Ceiling in a sorted array |  ; Function to get index of ceiling of x in arr [ low . . high ] ; If x is smaller than or equal to first element , then return the first element ; Otherwise , linearly search for ceil value ; if x lies between arr [ i ] and arr [ i + 1 ] including arr [ i + 1 ] , then return arr [ i + 1 ] ; If we reach here then x is greater than the last element of the array , return - 1 in this case ; Driver program to check above functions
class Main { static int ceilSearch ( int arr [ ] , int low , int high , int x ) { int i ; if ( x <= arr [ low ] ) return low ; for ( i = low ; i < high ; i ++ ) { if ( arr [ i ] == x ) return i ; if ( arr [ i ] < x && arr [ i + 1 ] >= x ) return i + 1 ; } return - 1 ; } public static void main ( String [ ] args ) { int arr [ ] = { 1 , 2 , 8 , 10 , 10 , 12 , 19 } ; int n = arr . length ; int x = 3 ; int index = ceilSearch ( arr , 0 , n - 1 , x ) ; if ( index == - 1 ) System . out . println ( " Ceiling ▁ of ▁ " + x + " ▁ doesn ' t ▁ exist ▁ in ▁ array " ) ; else System . out . println ( " ceiling ▁ of ▁ " + x + " ▁ is ▁ " + arr [ index ] ) ; } }