text: k | Java program for above approach ; Function to find kth missing number ; If the total missing number count is equal to k we can iterate backwards for the first missing number and that will be the answer . ; To further optimize we check if the previous element ' s ▁ ▁ missing ▁ number ▁ count ▁ is ▁ equal ▁ ▁ to ▁ k . ▁ Eg : ▁ arr ▁ = ▁ [ 4,5,6,7,8 ] ▁ ▁ If ▁ you ▁ observe ▁ in ▁ the ▁ example ▁ array , ▁ ▁ the ▁ total ▁ count ▁ of ▁ missing ▁ numbers ▁ for ▁ all ▁ ▁ the ▁ indices ▁ are ▁ same , ▁ and ▁ we ▁ are ▁ ▁ aiming ▁ to ▁ narrow ▁ down ▁ the ▁ ▁ search ▁ window ▁ and ▁ achieve ▁ O ( logn ) ▁ ▁ time ▁ complexity ▁ which ▁ ▁ otherwise ▁ would ' ve been O ( n ) . ; Else we return arr [ mid ] - 1. ; Here we appropriately narrow down the search window . ; In case the upper limit is - ve it means the missing number set is 1 , 2 , . . , k and hence we directly return k . ; Else we find the residual count of numbers which we 'd then add to  arr[u] and get the missing kth number. ; Return arr [ u ] + k ; Driver code ; Function Call
code: public class GFG { static int missingK ( int [ ] arr , int k ) { int n = arr . length ; int l = 0 , u = n - 1 , mid ; while ( l <= u ) { mid = ( l + u ) / 2 ; int numbers_less_than_mid = arr [ mid ] - ( mid + 1 ) ; if ( numbers_less_than_mid == k ) { if ( mid > 0 && ( arr [ mid - 1 ] - ( mid ) ) == k ) { u = mid - 1 ; continue ; } return arr [ mid ] - 1 ; } if ( numbers_less_than_mid < k ) { l = mid + 1 ; } else if ( k < numbers_less_than_mid ) { u = mid - 1 ; } } if ( u < 0 ) return k ; int less = arr [ u ] - ( u + 1 ) ; k -= less ; return arr [ u ] + k ; } public static void main ( String [ ] args ) { int [ ] arr = { 2 , 3 , 4 , 7 , 11 } ; int k = 5 ; System . out . println ( " Missing ▁ kth ▁ number ▁ = ▁ " + missingK ( arr , k ) ) ; } }