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
-
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
- if value of the element is not 0
-
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
-
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.
-
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
- Dictionary Method
- Time Complexity: O(n)
- Space Complexity: O(n)
- XOR Method
- Time Complexity: O(n)
- Space Complexity: O(n)
- Sort Method
- Time Complexity: O(nlogn)
- Space Complexity: O(n)
- 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()
Top comments (2)
Excellent! I was actually looking for this kind of program. Thanks for sharing :)
Java #Python #Challenge
Glad you liked it. 🙂