## Uber Interview Question for SDE1s

• 1
of 1 vote

Country: India

Comment hidden because of low score. Click to expand.
2
of 2 vote

For each element find closest index to left and right and then take closest of two.

``````// Calculate element closest to the left
for (i = 0; i < n; ++i) {
while (stack.top() < a[i]) {
stack.pop();
stack_pos.pop();
}

left[i] = stack.top();
left_pos[i] = stack_pos.top();
left_pos = stack_pos.top();
stack.push(a[i]);
stack_pos.push(i);
}

// Similarly populate right

// Take closest of left and right
for (i = 0; i < n; ++i)
{
closest[i] = closest(left[i], right[i], left_post[i], right_pos[i]);
}``````

Comment hidden because of low score. Click to expand.
0

Neat! Linear solution.

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

``````public void findIndex(int[] nums) {
//find index on the right side
int[] indexLeft = new int[nums.length];
for(int i = 0; i < nums.length; i++) {
int index = findIndex(indexLeft, nums, i, true);
indexLeft[i] = index;
}

int[] indexRight = new int[nums.length];
for(int i = nums.length - 1; i >= 0; i--) {
int index = findIndex(indexRight, nums, i, false);
indexRight[i] = index;
}
for(int i = 0; i< nums.length; i++) {
if(indexRight[i] == -1) System.out.print(indexLeft[i]+" ");
else if(indexLeft[i] == -1) System.out.print(indexRight[i]+" ");
else if(i - indexLeft[i] <= indexRight[i] - i) System.out.print(indexLeft[i]+" ");
else System.out.print(indexRight[i]+" ");
}
}

private int findIndex(int[] indexes, int[] nums, int i, boolean left) {
int index = left? i - 1: i + 1;
if(i == 4 && !left) System.out.println(index);
while(index >= 0 && index < nums.length) {
if(nums[index] <= nums[i]) index = indexes[index];
else return index;
}
return -1;
}``````

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

Python implementation -- worst case O(n^2), avg case can be better
start from i and move left and right till u find a number greater than that at index i

``````def nearestGreater(l):
lengthL = len(l)
k=[None]*lengthL

for i,num in enumerate(l):
j1 = j2 = i
while True:
j1 -= 1
j2 += 1
if j1>0 or j2<lengthL:
if j1>0 and l[j1]>num:
k[i]=j1
break
elif j2<lengthL-1 and l[j2]>num:
k[i]=j2
break
else:
break
return k``````

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

Python implementation:

``````def nearestGreater(l):
lengthL = len(l)
k=[None]*lengthL

for i,num in enumerate(l):
j1 = j2 = i
while True:
j1 -= 1
j2 += 1
if j1>0 or j2<lengthL:
if j1>0 and l[j1]>num:
k[i]=j1
break
elif j2<lengthL-1 and l[j2]>num:
k[i]=j2
break
else:
break
return k``````

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

``````// ZoomBA :: Simple but non-optimal way to solve it
def find_nearest ( arr , i ){
li = index ( [0:i] ) :: { arr[ \$.o ] > arr[i] }
ri = rindex ( [i+1: #|arr|] ) :: { arr[ \$.o ] > arr[i] }
if ( li == -1 && ri == -1 ) return -1
if ( li == -1 ) return ri
return li
}
lfold ( arr ){
printf ('%d -> %d\n', \$.i, find_nearest(arr, \$.i))
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

If we create a Binary search tree. Check for every i if there exists a right child. If there exists then return 1 or else return 0. The complexity would be nlogn.

The other way could be to sort the array using mergesort and then print 1's for all the elements in the sorted array except last element. Complexity= nlogn

Please correct me, I am not sure whether it is right.

Comment hidden because of low score. Click to expand.
0

This approach will not be correct. Using the approach of binary search, you will be able to find greater element than a[i], but that would not mean that it is the nearest maximum element for a[i].

Comment hidden because of low score. Click to expand.
0

Also it asks for index j to be returned if it exists, not 1. You return 1 if you can't find an index j that's value is greater than index i's value.

Comment hidden because of low score. Click to expand.
0
of 0 vote

What does nearest index means?
Can i assume for every index i, j > i & a[j] > a[i] else print 1?

Comment hidden because of low score. Click to expand.
0
of 0 vote

c# implementation.
O(n^2).

``````using System;

namespace NearestBiggerValue {

class Program {

private static int FindBigger( int[] arr, int i ) {

int indexL = i;
int indexR = i;
int? rightVal = null;
int? leftVal = null;

while ( indexL >= 0 || indexR < arr.Length ) {

indexR++;
indexL--;

if (indexR < arr.Length ) {
rightVal = arr[ indexR ];
} else {
rightVal = null;
}

if ( indexL >= 0 ) {
leftVal = arr[ indexL ];
} else {
leftVal = null;
}

if ( ( rightVal != null && rightVal > arr[ i ] ) || ( leftVal != null && leftVal > arr[ i ] ) ) {
break;
}
}

if ( ( rightVal == null && leftVal == null) || ( rightVal <= arr[ i ] && leftVal <= arr[ i ] ) ) {
return -1;
}
if ( rightVal != null && leftVal != null ) {
return rightVal > leftVal ? indexR : indexL;
}
return rightVal == null ? indexL: indexR;
}

private static void Process( int[] arr ) {

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

int val = FindBigger( arr, i );

string str = val == -1 ? -val + " (doesn't exist)" : val.ToString();

Console.WriteLine(\$"For index {i} nearest bigger value is at index {str}");
}
}

static void Main( string[] args ) {

int[] arr = new int[] { 77, 2, 15, 16, 5, 3, 19, 2, 0, -2, 77, 7, 77 };

Process(arr);
}
}
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

do a max R-B tree of values from given array and obviously print the max element's index. print 1 for the right most leaf node.

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

``#include <iostream>``

``#include <vector>``

``using namespace std;``

``void printNearestLargerIndex(const vector<int>& a)``

``{``

``for (int i = 1; i < a.size(); ++i)``

``{``

``int bestJ = 1;``

``int distance = 0;``

``for (int j = 0; j < a.size(); ++j)``

``{``

``if (j != i)``

``{``

``if (a[j] > a[i])``

``{``

``int currentDistance = abs(i-j);``

``if (distance == 0)``

``{``

``bestJ = j;``

``distance = currentDistance;``

``}``

``else``

``{``

``if (currentDistance < distance)``

``{``

``distance = currentDistance;``

``bestJ = j;``

``}``

``}``

``}``

``}``

``cout << bestJ << endl;``

``}``

``}``

``}``

``int main()``

``{``

``vector<int> a;``

``a.push_back(1);``

``a.push_back(77);``

``a.push_back(-4);``

``a.push_back(140);``

``printNearestLargerIndex(a);``

``return 0;``

``}``

Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

### Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

### Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.