Uber Interview Question for SDE-2s


Country: United States
Interview Type: Phone Interview




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0
of 0 vote

solution of approach 2; but still don't know how to do the improvement that i mentioned;

static int subArrays(int arr[], int k) {

        int count = 0;
        Set<Integer> set = new HashSet<>();

        //Count single length
        for (int i = 0; i < arr.length; i++) {
            count += set.contains(arr[i]) ? 0 : 1;
            set.add(arr[i]);
        }


        int s = 0;
        int oddCount = 0;

        int i = 0;
        for (; i < arr.length; i++) {

            if (arr[i] % 2 != 0)
                oddCount++;

            //count length > 1
            if (i - s > 0 && oddCount == k) {
                count++;
            }

            //Slide the window
            if (oddCount > k) {

                while (oddCount > k && s < i) {


                    if (arr[s] % 2 != 0)
                        oddCount--;
                    s++;

                    if (i - s > 0 && oddCount == k) {
                        count++;
                        break;

                    }

                }
            }
        }
        i--;

        //if left over array has more then 2 element, count them too
        while (i - s + 1 >= 2) {
            s++;
            if (i - s + 1 >= 2)
                count++;

        }


        return count;
    }

- nitinguptaiit April 20, 2019 | Flag Reply
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of 0 vote

Create a prefix array where the ith element of that array will denote the number of odd elements encountered up till ith index .
Then for every index i suppose the value of the pref[i]=p then use binary search to find the first occurence of a value greater than or equal to p-k . Suppose that index is j . Then j-i+1 is added to the answer .
Complexity O(Nlog2(N)).

- A2J007 April 30, 2019 | Flag Reply
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0
of 0 votes

I think u missed the last part of question or I may not understood ur solution, pls elobrate

- nitinguptaiit May 07, 2019 | Flag


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