Amazon
  • Introduction
  • Phone Interview I
    • 53.Reverse Words in a String
    • 31.Partition Array
  • Phone Interview II
    • 167.Add Two Numbers
    • 88.Lowest Common Ancestor
  • Onsite I
    • 655.Big Integer Addition
    • 221.Add Two Numbers II
  • Onsite II
    • 158.Two Strings Are Anagrams
    • 171.Anagrams
    • 386.Longest Substring with At Most K Distinct Characters
  • Onsite III
    • 479.Second Max of Array
    • 589.Connecting Graph
  • Onsite IV
    • 532.Reverse Pairs
  • 2022
    • OA
      • work simulation
      • Greyness
      • NearestRetailer
      • Sum of Scores of Subarray
      • StrengthOfPassword
      • ProductOf1
      • Move 0/1 InArray
      • Max deviation among all substrings
      • AWS power consumption
      • searchWordResultWord
      • maxOperationOfString
      • MinHealthGame
      • EarliestMonth
      • Package ship
      • RatingConsectiveDecresing
      • LinkedListSum
      • MovingBoxes
      • ValidString
      • MaxValueAfterRemovingFromString
      • Subtree with Maximum Average
    • VO
      • 2022.3
    • BQ
      • doc
      • 2022.4
      • Freq Question
      • 11大类BQ和Follow-ups
      • Page 1
      • BQ 100
      • 35 behavioral questions asked in 95% of Amazon interviews with examples
      • Page 2
      • 反向BQ
    • LP
      • LP-1
      • LP-2
    • SD
      • Design Amazon Prime Video Recommendation System
      • Amazon Order system
    • OOD
      • Linux Find Command
      • Amazon Locker
    • AWS Identity call
    • Interviews
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  1. 2022
  2. OA

Greyness

https://leetcode.com/discuss/interview-question/1753436/Amazon-OA

Previouswork simulationNextNearestRetailer

Last updated 3 years ago

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public int gScale(int [][] arr) {
int[] ones = new int[arr.length];
int[] zeros = new int[arr.length];

    int[] ones_c = new int[arr[0].length];
    int[] zeros_c = new int[arr[0].length];
    for (int i = 0; i< arr.length; i++) {
        for (int j = 0; j<arr[0].length; j++) {
            if (arr[i][j] ==1) {
                ones[i]++;
                ones_c[j]++;
            } else {
                zeros[i]++;
                zeros_c[j]++;
            }
        }
    }
    int max = 0;
    for (int i = 0; i< arr.length; i++) {
        for (int j = 0; j<arr[0].length; j++) {
            max =  Math.max(max, ones[i] + ones_c[j] - (zeros[i]+zeros_c[j]));
        }

    }
    return max;
}