Find Missing Element

Problem Statement: Write a function to find the missing element from given two arrays.

Difficulty level: Easy

Test Cases:

• [1, 2, 3, 4, 5, 6, 7],[3, 7, 2, 1, 4, 6] —> 5
• [5, 7, 7, 5, 7], [7, 7 ,5, 5] —> 7
• [9, 8, 7, 6, 5, 4, 3, 2, 1],[9, 8, 7, 5, 4, 3, 2, 1] —> 6
• [1, 2, 4, 7, 9], [7, 9, 4, 2] —> 1

There are number of ways this problem can be solved, below are 4 possible solutions

Algorithm

1. Dictionary Method

   • Initialise a dictionary.
   • Iterate through the first array,

   • If element at current iteration exists in dictionary,

     • Increment it’s value by 1.

   • Else,

     • Add the element as key and value as 1 in dictionary
   • Iterate through the second array,

   • If element at current iteration exists in dictionary,

     • Decrement it’s value by 1.

   • Else,

     • Add the element as key and value as 1 in dictionary.

   • for every element in dictionary

     • if value of the element is not 0

       • return the element

2. Exclusive OR (XOR) Method

   • Initialise variable result to 0
   • Combine both the arrays.
   • Iterate through the combined array
   • Do XOR operation of every element with result
   • Return the result, which will be the missing element

3. Sort Method

   • Sort both the given arrays.
   • Iterate over both the arrays simultaneously.

   • At a given iteration if value of both the iterators are different,

     • return the element from first array, as it is the missing element.

4. Sum Method

   • Calculate the sum of all the elements in both the arrays.
   • Subtract the sum of second array from the first array.
   • The result will be the missing element.

Time and Space complexity

1. Dictionary Method

   • Time Complexity: O(n)
   • Space Complexity: O(n)

2. XOR Method

   • Time Complexity: O(n)
   • Space Complexity: O(n)

3. Sort Method

   • Time Complexity: O(nlogn)
   • Space Complexity: O(n)

4. Sum Method

   • Time Complexity: O(n)
   • Space Complexity: O(1)

Code

class MissingElement(object):
    def missingElement(self, arr1, arr2):
        count = {}
        for element in arr1:
            if element in count:
                count[element] += 1
            else:
                count[element] = 1
        for element in arr2:
            if element in count:
                count[element] -= 1
            else:
                count[element] = 1
        for k in count:
            if count[k] != 0:
                return k

    def missingElement2(self, arr1, arr2):
        result = 0
        for element in arr1 + arr2:
            result ^= element
        return result

    def missingElement3(self, arr1, arr2):
        arr1.sort()
        arr2.sort()
        for ele1, ele2 in zip(arr1, arr2):
            if ele1 != ele2:
                return ele1
        return arr1[-1]

    def missingElement4(self, arr1, arr2):
        return abs(sum(arr1) - sum(arr2))

Unit Test

import unittest
from missingElement import MissingElement


class TestMisssingElement(unittest.TestCase):
    def test_ele(self, sol):
        self.assertEqual(sol([5, 5, 7, 7], [5, 7, 7]), 5)
        self.assertEqual(sol([1, 2, 3, 4, 5, 6, 7],
                             [3, 7, 2, 1, 4, 6]), 5)
        self.assertEqual(sol([9, 8, 7, 6, 5, 4, 3, 2, 1],
                             [9, 8, 7, 5, 4, 3, 2, 1]), 6)
        print("All test cases passed")


def main():
    test = TestMisssingElement()
    element = MissingElement()
    test.test_ele(element.missingElement)
    test.test_ele(element.missingElement2)
    test.test_ele(element.missingElement3)
    test.test_ele(element.missingElement4)


if __name__ == "__main__":
    main()

Github repo

Happy Coding !