You are given an integer limit and a 2D array queries of size n x 2.There are limit + 1 balls with distinct labels in the range [0, limit]. Initially, all balls are uncolored. For every query in queries that is of the form [x, y], you mark ball x with the color y. After each query, you need to find the number of distinct colors among the balls.Return an array result of length n, where result[i] denotes the number of distinct colors after ith query.Note that when answering a query, lack of a color will not be considered as a color. Example 1:Input: limit = 4, queries = [[1,4],[2,5],[1,3],[3,4]]Output: [1,2,2,3]Explanation:After query 0, ball 1 has color 4.After query 1, ball 1 has color 4, and ball 2 has color 5.After query 2, ball 1 has color 3, and ball 2 has color 5.After query 3, ball 1 has color 3, ball 2 has color 5, and ball 3 has color 4.Example 2:Input: limit = 4, queries = [[0,1],[1,2],[2,2],[3,4],[4,5]]Output: [1,2,2,3,4]Explanation:After query 0, ball 0 has color 1.After query 1, ball 0 has color 1, and ball 1 has color 2.After query 2, ball 0 has color 1, and balls 1 and 2 have color 2.After query 3, ball 0 has color 1, balls 1 and 2 have color 2, and ball 3 has color 4.After query 4, ball 0 has color 1, balls 1 and 2 have color 2, ball 3 has color 4, and ball 4 has color 5. Constraints:1 <= limit <= 1091 <= n == queries.length <= 105queries[i].length == 20 <= queries[i][0] <= limit1 <= queries[i][1] <= 109
Question
You are given an integer limit and a 2D array queries of size n x 2.There are limit + 1 balls with distinct labels in the range [0, limit]. Initially, all balls are uncolored. For every query in queries that is of the form [x, y], you mark ball x with the color y. After each query, you need to find the number of distinct colors among the balls.Return an array result of length n, where result[i] denotes the number of distinct colors after ith query.Note that when answering a query, lack of a color will not be considered as a color. Example 1:Input: limit = 4, queries = [[1,4],[2,5],[1,3],[3,4]]Output: [1,2,2,3]Explanation:After query 0, ball 1 has color 4.After query 1, ball 1 has color 4, and ball 2 has color 5.After query 2, ball 1 has color 3, and ball 2 has color 5.After query 3, ball 1 has color 3, ball 2 has color 5, and ball 3 has color 4.Example 2:Input: limit = 4, queries = [[0,1],[1,2],[2,2],[3,4],[4,5]]Output: [1,2,2,3,4]Explanation:After query 0, ball 0 has color 1.After query 1, ball 0 has color 1, and ball 1 has color 2.After query 2, ball 0 has color 1, and balls 1 and 2 have color 2.After query 3, ball 0 has color 1, balls 1 and 2 have color 2, and ball 3 has color 4.After query 4, ball 0 has color 1, balls 1 and 2 have color 2, ball 3 has color 4, and ball 4 has color 5. Constraints:1 <= limit <= 1091 <= n == queries.length <= 105queries[i].length == 20 <= queries[i][0] <= limit1 <= queries[i][1] <= 109
Solution
Para resolver este problema, podemos seguir los siguientes pasos:
- Crear un diccionario para almacenar el color de cada bola.
- Crear un conjunto para almacenar los colores distintos.
- Iterar sobre cada consulta y actualizar el diccionario y el conjunto de colores.
- Después de cada consulta, agregar el número de colores distintos al resultado.
Aquí está el código que implementa esta lógica:
def countDistinctColors(limit, queries):
# Diccionario para almacenar el color de cada bola
ball_colors = {}
# Conjunto para almacenar los colores distintos
distinct_colors = set()
# Lista para almacenar el resultado
result = []
# Iterar sobre cada consulta
for x, y in queries:
# Si la bola ya tiene un color, eliminar ese color del conjunto
if x in ball_colors:
distinct_colors.discard(ball_colors[x])
# Asignar el nuevo color a la bola
ball_colors[x] = y
# Agregar el nuevo color al conjunto
distinct_colors.add(y)
# Agregar el número de colores distintos al resultado
result.append(len(distinct_colors))
return result
# Ejemplo 1
limit = 4
queries = [[1, 4], [2, 5], [1, 3], [3, 4]]
print(countDistinctColors(limit, queries)) # Salida: [1, 2, 2, 3]
# Ejemplo 2
limit = 4
queries = [[0, 1], [1, 2], [2, 2], [3, 4], [4, 5]]
print(countDistinctColors(limit, queries)) # Salida: [1, 2, 2, 3, 4]
Explicación paso a paso:
- Inicializamos un diccionario
ball_colorspara almacenar el color de cada bola y un conjuntodistinct_colorspara almacenar los colores distintos. - Iteramos sobre cada consulta en
queries. - Para cada consulta
[x, y], si la bolaxya tiene un color, eliminamos ese color del conjuntodistinct_colors. - Asignamos el nuevo color
ya la bolaxen el diccionarioball_colors. - Agregamos el nuevo color
yal conjuntodistinct_colors. - Agregamos el número de colores distintos al resultado
result. - Finalmente, devolvemos el resultado
result.
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