human
Minimum number of jumps required to sort numbers placed on a number line | Function to find the minimum number of jumps required to sort the array.
Base Case.
Store the required result.
Stores the current position of elements and their respective maximum jump.
Used to check if a position is already taken by another element.
Stores the sorted array a [ ].
Traverse the array w [ ] & update positions jumps array a [ ].
Sort the array a [ ].
Traverse the array a [ ] over the range [ 1 , N - 1 ].
Store the index of current element and its just smaller element in array w [ ].
Iterate until current element position is at most its just smaller element position.
Update the position of the current element.
Print the result.
Driver Code"
Solution in Python:
 ... Can you provide a code to minimum number of jumps required to sort numbers placed on a number line and the other steps required?
assistant
#Here is code to minimum number of jumps required to sort numbers placed on a number line and the other steps required.
 def minJumps ( w , l , n ) :
	if ( n == 1 ) :
		print ( 0 )
		return
	ans = 0
	pos = { }
	jump = { }
	filled = { }
	a = [ 0 for i in range ( n ) ]
	for i in range ( n ) :
		pos [ w [ i ] ] = i
		filled [ i ] = True
		jump [ w [ i ] ] = l [ i ]
		a [ i ] = w [ i ]
	a . sort ( )
	for curr in range ( 1 , n , 1 ) :
		currElementPos = pos [ a [ curr ] ]
		prevElementPos = pos [ a [ curr - 1 ] ]
		if ( currElementPos > prevElementPos ) :
			continue
		while ( currElementPos <= prevElementPos or ( currElementPos in filled and filled [ currElementPos ] ) ) :
			currElementPos += jump [ a [ curr ] ]
			ans += 1
		pos [ a [ curr ] ] = currElementPos
		filled [ currElementPos ] = True
	print ( ans )
if __name__ == ' _ _ main _ _ ' :
	W = [ 2 , 1 , 4 , 3 ]
	L = [ 4 , 1 , 2 , 4 ]
	N = len ( W )
	minJumps ( W , L , N )