Optimal and efficient solution for the heavy number calculation?

Question

I need to find the number of heavy integers between two integers A and B , where A <= B at all times.

An integer is considered heavy whenever the average of it's digit is larger than 7 .

For example: 9878 is considered heavy, because (9 + 8 + 7 + 8)/4 = 8 , while 1111 is not, since (1 + 1 + 1 + 1)/4 = 1 .

I have the solution below, but it's absolutely terrible and it times out when run with large inputs. What can I do to make it more efficient?

int countHeavy(int A, int B) {
    int countHeavy = 0;

    while(A <= B){
        if(averageOfDigits(A) > 7){
            countHeavy++;
        }
        A++;
    }

    return countHeavy;
}

float averageOfDigits(int a) {
    float result = 0;
    int count = 0;

    while (a > 0) {
        result += (a % 10);
        count++;
        a = a / 10;
    }

    return result / count;
}

Answer 1

Counting the numbers with a look-up table

You can generate a table that stores how many integers with d digits have a sum of their digits that is greater than a number x . Then, you can quickly look up how many heavy numbers there are in any range of 10, 100, 1000 ... integers. These tables hold only 9×d values, so they take up very little space and can be quickly generated.

Then, to check a range AB where B has d digits, you build the tables for 1 to d -1 digits, and then you split the range AB into chunks of 10, 100, 1000 ... and look up the values in the tables, eg for the range A = 782, B = 4321:

   RANGE      DIGITS  TARGET     LOOKUP      VALUE

 782 -  789     78x    > 6    table[1][ 6]     3    <- incomplete range: 2-9
 790 -  799     79x    > 5    table[1][ 5]     4
 800 -  899     8xx    >13    table[2][13]    15
 900 -  999     9xx    >12    table[2][12]    21
1000 - 1999    1xxx    >27    table[3][27]     0
2000 - 2999    2xxx    >26    table[3][26]     1
3000 - 3999    3xxx    >25    table[3][25]     4
4000 - 4099    40xx    >24    impossible       0
4100 - 4199    41xx    >23    impossible       0
4200 - 4299    42xx    >22    impossible       0
4300 - 4309    430x    >21    impossible       0
4310 - 4319    431x    >20    impossible       0
4320 - 4321    432x    >19    impossible       0    <- incomplete range: 0-1
                                              --
                                              48

If the first and last range are incomplete (not *0 - *9), check the starting value or the end value against the target. (In the example, 2 is not greater than 6, so all 3 heavy numbers are included in the range.)

Generating the look-up table

For 1-digit decimal integers, the number of integers n that is greater than value x is:

x:  0  1  2  3  4  5  6  7  8  9
n:  9  8  7  6  5  4  3  2  1  0

As you can see, this is easily calculated by taking n = 9- x .

For 2-digit decimal integers, the number of integers n whose sum of digits is greater than value x is:

x:   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18
n:  99 97 94 90 85 79 72 64 55 45 36 28 21 15 10  6  3  1  0

For 3-digit decimal integers, the number of integers n whose sum of digits is greater than value x is:

x:   0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27
n: 999 996 990 980 965 944 916 880 835 780 717 648 575 500 425 352 283 220 165 120  84  56  35  20  10   4   1   0

Each of these sequences can be generated from the previous one: start with value 10 ^d and then subtract from this value the previous sequence in reverse (skipping the first zero). Eg to generate the sequence for 3 digits from the sequence for 2 digits, start with 10 ³ = 1000, and then:

 0. 1000 -   1      = 999
 1.  999 -   3      = 996
 2.  996 -   6      = 990
 3.  990 -  10      = 980
 4.  980 -  15      = 965
 5.  965 -  21      = 944
 6.  944 -  28      = 916
 7.  916 -  36      = 880
 8.  880 -  45      = 835
 9.  835 -  55      = 780
10.  780 -  64 +  1 = 717  <- after 10 steps, start adding the previous sequence again
11.  717 -  72 +  3 = 648
12.  648 -  79 +  6 = 575
13.  575 -  85 + 10 = 500
14.  500 -  90 + 15 = 425
15.  425 -  94 + 21 = 352
16.  352 -  97 + 28 = 283
17.  283 -  99 + 36 = 220
18.  220 - 100 + 45 = 165  <- at the end of the sequence, keep subtracting 10^(d-1)
19.  165 - 100 + 55 = 120
20.  120 - 100 + 64 =  84
21.   84 - 100 + 72 =  56
22.   56 - 100 + 79 =  35
23.   35 - 100 + 85 =  20
24.   20 - 100 + 90 =  10
25.   10 - 100 + 94 =   4
26.    4 - 100 + 97 =   1
27.    1 - 100 + 99 =   0

