8 hours ago

Rohit • 30

Hyderabad

Question 1: Maximize Zeroes

Problem Statement:

In an Amazon analytics team, the Analysts collectively have a preference for the number zero and a disapproval towards negative numbers. Their objective is to determine the maximum number of zeroes achievable after performing a series of operations (possibly zero) on an array, all while avoiding negative values.

Formally, given an array sequenceData of size n of positive integers, the Analysts can perform the following operation any number of times (possibly zero):

• Choose a prefix of size s (1 <= s <= n) that doesn't contain any zero (there is no index i such that sequenceData[i] = 0 and 1 \le i \le s).
• Decrease all values of this prefix by one, i.e., set sequenceData[i] = sequenceData[i] - 1 for all 1 <= i <= s.

Find the maximum number of zeroes that the array sequenceData can contain after performing any (possibly zero) number of operations on the array.

Question 2: Word Conjugates

Problem Statement:

Amazon has introduced an innovative tool to explore the fascinating world of "Word Conjugates", allowing users to analyze strings consisting of only characters: 'a', 'b', 'c', and 'd'.

The concept of conjugates follows a unique pairing mechanism where 'a' pairs with 'b', 'c' pairs with 'd' and vice-versa. As a result, there are only four possible valid conjugate pairs: ('a', 'b'), ('b', 'a'), ('c', 'd'), and ('d', 'c').

The objective is to calculate the total number of valid substrings in a given string s.

A string is considered valid if:

• There exists a permutation of the string such that the original string and its permutation are conjugates of each other.
• Two strings s and t of the same length say n are considered conjugates if, for every index i (0 <= i < n), the pair formed by the characters at that index (s[i], t[i]) must be a conjugate pair.

11 hours ago

Ayush Patel • 0

Raipur

using namespace std;

void solve(){
  int n; cin>>n;
  vector<int>a(n);
  for(int i=0;i<n;i++) cin>>a[i];
  vector<int>pre(n,1e9);
  pre[0]=a[0];
  int cnt_zeros=1;
  for(int i=1;i<n;i++){
    pre[i]=min(a[i],pre[i-1]);
    if(a[i]<=pre[i]){
      cnt_zeros++;
    }
  }
  cout<<cnt_zeros<<endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}