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Optimal and effective solution for calculating heavy numbers?

I need to find the number of heavy integers between two integers Aand B, where A <= Balways.

An integer is considered heavy whenever its average is greater than 7.

For example: 9878it is considered difficult because (9 + 8 + 7 + 8)/4 = 8, but 1111not, because (1 + 1 + 1 + 1)/4 = 1.

I have the solution below, but it is absolutely terrible, and it expires on startup 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;
}
+4
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6 answers

Counting numbers with lookup table

, d, , x. 10, 100, 1000... . 9 times; d , .

, AB, B d, 1 d -1 , AB 10, 100, 1000... , 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

( * 0 - * 9), . ( 2 6, 3 .)

1- n, x, :

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

, , n = 9- x.

2- n, x:

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

3- n, x:

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

: 10 d ( ). . 3 , 10 3= 1000, :

 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

, , "" , 7.

Javascript ( Java), . , 0 100 000 000 0,07 . , 7. Java, , .

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)");
Hide result
+5

@m69, , . , . n + 1 ( ) 10 n , 0-9.

, :

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

. , . . getResult - table , .

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);
    }
}

:

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

569484, 1us

+2

, . , base-10 , , , base-10. , Java -10, . , .

, . , , . ( ) ( ) , ( ), ( )?

Caveat - , , , , , .

:

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;
    }
}

, this SO Question java

100 000 000 :

569484

: 6985

EDIT: , 7x . 843,453 7025 .

+1

memoization, . A B, i .

(. 20 ).

JavaScript:

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
Hide result
+1

, , , , sum > 7 * length, Jeutnarg. , 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;
}


        if(isHeavy(i))


        if(isHeavyRV(i))

isHeavy(), , ( iMac) 12388 , 843453 .

, , 5416 .

, , , Integer.toString(i, 10), .

0

1 , . , , . , :

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;
    }
}

569484 1 100000000

: 31

, @JeutNarg, > 7, >= 7.

0

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