You are given an integer array nums consisting of 2 * n integers.
You need to divide nums into n pairs such that:
Return true if nums can be divided into n pairs, otherwise return false.
Example 1:
Input: nums = [3,2,3,2,2,2] Output: true Explanation: There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs. If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.
Example 2:
Input: nums = [1,2,3,4] Output: false Explanation: There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.
Constraints:
nums.length == 2 * n1 <= n <= 5001 <= nums[i] <= 500When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach to pairing elements in the array involves checking every single possible combination of pairs. Think of it as trying every imaginable pairing to see if it works. If we find a working set of pairs, we know we have found the solution.
Here's how the algorithm would work step-by-step:
def divide_array_into_equal_pairs(numbers):
numbers_list = numbers.copy()
while numbers_list:
first_number = numbers_list[0]
# Find a match for the first number
found_match = False
for index in range(1, len(numbers_list)):
if numbers_list[index] == first_number:
second_number_index = index
found_match = True
break
# If no match is found, return False.
if not found_match:
return False
# Remove the paired numbers
numbers_list.pop(second_number_index)
numbers_list.pop(0)
# If all numbers are paired, return True
return TrueThe core idea is to count how many times each number appears. If we can successfully pair all the numbers, each number should show up an even number of times.
Here's how the algorithm would work step-by-step:
def divide_array_into_equal_pairs(numbers) -> bool:
number_counts = {}
for number in numbers:
if number in number_counts:
number_counts[number] += 1
else:
number_counts[number] = 1
# Iterate through the counts to check for odd occurrences.
for number in number_counts:
# If any number appears an odd number of times, return false.
if number_counts[number] % 2 != 0:
return False
# If all numbers appear an even number of times, return true.
return True| Case | How to Handle |
|---|---|
| Null or undefined input array | Throw an IllegalArgumentException or return an appropriate error value like null or an empty array, depending on the requirements. |
| Empty array (length 0) | Return true (or an empty list of pairs), as vacuously true; an empty array is divisible into equal pairs. |
| Array with odd number of elements | Return false immediately because it cannot be divided into equal pairs. |
| Array with a large number of elements (close to maximum integer size) | Consider space complexity of data structures used (e.g., hash map) to ensure it doesn't exceed memory limits. |
| Array contains negative numbers | The hash map (or sorting-based approach) will work correctly with negative numbers. |
| Array contains zero values | The solution should correctly handle zero values like any other number. |
| Array where elements cannot be paired | The hash map or sorting approach will result in an unmatched element, leading to a return of false. |
| Integer overflow during calculations (if applicable) | Ensure calculations are done using data types that can accommodate the potential range of values, or implement checks to prevent overflow. |