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Maximum elements that can be made equal with k updates | Function to calculate the maximum number of equal elements possible with atmost K increment of values . Here we have done sliding window to determine that whether there are x number of elements present which on increment will become equal . The loop here will run in fashion like 0. . . x - 1 , 1. . . x , 2. . . x + 1 , ... . , n - x - 1. . . n - 1.
It can be explained with the reasoning that if for some x number of elements we can update the values then the increment to the segment ( i to j having length -> x ) so that all will be equal is ( x * maxx [ j ] ) this is the total sum of segment and ( pre [ j ] - pre [ i ] ) is present sum So difference of them should be less than k if yes , then that segment length ( x ) can be possible return true.
sort the array in ascending order.
Initializing the prefix array and maximum array.
Calculating prefix sum of the array.
Calculating max value upto that position in the array.
Binary search applied for computation here.
printing result.
Driver Code"
Solution in Python:
 ... What is a code to maximum elements that can be made equal with k updates and the other steps required?
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#Here is code to maximum elements that can be made equal with k updates and the other steps required.
 def ElementsCalculationFunc ( pre , maxx , x , k , n ) :
	i = 0
	j = x
	while j <= n :
		if ( x * maxx [ j ] - ( pre [ j ] - pre [ i ] ) <= k ) :
			return True
		i += 1
		j += 1
	return False
def MaxNumberOfElements ( a , n , k ) :
	a . sort ( )
	for i in range ( n + 1 ) :
		pre [ i ] = 0
		maxx [ i ] = 0
	for i in range ( 1 , n + 1 ) :
		pre [ i ] = pre [ i - 1 ] + a [ i - 1 ]
		maxx [ i ] = max ( maxx [ i - 1 ] , a [ i - 1 ] )
	l = 1
	r = n
	while ( l < r ) :
		mid = ( l + r ) // 2
		if ( ElementsCalculationFunc ( pre , maxx , mid - 1 , k , n ) ) :
			ans = mid
			l = mid + 1
		else :
			r = mid - 1
	print ( ans )
if __name__ == " _ _ main _ _ " :
	arr = [ 2 , 4 , 9 ]
	n = len ( arr )
	k = 3
	MaxNumberOfElements ( arr , n , k )