Question

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Easy interview question got harder: given numbers 1..100, find the missing number(s)
Find the missing and duplicate elements in an array in linear time and constant space

I saw an interesting Question on one forum.

you have 100 elements from 1 to 100 but byy mistake one of those number overlapped another by repeating itself.
E.g. 1,99,3,…,99,100
Array is not in sorted format , how to find the repeating number ?

I know Hash can do it O(n) time and O(n) space, I need O(1) space.

Answer 1

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;
}