Single Number - III - AlgoArena Witeso

Intuition:

The brute force way to solve this will be to utilize a frequency counting approach. By keeping track of the frequency of each element in the array, the element that appears only once can be easily identified.

Approach:

Use a hash map to store the frequency of each element in the array. The keys of the map are the elements of the array, and the values are their corresponding frequencies.
Traverse the entire array once, and for each element, update its frequency count in the map.
After populating the map with frequency counts, iterate over the map to find the elements with a frequency of 1. Add these elements in the result array.
Return the resulting array after sorting it.

Dry Run:

Solution:

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    /* Function to get the single 
    number in the given array */
    vector<int> singleNumber(vector<int>& nums){
        
        // Array to store the answer
        vector<int> ans;
        
        /* Map to store the elements 
        and their frequencies */
        unordered_map <int, int> mpp;
        
        // Iterate on the array
        for(int i=0; i < nums.size(); i++) {
            mpp[nums[i]]++; // Update the map
        }
        
        // Iterate on the map
        for(auto it : mpp) {
            // If frequency is 1
            if(it.second == 1) {
                /* Add the element to
                the result array */
                ans.push_back(it.first);
            }
        }   
        
        // Return the result after sorting
        sort(ans.begin(), ans.end());
        return ans;
    }
};

int main() {
    vector<int> nums = {1, 2, 1, 3, 5, 2};
    
    /* Creating an instance of 
    Solution class */
    Solution sol; 
    
    /* Function call to get the single 
    number in the given array */
    vector<int> ans = sol.singleNumber(nums);
    
    cout << "The single numbers in given array are: " << ans[0] << " and " << ans[1];
    
    return 0;
}
import java.util.*;

public class Solution {
    /* Function to get the single 
    number in the given array */
    public List<Integer> singleNumber(int[] nums) {
        
        // Array to store the answer
        List<Integer> ans = new ArrayList<>();
        
        /* Map to store the elements 
        and their frequencies */
        HashMap<Integer, Integer> mpp = new HashMap<>();
        
        // Iterate on the array
        for (int num : nums) {
            mpp.put(num, mpp.getOrDefault(num, 0) + 1); // Update the map
        }
        
        // Iterate on the map
        for (Map.Entry<Integer, Integer> entry : mpp.entrySet()) {
            // If frequency is 1
            if (entry.getValue() == 1) {
                /* Add the element to
                the result array */
                ans.add(entry.getKey());
            }
        }
        
        // Return the result after sorting
        Collections.sort(ans);
        return ans;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 1, 3, 5, 2};
        
        /* Creating an instance of 
        Solution class */
        Solution sol = new Solution(); 
        
        /* Function call to get the single 
        number in the given array */
        List<Integer> ans = sol.singleNumber(nums);
        
        System.out.println("The single numbers in given array are: " + ans.get(0) + " and " + ans.get(1));
    }
}
class Solution:
    # Function to get the single 
    # number in the given array 
    def singleNumber(self, nums):
        
        # Array to store the answer
        ans = []
        
        # Map to store the elements 
        # and their frequencies
        mpp = {}
        
        # Iterate on the array
        for num in nums:
            mpp[num] = mpp.get(num, 0) + 1 # Update the map
        
        # Iterate on the map
        for key, value in mpp.items():
            # If frequency is 1
            if value == 1:
                # Add the element to
                # the result array 
                ans.append(key)
        
        # Return the result after sorting
        ans.sort()
        return ans

# Creating an instance of Solution class
sol = Solution()

# Function call to get the single number in the given array
nums = [1, 2, 1, 3, 5, 2]
ans = sol.singleNumber(nums)

print("The single numbers in given array are:", ans[0], "and", ans[1])
class Solution {
    /* Function to get the single 
    number in the given array */
    singleNumber(nums) {
        
        // Array to store the answer
        let ans = [];
        
        /* Map to store the elements 
        and their frequencies */
        let mpp = new Map();
        
        // Iterate on the array
        for (let num of nums) {
            mpp.set(num, (mpp.get(num) || 0) + 1); // Update the map
        }
        
        // Iterate on the map
        for (let [key, value] of mpp.entries()) {
            // If frequency is 1
            if (value === 1) {
                /* Add the element to
                the result array */
                ans.push(key);
            }
        }
        
        // Return the result after sorting
        ans.sort((a, b) => a - b);
        return ans;
    }
}

// Creating an instance of Solution class
let sol = new Solution();

// Function call to get the single number in the given array
let nums = [1, 2, 1, 3, 5, 2];
let ans = sol.singleNumber(nums);

console.log("The single numbers in given array are: " + ans[0] + " and " + ans[1]);

Complexity Analysis:

Time Complexity: O(N) (where N is the size of the array)

Traversing the array to update the Hash Map - O(N).

