Sum of elements in 1 st array such that number of elements less than or equal to them in 2 nd array is maximum | Java implementation of the approach ; Function to return the required sum ; Creating hash array initially filled with zero ; Calculate the frequency of elements of arr2 [ ] ; Running sum of hash array such that hash [ i ] will give count of elements less than or equal to i in arr2 [ ] ; To store the maximum value of the number of elements in arr2 [ ] which are smaller than or equal to some element of arr1 [ ] ; Calculate the sum of elements from arr1 [ ] corresponding to maximum frequency ; Return the required sum ; Driver code
class GFG { static int MAX = 100000 ; static int findSumofEle ( int arr1 [ ] , int m , int arr2 [ ] , int n ) { int hash [ ] = new int [ MAX ] ; for ( int i = 0 ; i < n ; i ++ ) { hash [ arr2 [ i ] ] ++ ; } for ( int i = 1 ; i < MAX ; i ++ ) { hash [ i ] = hash [ i ] + hash [ i - 1 ] ; } int maximumFreq = 0 ; for ( int i = 0 ; i < m ; i ++ ) { maximumFreq = Math . max ( maximumFreq , hash [ arr1 [ i ] ] ) ; } int sumOfElements = 0 ; for ( int i = 0 ; i < m ; i ++ ) { sumOfElements += ( maximumFreq == hash [ arr1 [ i ] ] ) ? arr1 [ i ] : 0 ; } return sumOfElements ; } public static void main ( String [ ] args ) { int arr1 [ ] = { 2 , 5 , 6 , 8 } ; int arr2 [ ] = { 4 , 10 } ; int m = arr1 . length ; int n = arr2 . length ; System . out . println ( findSumofEle ( arr1 , m , arr2 , n ) ) ; } }