human
Generate an array consisting of most frequent greater elements present on the right side of each array element | Function to generate an array containing the most frequent greater element on the right side of each array element.
Stores the generated array.
Traverse the array arr [ ].
Store the result for the current index and its frequency.
Iterate over the right subarray.
Store the frequency of the current array element.
If the frequencies are equal.
Update ans to smaller of the two elements.
If count of new element is more than count of ans.
Insert answer in the array.
Print the resultant array.
Driver Code.
Given Input"
Solution in Python:
assistant
#Here is code to generate an array consisting of most frequent greater elements present on the right side of each array element and the other steps required.
 def findArray ( arr , n ) :
	v = [ ]
	for i in range ( n ) :
		ans = - 1
		old_c = 0
		for j in range ( i + 1 , n ) :
			if ( arr [ j ] > arr [ i ] ) :
				curr_c = arr [ j : n + 1 ] . count ( arr [ j ] )
				if ( curr_c == old_c ) :
					if ( arr [ j ] < ans ) :
						ans = arr [ j ]
				if ( curr_c > old_c ) :
					ans = arr [ j ]
					old_c = curr_c
		v . append ( ans )
	for i in range ( len ( v ) ) :
		print ( v [ i ] , end = " ▁ " )
if __name__ == ' _ _ main _ _ ' :
	arr = [ 4 , 5 , 2 , 25 , 10 , 5 , 10 , 3 , 10 , 5 ]
	size = len ( arr )
	findArray ( arr , size )