By the way, you can use the same tables if "heavy" numbers are defined with a value other than 7.

---

Code example

Below is a Javascript code snippet (I don't speak Java) that demonstrates the method. It is very much unoptimised, but it does the 0→100,000,000 example in less than 0.07ms. It also works for weights other than 7. Translated to Java, it should easily beat any algorithm that actually runs through the numbers and checks their weight.

 function countHeavy(A, B, weight) { var a = decimalDigits(A), b = decimalDigits(B); // create arrays while (a.length < b.length) a.push(0); // add leading zeros var digits = b.length, table = weightTable(); // create table var count = 0, diff = B - A + 1, d = 0; // calculate range for (var i = digits - 1; i >= 0; i--) if (a[i]) d = i; // lowest non-0 digit while (diff) { // increment a until a=b while (a[d] == 10) { // move to higher digit a[d++] = 0; ++a[d]; // carry 1 } var step = Math.pow(10, d); // value of digit d if (step <= diff) { diff -= step; count += increment(d); // increment digit d } else --d; // move to lower digit } return count; function weightTable() { // see above for details var t = [[],[9,8,7,6,5,4,3,2,1,0]]; for (var i = 2; i < digits; i++) { var total = Math.pow(10, i), final = total / 10; t[i] = []; for (var j = 9 * i; total > 0; --j) { if (j > 9) total -= t[i - 1][j - 10]; else total -= final; if (j < 9 * (i - 1)) total += t[i - 1][j]; t[i].push(total); } } return t; } function increment(d) { var sum = 0, size = digits; for (var i = digits - 1; i >= d; i--) { if (a[i] == 0 && i == size - 1) size = i; // count used digits sum += a[i]; // sum of digits } ++a[d]; var target = weight * size - sum; if (d == 0) return (target < 0) ? 1 : 0; // if d is lowest digit if (target < 0) return table[d][0] + 1; // whole range is heavy return (target > 9 * d) ? 0 : table[d][target]; // use look-up table } function decimalDigits(n) { var array = []; do {array.push(n % 10); n = Math.floor(n / 10); } while (n); return array; } } document.write("0 &rarr; 100,000,000 = " + countHeavy(0, 100000000, 7) + "<br>"); document.write("782 &rarr; 4321 = " + countHeavy(782, 4321, 7) + "<br>"); document.write("782 &rarr; 4321 = " + countHeavy(782, 4321, 5) + " (weight: 5)"); 

Answer 2

I really liked the post of @m69 so I wrote implementation inspired by it. The table creation is not that elegant, but works. For n+1 digits long integer I sum (at most) 10 values from n digits long integer, one for every digit 0-9.

I use this simplification to avoid arbitrary range calculation:

countHeavy(A, B) = countHeavy(0, B) - countHeavy(0, A-1)

The result is calculated in two loops. One for numbers shorter than the given number and one for the rest. I was not able to merge them easily. getResult is just lookup into the table with range checking, the rest of the code should be quite obvious.

