time limit in Sorted Permutation Rank with Repeats

Question

First I want to mention that I am not very expert in programming. I am trying to solve the repeated permutation problem but I dont understand why i always get time limit. My code is in java and I am using BigInteger for factorial and other calculation.

please help me to find out what is the reason i got time limit. I found same approach to solve the problem but that was in python. Here I provided my code. Thanks in advance.

import java.math.BigInteger;
public class Solution {
    BigInteger  factorial(int n){
        BigInteger sum=new BigInteger(String.valueOf(1));

        if(n<2) return new BigInteger(String.valueOf(n));
        for(int j=2;j<=n;j++){
            sum=sum.multiply(new BigInteger(String.valueOf(j)));

        }
        return sum;
    }
    public int findRank(String a) {
        if(a.length()<2) return 1;
        Map<Character,Integer> map1=new HashMap<Character,Integer>();
        BigInteger sum=new BigInteger(String.valueOf(1));

        for(int i=0;i<a.length();i++){
            if(!map1.containsKey(a.charAt(i))){
                map1.put(a.charAt(i),1);
            }
             else{
                int cc=map1.get(a.charAt(i));
                cc=cc+1;
                map1.put(a.charAt(i),cc);
            }
        }
        BigInteger temp1=new BigInteger(String.valueOf(1));
        for (Map.Entry<Character, Integer> entry : map1.entrySet())
        {

            temp1=temp1.multiply(new BigInteger(String.valueOf(factorial( entry.getValue()))));

        }
        temp1=temp1.pow(1000001);
        temp1=temp1.mod(new BigInteger(String.valueOf(1000003)));
         for(int i=0;i<a.length();i++){

                BigInteger rank=new BigInteger(String.valueOf(0));

                for(int j=i+1;j<a.length();j++){
                    if(a.charAt(i)>a.charAt(j)){
                        rank=rank.add(new BigInteger(String.valueOf(1)));
                    }
                }
                BigInteger temp=new BigInteger(String.valueOf(factorial(a.length()-i-1)));

                rank=rank.multiply(temp1);
                rank=rank.multiply(temp);

                sum=sum.add(rank);
                sum=sum.mod(new BigInteger(String.valueOf(1000003)));
         }
        return sum.intValue();
    }
}

Answer 1

You using excessive string parsing and object creation. This all are expensive for runtime.

With optimized usage of BigInteger the method will return in ~1ms on my maschine.

public class Solution {

        private BigInteger factorial(final int n) {
            BigInteger sum = BigInteger.ONE;

            if (n < 2) {
                return BigInteger.valueOf(n);
            }

            for (int j = 2; j <= n; j++) {
                sum = sum.multiply(BigInteger.valueOf(j));

            }
            return sum;
        }

        public int findRank(final String a) {

            if (a.length() < 2) {
                return 1;
            }

            final Map<Character, Integer> map1 = new HashMap<>();
            BigInteger sum = BigInteger.ONE;

            for (int i = 0; i < a.length(); i++) {
                final char charAt = a.charAt(i);

                if (!map1.containsKey(charAt)) {
                    map1.put(charAt, 1);
                } else {
                    int cc = map1.get(charAt);
                    map1.put(charAt, cc++);
                }
            }

            BigInteger temp1 = BigInteger.ONE;

            for (final Map.Entry<Character, Integer> entry : map1.entrySet()) {
                temp1 = temp1.multiply(factorial(entry.getValue()));
            }

            temp1 = temp1.pow(1000001);
            temp1 = temp1.mod(BigInteger.valueOf(1000003));

            for (int i = 0; i < a.length(); i++) {

                BigInteger rank = BigInteger.ZERO;

                for (int j = i + 1; j < a.length(); j++) {
                    if (a.charAt(i) > a.charAt(j)) {
                        rank = rank.add(BigInteger.ONE);
                    }
                }

                final BigInteger temp = factorial(a.length() - i - 1);

                rank = rank.multiply(temp1);
                rank = rank.multiply(temp);

                sum = sum.add(rank);
                sum = sum.mod(BigInteger.valueOf(1000003));
            }
            return sum.intValue();
        }
    }