You are given an array nums, where each number in the array appears either once or twice.Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice. Example 1:Input: nums = [1,2,1,3]Output: 1Explanation:The only number that appears twice in nums is 1.Example 2:Input: nums = [1,2,3]Output: 0Explanation:No number appears twice in nums.Example 3:Input: nums = [1,2,2,1]Output: 3Explanation:Numbers 1 and 2 appeared twice. 1 XOR 2 == 3. Constraints:1 <= nums.length <= 501 <= nums[i] <= 50Each number in nums appears either once or twice.
Question
You are given an array nums, where each number in the array appears either once or twice.Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice. Example 1:Input: nums = [1,2,1,3]Output: 1Explanation:The only number that appears twice in nums is 1.Example 2:Input: nums = [1,2,3]Output: 0Explanation:No number appears twice in nums.Example 3:Input: nums = [1,2,2,1]Output: 3Explanation:Numbers 1 and 2 appeared twice. 1 XOR 2 == 3. Constraints:1 <= nums.length <= 501 <= nums[i] <= 50Each number in nums appears either once or twice.
Solution
Para resolver este problema, podemos seguir los siguientes pasos:
- Crear un diccionario para contar la frecuencia de cada número en el array.
- Recorrer el array y actualizar el diccionario con las frecuencias.
- Inicializar una variable para almacenar el resultado del XOR.
- Recorrer el diccionario y hacer el XOR de los números que aparecen dos veces.
- Devolver el resultado del XOR.
Aquí está el código paso a paso:
def findXOR(nums):
# Paso 1: Crear un diccionario para contar la frecuencia de cada número
frequency = {}
# Paso 2: Recorrer el array y actualizar el diccionario con las frecuencias
for num in nums:
if num in frequency:
frequency[num] += 1
else:
frequency[num] = 1
# Paso 3: Inicializar una variable para almacenar el resultado del XOR
result = 0
# Paso 4: Recorrer el diccionario y hacer el XOR de los números que aparecen dos veces
for num, count in frequency.items():
if count == 2:
result ^= num
# Paso 5: Devolver el resultado del XOR
return result
# Ejemplos de prueba
print(findXOR([1, 2, 1, 3])) # Salida: 1
print(findXOR([1, 2, 3])) # Salida: 0
print(findXOR([1, 2, 2, 1])) # Salida: 3
Este código sigue los pasos descritos y resuelve el problema de encontrar el XOR de los números que aparecen dos veces en el array.
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