Find the missing and duplicate elements in an array in linear time and constant space

If all numbers were present in the array the sum would be N(N+1)/2.

Determine the actual sum by summing up all numbers in the array in O(n), let this be Sum(Actual).

One number is missing, let this be j and one number is duplicated, let this be k. That means that

Sum(Actual) = N(N+1)/2 + k - j

derived from that

k = Sum(Actual) -N(N+1)/2 + j

Also we can calculate the sum of squares in the array, which would sum up to n³/3 + n²/2 + n/6 if all numbers were present.

Now we can calculate the actual sum of squares in O(n), let this be Sum(Actual Squares).

Sum(Actual Squares) =n³/3 + n²/2 + n/6 + k² - j²

Now we have two equations with which we can determine j and k.

The XOR trick works in two passes with a read-only array.

This avoids the problem of possible integer overflows which the sum and sum of squares solution has.

Let the two numbers be x and y, one of which is the missing number and the other repeated.

XOR all the elements of the array, along with 1,2,...,N.

The result is w = x XOR y.

Now since x and y are distinct, w is non-zero.

Pick any non-zero bit of w. x and y differ in this bit. Say the position of the bit is k.

Now consider a split of the array (and the numbers 1,2,...,N) into two sets, based on whether the bit at position k is 0 or 1.

Now if we compute the XOR (separately) of the elements of the two sets, the result has to be x and y.

Since the criteria for splitting is just checking if a bit is set of not, we can compute the two XORs of the two sets by making another pass through the array and having two variables, each of which holds the XOR of the elements seen so far (and 1,2,...N), for each set. At the end, when we are done, those two variables will hold x and y.

Related:

• Finding missing elements in an array which can be generalized to m appearing twice and m missing.

• Find three numbers appeared only once which is about three missing.

Using the basic idea from a related interview question you could do:

• Sum up all the numbers (we shall call this S1) and their squares (S2)
• Compute the expected sum of the numbers, without modifications, i.e. E1 = n*(n+1)/2 and E2 = n*(n+1)*(2n+1)/6
• Now you know that E1 - S1 = d - m and E2 - S2 = d^2 - m^2, where d is the duplicated number and m the missing one.

• Solve this system of equations and you'll find that:

  m = 1/2 ((E2 - S2)/(E1 - S1) - (E1 - S1))
  d = 1/2 ((E2 - S2)/(E1 - S1) + (E1 - S1)) // or even simpler: d = m + (E1 - S1)
  

.

$S1 = $S2 = 0;
foreach ($nums as $num) {
    $S1 += $num;
    $S2 += $num * $num;
}

$D1 = $n * ($n + 1) / 2                - $S1;
$D2 = $n * ($n + 1) * (2 * $n + 1) / 6 - $S2;

$m = 1/2 * ($D2/$D1 - $D1);
$d = 1/2 * ($D2/$D1 + $D1);