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Minimum removals from array to make max | Function to find rightmost index which satisfies the condition arr [ j ] - arr [ i ] <= k.
Initialising start to i + 1.
Initialising end to n - 1.
Binary search implementation to find the rightmost element that satisfy the condition.
Check if the condition satisfies.
Store the position.
Make start = mid + 1.
Make end = mid.
Return the rightmost position.
Function to calculate minimum number of elements to be removed.
Sort the given array.
Iterate from 0 to n - 1.
Find j such that arr [ j ] - arr [ i ] <= k.
If there exist such j that satisfies the condition.
Number of elements to be removed for this particular case is ( j - i + 1 ).
Return answer.
Driver Code"
Solution in Python:
 ... What is a code to minimum removals from array to make max and the other steps required?
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#Here is code to minimum removals from array to make max and the other steps required.
 def findInd ( key , i , n , k , arr ) :
	ind = - 1
	start = i + 1
	end = n - 1 ;
	while ( start < end ) :
		mid = int ( start + ( end - start ) / 2 )
		if ( arr [ mid ] - key <= k ) :
			ind = mid
			start = mid + 1
		else :
			end = mid
	return ind
def removals ( arr , n , k ) :
	ans = n - 1
	arr . sort ( )
	for i in range ( 0 , n ) :
		j = findInd ( arr [ i ] , i , n , k , arr )
		if ( j != - 1 ) :
			ans = min ( ans , n - ( j - i + 1 ) )
	return ans
a = [ 1 , 3 , 4 , 9 , 10 , 11 , 12 , 17 , 20 ]
n = len ( a )
k = 4
print ( removals ( a , n , k ) )