Traversing on the map - O(N) (in worst case).

Sorting the answer array - O(2*log(2)) ~ O(1).
Hence, O(N) + O(N) + O(1) ~ O(N).

Space Complexity: O(N), Using a hashmap data structure and in the worst-case (when all elements in the array are unique), it will store N key-value pairs.

Intuition:

An optimal approach to solve this problem will be possible if we can divide the elements in array into two groups such that each group only contains one unique element. This way, the problem boils down to Single Number-I and both the unique elements can be identified with ease.

Now, to divide the numbers into two groups(buckets), the rightmost set bit in the overall XOR of all elements can be used. The overall XOR of all elements is equivalent to the XOR of the two unique numbers.

Why Divide using the Rightmost Set Bit?

This is because of the following reasons:

The two unique numbers needs to be separated into two different groups, and the rightmost set bit in the overall XOR will indicate the bit position where the the two unique numbers differ.
Using this bit, all the elements in the array can be divided into two groups (buckets):
- One group where this bit is set.
- Another group where this bit is not set.

Approach:

Traverse the entire array, performing an XOR operation on all numbers. This will effectively cancel out all the numbers that appear twice, leaving us with the XOR of the two unique numbers.
Determine the rightmost set bit (bit that is 1) in the result from the first step. This set bit can be used to differentiate the two unique numbers since they must differ at this bit position.
Traverse the array again, but this time divide the numbers into two groups:
- One group where the numbers have the rightmost set bit.
- Another group where the numbers do not have this bit set.
- Perform XOR operations while adding numbers in each group. This will cancel out the duplicate numbers, leaving only the unique numbers in each group.
Sort the two unique numbers in ascending order and return them.

Dry Run:

Solution:

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    /* Function to get the single 
    numbers in the given array */
    vector<int> singleNumber(vector<int>& nums){
        // Variable to store size of array
        int n = nums.size();
        
        // Variable to store XOR of all elements
        long XOR = 0;
        
        // Traverse the array
        for(int i=0; i < n; i++) {
            
            // Update the XOR
            XOR = XOR ^ nums[i];
        }
        
        /* Variable to get the rightmost 
        set bit in overall XOR */
        int rightmost = (XOR & (XOR - 1)) ^ XOR;
        
        /* Variables to stores XOR of
        elements in bucket 1 and 2 */
        int XOR1 = 0, XOR2 = 0;
        
        // Traverse the array
        for(int i=0; i < n; i++) {
            
            /* Divide the numbers among bucket 1
             and 2 based on rightmost set bit */
            if(nums[i] & rightmost) {
                XOR1 = XOR1 ^ nums[i];
            }
            else {
                XOR2 = XOR2 ^ nums[i];
            }
        }
        
        // Return the result in sorted order
        if(XOR1 < XOR2) return {XOR1, XOR2};
        return {XOR2, XOR1};
    }
};

int main() {
    vector<int> nums = {1, 2, 1, 3, 5, 2};
    
    /* Creating an instance of 
    Solution class */
    Solution sol; 
    
    /* Function call to get the single 
    numbers in the given array */
    vector<int> ans = sol.singleNumber(nums);
    
    cout << "The single numbers in given array are: " << ans[0] << " and " << ans[1];
    
    return 0;
}
import java.util.*;

class Solution {
    /* Function to get the single 
    numbers in the given array */
    public int[] singleNumber(int[] nums) {
        // Variable to store size of array
        int n = nums.length;
        
        // Variable to store XOR of all elements
        long XOR = 0;
        
        // Traverse the array
        for(int i=0; i < n; i++) {
            // Update the XOR
            XOR = XOR ^ nums[i];
        }
        
        /* Variable to get the rightmost 
        set bit in overall XOR */
        int rightmost = (int)(XOR & (XOR - 1)) ^ (int)XOR;
        
