' m ' appears ' num ' times If the count is even we have to perform XOR operation on alternate ' m ' starting from the 0 th index because count is even and we have to perform XOR operations starting with initial ' m ' Hence there will be ceil ( num / 2 ) operations on ' m ' that will change ' m ' to xor_val [ m ] i . e . m ^ x ; Decrease the frequency of ' m ' from arr1 [ ] ; Increase the frequency of ' m ^ x ' in arr2 [ ] ; If the count is odd we have to perform XOR operation on alternate ' m ' starting from the 1 st index because count is odd and we have to leave the 0 th ' m ' Hence there will be ( num / 2 ) XOR operations on ' m ' that will change ' m ' to xor_val [ m ] i . e . m ^ x ; Updating the count by frequency of the current elements as we have processed that many elements ; Updating arr1 [ ] which will now store the next sequence of elements At this time , arr1 [ ] stores the remaining ' m ' on which XOR was not performed and arr2 [ ] stores the frequency of ' m ^ x ' i . e . those ' m ' on which operation was performed Updating arr1 [ ] with frequency of remaining ' m ' & frequency of ' m ^ x ' from arr2 [ ] With help of arr2 [ ] , we prevent sorting of the array again and again ; Resetting arr2 [ ] for next iteration ; Finding the maximum and the minimum element from the modified array after the operations ; Printing the max and the min element ; Driver code
class GFG { static int MAX = 100000 ; public static void xorOnSortedArray ( int [ ] arr , int n , int k , int x ) { int [ ] arr1 = new int [ MAX + 1 ] ; int [ ] arr2 = new int [ MAX + 1 ] ; int [ ] xor_val = new int [ MAX + 1 ] ; for ( int i = 0 ; i < n ; i ++ ) arr1 [ arr [ i ] ] ++ ; for ( int i = 0 ; i <= MAX ; i ++ ) xor_val [ i ] = i ^ x ; while ( k -- > 0 ) { int count = 0 ; for ( int i = 0 ; i <= MAX ; i ++ ) { int store = arr1 [ i ] ; if ( arr1 [ i ] > 0 ) { if ( count % 2 == 0 ) { int div = ( int ) Math . ceil ( arr1 [ i ] / 2 ) ; arr1 [ i ] = arr1 [ i ] - div ; arr2 [ xor_val [ i ] ] += div ; } else if ( count % 2 != 0 ) { int div = arr1 [ i ] / 2 ; arr1 [ i ] = arr1 [ i ] - div ; arr2 [ xor_val [ i ] ] += div ; } } count = count + store ; } for ( int i = 0 ; i <= MAX ; i ++ ) { arr1 [ i ] = arr1 [ i ] + arr2 [ i ] ; arr2 [ i ] = 0 ; } } int min = Integer . MAX_VALUE ; int max = Integer . MIN_VALUE ; for ( int i = 0 ; i <= MAX ; i ++ ) { if ( arr1 [ i ] > 0 ) { if ( min > i ) min = i ; if ( max < i ) max = i ; } } System . out . println ( min + " ▁ " + max ) ; } public static void main ( String [ ] args ) { int [ ] arr = { 605 , 986 } ; int n = arr . length ; int k = 548 , x = 569 ; xorOnSortedArray ( arr , n , k , x ) ; } }