Suppose we have a binary string called boxes, where boxes[i] is '0' indicates the ith box is empty, and '1' indicates it contains one ball. Now, in one operation, we can move one ball from a box to an adjacent box. After doing so, there may be more than one ball in some boxes. We have to find an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the ith box.
So, if the input is like boxes = "1101", then the output will be [4, 3, 4, 5]
To put all balls on first box, we have to take from box2 with one operation and from last ball by three operations, so total 4 operations.
To put all balls on second box, we have to take from box1 with one operation and from last ball by two operations, so total 3 operations.
To put all balls on third box, we have to take from box2 and last with one operation each and from box1 by two operations, so total 4 operations.
To put all balls on last box, we have to take from box1 with three operations and from box2 with two operations, so total 5 operations.
To solve this, we will follow these steps −
left := 0, right := 0, dist := 0
for i in range 0 to size of boxes - 1, do
if boxes[i] is same as "1", then
dist := dist + i
if i is same as 0, then
left := left + 1
otherwise,
right := right + 1
arr := a list and initially put dist into it
for i in range 1 to size of boxes - 1, do
insert arr[i-1] + left - right at the end of arr
if boxes[i] is same as "1", then
left := left + 1
right := right - 1
return arr
Example
Let us see the following implementation to get better understanding −
def solve(boxes):
left = 0
right = 0
dist = 0
for i in range(len(boxes)):
if boxes[i] == "1":
dist += i
if i == 0:
left += 1
else:
right += 1
arr = [dist]
for i in range(1, len(boxes)):
arr.append(arr[i-1] + left - right)
if boxes[i] == "1":
left += 1
right -= 1
return arr
boxes = "1101"
print(solve(boxes))Input
"1101"
Output
[4, 3, 4, 5]