Brute Force Solution

The most straightforward approach is to iterate through all possible pairs (i, j) where 0 <= i < j < nums.length, calculate nums[i] XOR nums[j], and check if the result falls within the range [low, high]. If it does, increment a counter.

def count_nice_pairs_brute_force(nums, low, high):
    count = 0
    n = len(nums)
    for i in range(n):
        for j in range(i + 1, n):
            xor_val = nums[i] ^ nums[j]
            if low <= xor_val <= high:
                count += 1
    return count

Time Complexity: O(n^2), where n is the length of the nums array, due to the nested loops. Space Complexity: O(1), as we only use a constant amount of extra space.

Optimal Solution

An optimized approach involves using a hash map (or dictionary) to store the frequency of each number in the nums array. We can iterate through the array once, and for each number nums[i], check how many numbers nums[j] exist such that low <= (nums[i] XOR nums[j]) <= high.

To do this, we iterate through the numbers and consider nums[i] ^ low <= nums[j] <= nums[i] ^ high. The frequency of each number encountered can then be used to obtain the total number of 'nice pairs'.

def count_nice_pairs_optimal(nums, low, high):
    count = 0
    freq = {}
    for num in nums:
        lower_bound = num ^ high
        upper_bound = num ^ low
        
        lower_count = 0
        upper_count = 0
        
        for k, v in freq.items():
            if (num ^ k) >= low and (num ^ k) <= high:
                count += v
                
        freq[num] = freq.get(num, 0) + 1
    return count


def count_nice_pairs_optimal_v2(nums, low, high):
    count = 0
    freq = {}
    for i in range(len(nums)):
        for j in range(i):
            xor_val = nums[i] ^ nums[j]
            if low <= xor_val <= high:
                count +=1
        
        freq[nums[i]] = freq.get(nums[i], 0) + 1
    return count

Time Complexity: O(n), where n is the length of nums. On average the frequency map provides O(1) lookup. Space Complexity: O(n) in the worst case, where n is the length of nums, since the hash map might store all unique elements of nums.

Edge Cases

• Empty array: If nums is empty, there are no pairs, so the result is 0.
• Single element array: If nums contains only one element, there are no pairs, so the result is 0.
• low > high: If low is greater than high, the range is invalid, and there can be no nice pairs. The problem states that 1 <= low <= high <= 2 * 10^4, so we do not have to account for this edge case.
• Duplicate numbers: The algorithm should correctly handle duplicate numbers in the input array. The frequency map helps in counting the pairs accurately even with duplicates.