Divide array into equal pairs

You are given an integer array nums consiting of 2 * n integers.

you need to divide nums into n pairs such that:

Each element belongs to exactly one pair.
The element present in a pair are equal.

Return true if nums can be divided into 'n' pairs, otherwise return 'false'.

Constraints:

nums.length == 2 * n
1 <= n <= 500
1 <= nums[i] <= 500

Approach: Hash Map

func divideArray(nums []int) bool {
    hash := make(map[int]int)
    for i := range nums {
        hash[nums[i]]++
    }
    for _, val := range hash {
        if val % 2 != 0 {
            return false
        }
    }
    return true 
}

Intuition

The main idea is to check if every number in the array appears an even number of times. If a number appears an odd number of times, it cannot be completely paired, so the function returns false. Using a hash map to count occurrences makes it easy to verify this condition.

Algorithm

Initialize a hash map:
Create an empty hash map to store the frequency of each number.
Count frequencies:
Loop through each element in the array and update its count in the hash map.
Check for even counts:
Iterate over the hash map values. If any count is odd, return false.
Return true:
If all counts are even, return true, meaning the array can be divided into equal pairs.

Complexity Analysis

Time Complexity:
O(n)
The algorithm iterates through the array once to build the frequency map and then iterates through the map's values. Both operations are linear with respect to the number of elements.
Space Complexity:
O(n)
In the worst case, if all elements are unique, the hash map will store n elements.

Wrap up

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