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First completed : June 20, 2024
Last updated : July 04, 2024
Related Topics : Array, Binary Search, Sorting
Acceptance Rate : 71.09 %
Notes In essence, we have x positions and m balls, and we need to find AN ideal way where we can place the m balls into m positions so that the min distance between the balls is minimized The ideal case will be (max(position) - min(position)) / (m - 1)
class Solution:
def maxDistance(self, position: List[int], m: int) -> int:
position.sort()
minn = maxx = position[0]
for p in position :
if p < minn :
minn = p
elif p > maxx :
maxx = p
left = 1 # 1 = worst case since m=len(positions) += 1 each indx worst
right = (maxx - minn) // (m - 1) # m - 1 = number of gaps so this is aideal case
while left < right :
mid = (left + right + 1) // 2
worked = False
counter = 1
prevVal = position[0]
for i in range(1, len(position)) :
if position[i] - prevVal >= mid :
counter += 1
prevVal = position[i]
if counter >= m :
worked = True
break
if worked :
left = mid
else :
right = mid - 1
return left