public class HeavyNumbers {
    private static int maxDigits = String.valueOf(Long.MAX_VALUE).length();
    private int[][] table = null;

    public HeavyNumbers(){
        table = new int[maxDigits + 1][];
        table[0] = new int[]{1};

        for (int s = 1; s < maxDigits + 1; ++s) {
            table[s] = new int[s * 9 + 1];
            for (int k = 0; k < table[s].length; ++k) {
                for (int d = 0; d < 10; ++d) {
                    if (table[s - 1].length > k - d) {
                        table[s][k] += table[s - 1][Math.max(0, k - d)];
                    }
                }
            }
        }
    }

    private int[] getNumberAsArray(long number) {
        int[] tmp = new int[maxDigits];
        int cnt = 0;

        while (number != 0) {
            int remainder = (int) (number % 10);
            tmp[cnt++] = remainder;
            number = number / 10;
        }

        int[] ret = new int[cnt];
        for (int i = 0; i < cnt; ++i) {
            ret[i] = tmp[i];
        }
        return ret;
    }

    private int getResult(int[] sum, int digits, int fixDigitSum, int heavyThreshold) {
        int target = heavyThreshold * digits - fixDigitSum + 1;
        if (target < sum.length) {
            return sum[Math.max(0, target)];
        }
        return 0;
    }

    public int getHeavyNumbersCount(long toNumberIncl, int heavyThreshold) {
        if (toNumberIncl <= 0) return 0;

        int[] numberAsArray = getNumberAsArray(toNumberIncl);

        int res = 0;

        for (int i = 0; i < numberAsArray.length - 1; ++i) {
            for (int d = 1; d < 10; ++d) {
                res += getResult(table[i], i + 1, d, heavyThreshold);
            }
        }

        int fixDigitSum = 0;
        int fromDigit = 1;
        for (int i = numberAsArray.length - 1; i >= 0; --i) {
            int toDigit = numberAsArray[i];
            if (i == 0) {
                toDigit++;
            }
            for (int d = fromDigit; d < toDigit; ++d) {
                res += getResult(table[i], numberAsArray.length, fixDigitSum + d, heavyThreshold);
            }

            fixDigitSum += numberAsArray[i];
            fromDigit = 0;
        }

        return res;
    }

    public int getHeavyNumbersCount(long fromIncl, long toIncl, int heavyThreshold) {
        return getHeavyNumbersCount(toIncl, heavyThreshold) -
                getHeavyNumbersCount(fromIncl - 1, heavyThreshold);
    }
}

It is used like this:

HeavyNumbers h = new HeavyNumbers();
System.out.println( h.getHeavyNumbersCount(100000000,7));

prints out 569484, the repeated calculation time without initialization of the table is under 1us

Answer 3

I looked at the problem differently than you did. My perception is that the problem is based on the base-10 representation of a number, so the first thing you should do is to put the number into a base-10 representation. There may be a nicer way of doing it, but Java Strings represent Integers in base-10, so I used those. It's actually pretty fast to turn a single character into an integer, so this doesn't really cost much time.

Most importantly, your calculations in this matter never need to use division or floats. The problem is, at its core, about integers only. Do all the digits (integers) in the number (integer) add up to a value greater than or equal to seven (integer) times the number of digits (integer)?

Caveat - I don't claim that this is the fastest possible way of doing it, but this is probably faster than your original approach.

Here is my code:

package heavyNum;

public class HeavyNum
{
    public static void main(String[] args)
    {
        HeavyNum hn = new HeavyNum();
        long startTime = System.currentTimeMillis();
        hn.countHeavy(100000000, 1);
        long endTime = System.currentTimeMillis();
        System.out.println("Time elapsed: "+(endTime- startTime));
    }

    private void countHeavy(int A, int B)
    {
        int heavyFound = 0;
        for(int i = B+1; i < A; i++)
        {
            if(isHeavy(i))
                heavyFound++;
        }
        System.out.println("Found "+heavyFound+" heavy numbers");
    }

    private boolean isHeavy(int i)
    {
        String asString = Integer.valueOf(i).toString();
        int length = asString.length();
        int dividingLine = length * 7, currTotal = 0, counter = 0;
        while(counter < length)
        {
            currTotal += Character.getNumericValue(asString.charAt(counter++));
        }
        return currTotal > dividingLine;
    }
}

Credit goes to this SO Question for how I get the number of digits in an integer and this SO Question for how to quickly convert characters to integers in java

Running on a powerful computer with no debugger for numbers between one and 100,000,000 resulted in this output:

Found 569484 heavy numbers

Time elapsed: 6985

EDIT: I initially was looking for numbers whose digits were greater than or equal to 7x the number of digits. I previously had results of 843,453 numbers in 7025 milliseconds.

