Find repeating in O(n) and constant space [duplicate]

Answer 1

New approach. Let m be the missing number and r be the repeated number. Passing through the array once, let X be the result of XORing the entries of the array along with the indices 1 to n. Then X = m XOR r. In particular, it isn't 0. Let b be any nonzero bit of X (you only need to choose one, and one exists since X is not 0). Passing through the array, let Y be the result of XORing the entries of the array and the indices 1 to n where the bit b is 0 and let Z be the result of XORing the entries of the array and the indices 1 to n where the bit b is 1. Then Y and Z hold m and r, so all that remains is to make a final pass to see which is in the array.

Total space: 4 (or 3 if you reuse X for b)

Total time: 7 passes (or 3 if you do indices at the same time as the array and compute Y and Z at the same time.

Hence O(1) space and O(n) time.

Answer 2

Calculate Sum of all integers Calculate int p = sum % 100 100 - p is the answer

Answer 3

We can do it in O(n) and constant space:

1. For every element, calculate index = Math.abs(a[i]) - 1
2. Check the value at index
   • If it is positive, multiply by -1, i.e., make it negative.
   • if it is negative, return (index+1) as answer, as it means we have seen this index before.

""

static int findDup(int[] a){
    for(int i=0;i<a.length;i++){
        int index = Math.abs(a[i]) - 1;
        if(a[index] < 0)
            return index+1;
        else
            a[index] = -1 * a[index];
    }
    return -1;
}

Answer 4

Calculate two sums: sum and square sum.

In your example:

sum = 1+99+3...+100

sq_sum = 1^2+99^2+3^2+...+100^2

Assume y replaced x in the sequence.

sum = n(n+1)/2 -y+x.
sq_sum = n(n+1)(2n+1)/6 -x^2 +y^2

Now, solve for x & y.

Constant space and O(n) time.

How to solve for x and y

From equation:

x = sum - n(n+1)/2 +y

Substitute this in the second equation:

sq_sum = n(n+1)(2n+1)/6 -(sum - n(n+1)/2 +y)^2 +y^2

Note that y^2 cancels and you are left with a linear equation with only one unknown:y. Solve it!