1234567891011121314151617181920212223242526272829303132

class Solution:
    # Function to find the element that appears only once
    def findSingleNumber(self, nums):
        # Initialize dictionary to store frequency of elements
        freq = {}

        # Populate the dictionary with frequency of each element
        for num in nums:
            freq[num] = freq.get(num, 0) + 1

        # Iterate through dictionary to find element with frequency 1
        for key, value in freq.items():
            if value == 1:
                return key

        # Return -1 if no such element is found
        return -1

# Driver code
if __name__ == "__main__":
    nums = [4, 1, 2, 1, 2]

    # Print the input array
    print("Initial array:", nums)

    # Create solution object and call function
    sol = Solution()
    ans = sol.findSingleNumber(nums)

    # Print the result
    print("Element that appears only once:", ans)

123456789101112131415161718192021222324252627282930313233343536

class Solution {
    // Function to find the element that appears only once
    findSingleNumber(nums) {
        // Initialize a map to store frequency of elements
        let freq = new Map();

        // Populate the map with frequency of each element
        for (let num of nums) {
            freq.set(num, (freq.get(num) || 0) + 1);
        }

        // Iterate through the map to find element with frequency 1
        for (let [key, value] of freq.entries()) {
            if (value === 1) {
                return key;
            }
        }

        // Return -1 if no such element is found
        return -1;
    }
}

// Driver code
let nums = [4, 1, 2, 1, 2];

// Print the input array
console.log("Initial array:", nums);

// Create solution object and call function
let sol = new Solution();
let ans = sol.findSingleNumber(nums);

// Print the result
console.log("Element that appears only once:", ans);

Complexity Analysis

Time Complexity: O(N x log₂M + M) where N is the number of elements in the input array and M = (N/3 + 1).

Inserting into the hashmap with the frequency of each element in the input array requires iterating through the entir