Answer 4

Here's a pretty barebones recursion with memoization that enumerates the digit possibilities one by one for a fixed-digit number. You may be able to set A and B by controlling the range of i when calculating the corresponding number of digits.

Seems pretty fast (see the result for 20 digits).

JavaScript code:

 var hash = {} function f(k,soFar,count){ if (k == 0){ return 1; } var key = [k,soFar].join(","); if (hash[key]){ return hash[key]; } var res = 0; for (var i=Math.max(count==0?1:0,7*(k+count)+1-soFar-9*(k-1)); i<=9; i++){ res += f(k-1,soFar+i,count+1); } return hash[key] = res; } // Output: console.log(f(3,0,0)); // 56 hash = {}; console.log(f(6,0,0)); // 12313 hash = {}; console.log(f(20,0,0)); // 2224550892070475 

Answer 5

You can indeed use strings to get the number of digits and then add the values of the individual digits to see if their sum > 7 * length , as Jeutnarg seems to do. I took his code and added my own, simple isHeavyRV(int) :

private boolean isHeavyRV(int i)
{
    int sum = 0, count = 0;
    while (i > 0)
    {
        sum += i % 10;
        count++;
        i = i / 10;
    }
    return sum >= count * 7;
}

Now, instead of

        if(isHeavy(i))

I tried

        if(isHeavyRV(i))

I actually first tested his implementation of isHeavy() , using strings, and that ran in 12388 milliseconds on my machine (an older iMac), and it found 843453 heavy numbers.

Using my implementation, I found exactly the same number of heavy numbers, but in a time of a mere 5416 milliseconds.

Strings may be fast, but they can't beat a simple loop doing basically what Integer.toString(i, 10) does as well, but without the string detour.

Answer 6

When you add 1 to a number, you are incrementing one digit, and changing all the smaller digits to zero. If incrementing changes from a heavy to a non-heavy number, its because too many low-order digits were zeroed. In this case, it's pretty easy to find the next heavy number without checking all the numbers in between:

public class CountHeavy
{
    public static void main(String[] args)
    {
        long startTime = System.currentTimeMillis();
        int numHeavy = countHeavy(1, 100000000);
        long endTime = System.currentTimeMillis();
        System.out.printf("Found %d heavy numbers between 1 and 100000000\n", numHeavy);
        System.out.println("Time elapsed: "+(endTime- startTime)+" ms");
    }

    static int countHeavy(int from, int to)
    {
        int numdigits=1;
        int maxatdigits=9;
        int numFound = 0;
        if (from<1)
        {
            from=1;
        }
        for(int i = from; i < to;)
        {
            //keep track of number of digits in i
            while (i > maxatdigits)
            {
                long newmax = 10L*maxatdigits+9;
                maxatdigits = (int)Math.min(Integer.MAX_VALUE, newmax);
                ++numdigits;
            }
            //get sum of digits
            int digitsum=0;
            for(int digits=i;digits>0;digits/=10)
            {
                digitsum+=(digits%10);
            }

            //calculate a step size that increments the first non-zero digit
            int step=1;
            int stepzeros=0;
            while(step <= (Integer.MAX_VALUE/10) && to-i >= step*10 && i%(step*10) == 0)
            {
                step*=10;
                stepzeros+=1;
            }
            //step is a 1 followed stepzeros zeros

            //how much is our sum too small by?
            int need = numdigits*7+1 - digitsum;
            if (need <= 0)
            {
                //already have enough.  All the numbers between i and i+step are heavy
                numFound+=step;
            }
            else if (need <= stepzeros*9)
            {
                //increment to the smallest possible heavy number. This puts all the
                //needed sum in the lowest-order digits
                step = need%9;
                for(;need >= 9;need-=9)
                {
                    step = step*10+9;
                }
            }
            //else there are no heavy numbers between i and i+step
            i+=step;
        }
        return numFound;
    }
}

Found 569484 heavy numbers between 1 and 100000000

Time elapsed: 31 ms

Note that the answer is different from @JeutNarg's, because you asked for average > 7, not average >= 7.