In this approach, I used a dictionary ball_colors to keep track of the latest color assigned to each ball. Also, I added a set called distinct_colors to store all unique colors in use.
For each query [x, y], I first check if ball x has already been assigned a color. If it has and the color is different from y, I decrement the count of the old color in a dictionary color_count. If the count reaches zero, I remove the old color from distinct_colors. Then, I update ball x with the new color y and update its count in color_count. Finally, I add y to distinct_colors and append the size of distinct_colors to my result list.
class Solution:
def queryResults(self, limit: int, queries: List[List[int]]) -> List[int]:
ball_colors = {}
distinct_colors = set()
color_count = {}
result = []
for x, y in queries:
if x in ball_colors:
old_color = ball_colors[x]
if old_color != y:
color_count[old_color] -= 1
if color_count[old_color] == 0:
distinct_colors.remove(old_color)
ball_colors[x] = y
if y not in color_count:
color_count[y] = 0
color_count[y] += 1
distinct_colors.add(y)
result.append(len(distinct_colors))
return result- Time: O(n)
- Space: O(n)
The code didn't pass all testcases so I gave up and checked the editorial and it worked!
