Find the Missing Value

See the original problem on HackerRank.

Given an unsorted array of \( n-1 \) elements, it is missing exactly one element from the sequence of \( 1 \) to \( n\). Find the missing element.

Input Format

An integer \( n \) followed by \( n-1 \) space separated numeric values.

Constraints

\( 1<n<10^6 \)

Output Format

The missing value.

Solutions

This is an example of problems that we call manifold at Coding Gym, since they give us room to find several alternative solutions.

Sort + Search

We sort the input array and we linearly scan the sequence to find the missing value. For example: In C++:

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#include <iterator>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() 
{
    int n; cin >> n;
    vector<int> v(n-1);
    copy_n(istream_iterator<int>(cin), n-1, begin(v));
    sort(begin(v), end(v));
    auto missing = n;
    for (auto i=1; i<n; ++i)
    {
        if (i != v[i-1])
        {
            missing = i;
            break;
        }
    }
    cout << missing;
}

In Haskell:

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import Data.List (sort)
findTheMissingValue (_:xs) = head [ i | (x, i) <- zip (sort xs) [1..], i /= x ]
main = interact $ show . findTheMissingValue . fmap read . words

Pros:

easy to extend to a more generic domain (e.g. if we have a baseline array, we can easily adapt this solution to locate the element missing in the second array)

Cons:

we pay sort - \( O(N \cdot logN) \)

Notes:

we change the original array (what if not permitted?)

Variant: use binary search to locate the missing value, since the array is sorted.

Additional Storage

We could keep track of elements into a support storage (e.g. a flag array). This is just an example:

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vector<bool> hits(n, false); // C++ addicted will disapprove this!
for (auto val : v)
    hits[val-1] = true;
    
cout << (distance(begin(hits), find(begin(hits), end(hits), false))+1);

Pros:

it’s linear
we do not modify the input array
easy to extend to a more generic domain and also to duplicates (e.g. turn the vector into a frequency array)

Cons:

we pay additional storage

Mathematics

We can exploit the problem domain to the limit.

calculate the sum of all the numbers from \( 1 \) to \( N \). Call it \( expected \)
calculate the sum of input numbers. Call if \( sum \)
the missing value is just \( expected - sum \)

It’s interesting to note that the sum from \( 1 \) to \( N \) is the Gauss Formula:

\( \dfrac{n(n+1)}{2} \)

In C++:

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auto sum = accumulate(begin(v), end(v), 0ull); // aka: reduce
auto expected = n*(n+1)/2;
cout << (expected - sum);

Pros:

efficient
no extra space
we do not modify the input array
the same idea (summation, not Gauss) can be reused to solve the same problem in more generic domains

Cons:

what if \( \dfrac{n(n+1)}{2} \) overflows?

In Haskell there is no overflow, because the type Integer is unbounded (also known as bignum).

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findTheMissingValue :: [Integer] -> Integer
findTheMissingValue (n:xs) = (n * (n + 1) `div` 2) - sum xs
main = interact $ show . findTheMissingValue . fmap read . words

XOR

The following solution is very similar to the previous one but it never overflows.

It’s based on XOR property:

\( A \oplus B = C \)
\( A \oplus C = B \)
\( B \oplus C = A \)

That’s the idea:

suppose \( A \) is the result of XORing all the input elements
suppose \( B \) is the missing value
the result of XORing \( A \oplus B \) is \( C \)
then \( C \) is the result of XORing all the elements from 1 to N
we can easily find \( B \) as \( A \oplus C \)

Talk is cheap, let’s look at some code:

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auto xorVals = accumulate(begin(v), end(v), 0, bit_xor<>{}); // aka: reduce
auto expected = 0;
// naive approach, for an efficient method see, for instance,
// here https://www.geeksforgeeks.org/calculate-xor-1-n/
for (auto i=1; i<=n; ++i)
    expected ^= i;
    
cout << (expected ^ xorVals);

In Haskell, but reducing with xor on a single list: the concatenation of the original list plus numbers from 1 to n.

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import Data.Bits (xor)
findTheMissingValue :: [Integer] -> Integer
findTheMissingValue (n:xs) = foldl xor 0 ([1..n] ++ xs)
main = interact $ show . findTheMissingValue . fmap read . words

Pros:

never overflows
the same idea can be reused to solve the same problem in more generic domains

Cons:

tricky ;)

manifold