Facebook Interview Question for Software Developers


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The output seems to be wrong I guess

After first transformation , we get (3,-1,4,-1,2) , the output is 7
After second transformation , we get (3,4,-1,2,-1) , the output is 8

We can use Kadane's algorithm to find the max sum subarray

- Anonymous October 29, 2018 | Flag Reply
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I thought same at first but I guess they are considering array from position 1 so the case will be L-1 and R-1.
(I guessed that from the test case given)

- Anonymous October 29, 2018 | Flag
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Using Kadane's for each query, we will end up with time O(q*n). What if q and n are of order 10^5. Can we do better in time?

- Anonymous November 03, 2018 | Flag
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Output given seems to be wrong 

After first query the array is (3,-1,4,-1,2), output is 7
After second query the array is (3,4,-1,2,-1), output is 8

- Anonymous October 29, 2018 | Flag Reply
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The Li and Ri don't start at 0. They start at 1. So the indices are Li-1 and Ri-1. After that it's a normal 'return the maximum continuous sum of an array'.

- Josh October 29, 2018 | Flag Reply
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This is very good exercise to learn about segment trees.
once you build segment tree, for each query, do following;
L = query(0, l-1), M = query(l, r), R = query(r+1, n-1);
since the subarray between 'l' & 'r' is being reversed, all you got to do is swap, prefix and suffix sum for M.
answer will be L + M + R.

- Anonymous December 23, 2018 | Flag Reply
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Here's an improved version of the question along with the answer in PHP :

//You are given an array A of size N and Q queries. 
//For each query, you are given two indices of the array L and R.
//The subarray generated from L to R is reversed.
//Your task is to determine the maximum contiguous sum of the subarrays.
//Ex : 
//5 2
//3 -1 4 2 -1
//3 4
//1 2
//Given array is {3,-1,4,2,-1}.
//For first query L=3 and R=4. Now the array becomes {3,-1,2,4,-1}.
//Maximum sub-array sum is 8 and the sub-array is {3,-1,2,4}.
//For second query L=1 and R=2. Now the array becomes {-1,3,4,2,-1}.
//Maximum sub-array sum is 9 and the sub-array is {3,4,2}.
function maxSubArray($array,$startTrans,$endTrans){
    $newArray=array();
    for($i=0;$i<sizeof($array);$i++){
        if($i == $startTrans-1){
            for($j=$endTrans-1;$j>=$i;$j--)
                $newArray[]=$array[$j];
            $i=$endTrans-1;
        }else{
            $newArray[]=$array[$i];
        }
    }
    $max=0;$start=0;$end=0;$sum=0;
    for($i=0;$i<sizeof($newArray);$i++){
       $sum+=$newArray[$i];
       if($sum>$max){
            $max=$sum;
            $end=$i;
       }
       if($sum<0){
           $sum=0;
           $start=$i+1;
       }
    }
    return $max;
}

- Mathboy January 27, 2019 | Flag Reply
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build segment tree geeksforgeeks/maximum-subarray-sum-given-range/
As pointed out earlier, do following query of segment tree
L = query(0, l-1), M = query(l, r), R = query(r+1, n-1);
since the subarray between 'l' & 'r' is being reversed, all you got to do is swap, prefix and suffix sum for M.
answer will be merge(merge(L,M),R).

- Anonymous September 07, 2019 | Flag Reply
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The problem statement says left inclusive and right inclusive.
The example works with zero indexed too.
Another note, the array goes back to normal after each query, so we can iterate array from 0->Li, Ri->Li, (Ri+1)->length.
It looks to be Kadane's algorithm with small modification.

import java.util.*;
public class Solution {

  static int subArraySumWithFlip(int arr[], int l, int r) {
      int max_so_far = Arrays.stream(arr).max().getAsInt();
      int max_sum = max_so_far;
      for (int i=0; i<l; i++) {
          max_so_far = Math.max(arr[i], max_so_far + arr[i]);
          max_sum = Math.max(max_sum, max_so_far);
      }
      for (int i=r; i>=l; i--) {
          max_so_far = Math.max(arr[i], max_so_far + arr[i]);
          max_sum = Math.max(max_sum, max_so_far);
      }
      for (int i=r+1; i<arr.length; i++) {
          max_so_far = Math.max(arr[i], max_so_far + arr[i]);
          max_sum = Math.max(max_sum, max_so_far);
      }
      return max_sum;
  }
  public static void main(String args[]){
      int[] arr = new int[]{5,-5,-2,4,1};
      System.out.println(subArraySumWithFlip(arr,0,2));
  }
}

- Kiran T October 23, 2019 | Flag Reply


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