        /* Variables to stores XOR of
        elements in bucket 1 and 2 */
        int XOR1 = 0, XOR2 = 0;
        
        // Traverse the array
        for(int i=0; i < n; i++) {
            /* Divide the numbers among bucket 1
             and 2 based on rightmost set bit */
            if((nums[i] & rightmost) != 0) {
                XOR1 = XOR1 ^ nums[i];
            }
            else {
                XOR2 = XOR2 ^ nums[i];
            }
        }
        
        // Return the result in sorted order
        if(XOR1 < XOR2) return new int[]{XOR1, XOR2};
        return new int[]{XOR2, XOR1};
    }
    
    public static void main(String[] args) {
        int[] nums = {1, 2, 1, 3, 5, 2};
        
        /* Creating an instance of 
        Solution class */
        Solution sol = new Solution();
        
        /* Function call to get the single 
        numbers in the given array */
        int[] ans = sol.singleNumber(nums);
        
        System.out.println("The single numbers in given array are: " + ans[0] + " and " + ans[1]);
    }
}
class Solution:
    # Function to get the single 
    # numbers in the given array 
    def singleNumber(self, nums):
        # Variable to store size of array
        n = len(nums)
        
        # Variable to store XOR of all elements
        XOR = 0
        
        # Traverse the array
        for i in range(n):
            # Update the XOR
            XOR = XOR ^ nums[i]
        
        # Variable to get the rightmost 
        # set bit in overall XOR 
        rightmost = (XOR & (XOR - 1)) ^ XOR
        
        # Variables to stores XOR of
        # elements in bucket 1 and 2
        XOR1, XOR2 = 0, 0
        
        # Traverse the array
        for i in range(n):
            # Divide the numbers among bucket 1
            # and 2 based on rightmost set bit 
            if nums[i] & rightmost:
                XOR1 = XOR1 ^ nums[i]
            else:
                XOR2 = XOR2 ^ nums[i]
        
        # Return the result in sorted order
        return [XOR1, XOR2] if XOR1 < XOR2 else [XOR2, XOR1]

# Example usage
nums = [1, 2, 1, 3, 5, 2]

# Creating an instance of Solution class
sol = Solution()

# Function call to get the single numbers in the given array
ans = sol.singleNumber(nums)

print("The single numbers in given array are:", ans[0], "and", ans[1])
class Solution {
    /* Function to get the single 
    numbers in the given array */
    singleNumber(nums) {
        // Variable to store size of array
        let n = nums.length;
        
        // Variable to store XOR of all elements
        let XOR = 0;
        
        // Traverse the array
        for(let i = 0; i < n; i++) {
            // Update the XOR
            XOR = XOR ^ nums[i];
        }
        
        /* Variable to get the rightmost 
        set bit in overall XOR */
        let rightmost = (XOR & (XOR - 1)) ^ XOR;
        
        /* Variables to stores XOR of
        elements in bucket 1 and 2 */
        let XOR1 = 0, XOR2 = 0;
        
        // Traverse the array
        for(let i = 0; i < n; i++) {
            /* Divide the numbers among bucket 1
             and 2 based on rightmost set bit */
            if(nums[i] & rightmost) {
                XOR1 = XOR1 ^ nums[i];
            }
            else {
                XOR2 = XOR2 ^ nums[i];
            }
        }
        
        // Return the result in sorted order
        return XOR1 < XOR2 ? [XOR1, XOR2] : [XOR2, XOR1];
    }
}

// Example usage
let nums = [1, 2, 1, 3, 5, 2];

/* Creating an instance of 
Solution class */
let sol = new Solution();

/* Function call to get the single 
numbers in the given array */
let ans = sol.singleNumber(nums);

console.log("The single numbers in given array are:", ans[0], "and", ans[1]);

Complexity Analysis:

Time Complexity: O(N), Traversing the array twice results in O(2*N) TC.

Space Complexity: O(1), Using a couple of variables i.e., constant space.

Problem List

Single Number - III

Examples:

Constraints

Hints

Company Tags

Editorial

Intuition:

Approach:

Dry Run:

Solution:

Complexity Analysis:

Intuition:

Why Divide using the Rightmost Set Bit?

Approach:

Dry Run:

Solution:

Complexity Analysis:

Frequently Occurring Doubts

Interview Followup Questions

Notes

Code