Adobe Interview Question for Interns


Country: India
Interview Type: Written Test




Comment hidden because of low score. Click to expand.
1
of 1 vote

solution O(n*lg(n)) time O(n) space:
1) iterate j from 0 to n-1
2) maintain a vector, e.g dp. That vector gets extended if element a[j] is smaller than
last element of vector
3) if element doesn't get extended, binary search in vector for the element that is smaller
than a[j] it's index will be i, max on j-i

#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;

// returns index of largest element < value in a[dp[i]]
int largest_smaller_idx(const vector<int>&a, const vector<int>& dp, int value)
{
	int l = 0;
	int r = dp.size() + 1;
	while (l + 1 < r) {
		int m = (l + r - 2) / 2;
		if (a[dp[m]] < value) {
			r = m + 1;
		} else {
			l = m + 1;
		}
	}
	return l;
}

int max_dist_larger_right(const vector<int>& a)
{
	if (a.size() < 2) return -1;
	
	int max_d = -1;
	vector<int> dp(1, 0);
	for (int j = 1; j < a.size(); ++j) {
		if (a[j] < a[dp.back()]) {
			dp.push_back(j);
		} else if (a[j] > a[dp.back()]) { 
			int i = largest_smaller_idx(a, dp, a[j]);
			max_d = max(j - dp[i], max_d);
		}
	}
	return max_d;
}

int main()
{
	cout << "TC1: " << (max_dist_larger_right({ 9, 11, 7, 8, 3, 6, 6, 11, 9, 8, 7 }) == 7) << endl;
	cout << "TC2: " << (max_dist_larger_right({ 1, 2 }) == 1) << endl;
	cout << "TC3: " << (max_dist_larger_right({ 1, 2, 1 }) == 1) << endl;
	cout << "TC4: " << (max_dist_larger_right({ 1, 1, 2, 1 }) == 2) << endl;
	cout << "TC5: " << (max_dist_larger_right({ 1, 2, 1, 3 }) == 3) << endl;
	cout << "TC6: " << (max_dist_larger_right({ 9, 8, 1, 2, 1, 2, 1, 3, 2, 1, 0 }) == 6) << endl;
	cout << "TC7: " << (max_dist_larger_right({ 1, 1 }) == -1) << endl;
	cout << "TC8: " << (max_dist_larger_right({ 2, 1 }) == -1) << endl;
}

- Chris December 01, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

Variant of Longest Increasing Subsequence problem.

- Parbays January 02, 2018 | Flag Reply


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