Count all distinct pairs with difference equal to k | A sorting base java program to count pairs with difference k ; Standard binary search function ; Returns count of pairs with difference k in arr [ ] of size n . ; Sort array elements ; Pick a first element point ; Driver code
import java . util . * ; import java . io . * ; class GFG { static int binarySearch ( int arr [ ] , int low , int high , int x ) { if ( high >= low ) { int mid = low + ( high - low ) / 2 ; if ( x == arr [ mid ] ) return mid ; if ( x > arr [ mid ] ) return binarySearch ( arr , ( mid + 1 ) , high , x ) ; else return binarySearch ( arr , low , ( mid - 1 ) , x ) ; } return - 1 ; } static int countPairsWithDiffK ( int arr [ ] , int n , int k ) { int count = 0 , i ; Arrays . sort ( arr ) ; for ( i = 0 ; i < n - 1 ; i ++ ) if ( binarySearch ( arr , i + 1 , n - 1 , arr [ i ] + k ) != - 1 ) count ++ ; return count ; } public static void main ( String args [ ] ) { int arr [ ] = { 1 , 5 , 3 , 4 , 2 } ; int n = arr . length ; int k = 3 ; System . out . println ( " Count ▁ of ▁ pairs ▁ with ▁ given ▁ diff ▁ is ▁ " + countPairsWithDiffK ( arr , n , k ) ) ; } }