Amazon Max deviation among all substrings

很值得思考的一个问题.

引用自: https://leetcode.com/discuss/interview-question/1742621/Amazon-or-OA-or-Max-deviation-among-all-substrings

Let's have a string: abbbcacbcdce

 For substring abbb, you have most frequent letter b -> 3 and least frequent letter a -> 1.

So the deviation is = most frequent - least frequent = 3 - 1 = 2. You need to look at all the substrings and find the max deviation.

 Here substring cacbcdc has the max deviation.

Frequency is like below:
c -> 4, a ->1, b->1, d->1.
So max freq - min freq = 4 - 1 = 3.

Among all substrings deviation, this is the max. So need to return it. 

String length is 10^4. So you can't check each substring.

#include<bits/stdc++.h>
using namespace std;

int modifiedKadane( vector<int> &arr , int k ){
    if( arr.size() < k )
        return 0;
    
    int n = arr.size();
    vector<int> maxSum(n);

    // use kadane's
    maxSum[0] = arr[0];
    for (int i = 1 ; i < arr.size(); i++) {
      maxSum[i] = max(arr[i], maxSum[i - 1] + arr[i]);
    }

    int sum = 0 ;
    for (int i = 0 ; i < k; i++) {
      sum += arr[i];
    }

    int ans = sum;
    for (int i = k ; i < arr.size(); i++) {
      sum = sum + arr[i] - arr[i - k];
      ans = max(ans, sum);
      ans = max(ans, sum + maxSum[i - k]);
    }

    return ans;
}

int maxDeviation( string str ){
    int ans = 0 ;

    for( char c1 = 'a' ; c1<='z' ; c1++ ){
        for( char c2 = 'a' ; c2<='z' ; c2++ ){
            if ( c1 == c2 )
                continue;

            vector<int> arr;
            // we consider c1 as character with maxFreq and c2 with minFreq
            for( auto c : str ){
                if( c==c1 ){
                    // We shall include all consecutive c1's in our array so we add their frequency
                    if( arr.size() && arr.back() != -1 ){
                        arr.back() += 1;
                    }
                    else{
                        arr.push_back( 1 );
                    }
                }
                else if( c==c2 ){
                    // we take distinct c2
                    arr.push_back( -1 );
                }
            }
            ans= max( ans , modifiedKadane(arr, 2) );
        }
    }
    return ans;
}
int main(){
    // string str = "abacccabab";
    string str = "baaa";
    
    cout<<maxDeviation(str)<<"\n";
    return 0;
}

标签：deviation,arr,int,Max,sum,Amazon,ans,max
来源： https://www.cnblogs.com/micoblog/p/16199296.html