human
Minimum cost to modify a string | Function to return the minimum cost.
Initialize result.
To store the frequency of characters of the string.
Update the frequencies of the characters of the string.
Loop to check all windows from a - z where window size is K.
Starting index of window.
Ending index of window.
Check if the string contains character.
Check if the character is on left side of window find the cost of modification for character add value to count calculate nearest distance of modification.
Check if the character is on right side of window find the cost of modification for character add value to count calculate nearest distance of modification.
Find the minimum of all costs for modifying the string.
Loop to check all windows Here window contains characters before z and after z of window size K.
Starting index of window.
Ending index of window.
Check if the string contains character.
If characters are outside window find the cost for modifying character add value to count.
Find the minimum of all costs for modifying the string.
Driver code"
Solution in Python:
 Give me a code to minimum cost to modify a string and the other steps required?
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#Here is code to minimum cost to modify a string and the other steps required.
 def minCost ( str1 , K ) :
	n = len ( str1 )
	res = 999999999
	count = 0
	cnt = [ 0 for i in range ( 27 ) ]
	for i in range ( n ) :
		cnt [ ord ( str1 [ i ] ) - ord ( ' a ' ) + 1 ] += 1
	for i in range ( 1 , 26 - K + 1 , 1 ) :
		a = i
		b = i + K
		count = 0
		for j in range ( 1 , 27 , 1 ) :
			if ( cnt [ j ] > 0 ) :
				if ( j >= a and j >= b ) :
					count = count + ( min ( j - b , 25 - j + a + 1 ) ) * cnt [ j ]
				elif ( j <= a and j <= b ) :
					count = count + ( min ( a - j , 25 + j - b + 1 ) ) * cnt [ j ]
		res = min ( res , count )
	for i in range ( 26 - K + 1 , 27 , 1 ) :
		a = i
		b = ( i + K ) % 26
		count = 0
		for j in range ( 1 , 27 , 1 ) :
			if ( cnt [ j ] > 0 ) :
				if ( j >= b and j <= a ) :
					count = count + ( min ( j - b , a - j ) ) * cnt [ j ]
		res = min ( res , count )
	return res
if __name__ == ' _ _ main _ _ ' :
	str1 = " abcdefghi "
	K = 2
	print ( minCost ( str1 , K ) )