HackerRank Lucky Numbers problem solution

YASH PAL, 31 July 202425 January 2026

In this HackerRank Lucky Numbers problem solution, A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between b and inclusive, are lucky?

Function Description

Complete the luckyNumbers function in the editor below. It should return an integer that represents the number of lucky numbers in the given range.

luckyNumbers has the following parameter(s):

a: an integer, the lower range bound
b: an integer, the higher range bound

HackerRank Lucky Numbers problem solution

HackerRank Lucky Numbers problem solution in Java.

import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

public class Solution {

    /*
     * Complete the luckyNumbers function below.
     */
    static long luckyNumbers(long a, long b) {
        return Method1.luckyNumbers(a, b);
    }

    static class Method1 {
        static final int max_bit = 19;
        static final int max_digit_sum = 9 * 18 + 1;
        static final int max_squre_digit_sum = 9 * 9 * 18 + 1;

        static long[][][] sum = new long[max_bit][max_digit_sum][max_squre_digit_sum];
        static long[][][] left_sum = new long[max_bit][max_digit_sum][max_squre_digit_sum];
        static boolean[] is_prime = new boolean[max_squre_digit_sum];
        static long[] bit_sum = new long[max_bit];
        static int tot = 231, max1, max2;
        static int[] prime = new int[tot];
        static {
            Arrays.fill(is_prime, true);
            int ct = 0;
            is_prime[0] = is_prime[1] = false;
            for (int i = 2; i < max_squre_digit_sum; i++)
                if (is_prime[i]) {
                    for (int j = i + i; j < max_squre_digit_sum; j += i)
                        is_prime[j] = false;
                    prime[ct++] = i;
                    if (i < max_digit_sum)
                        max1 = Math.max(max1, i);
                    max2 = Math.max(max2, i);
                }
            sum[0][0][0] = 1;
            for (int i = 0; prime[i] <= max1; i++)
                for (int j = 0; j < tot && prime[j] <= max2; j++) {
                    left_sum[0][prime[i]][prime[j]] += 1;
                }

            for (int i = 0; i < max_bit - 1; i++) {
                for (int next = 0; next < 10; next++) {

                    for (int j = 0; j + next < max_digit_sum; j++)
                        for (int k = 0; k + next * next < max_squre_digit_sum; k++) {
                            sum[i + 1][j + next][k + next * next] += sum[i][j][k];
                            if (next > 0 && is_prime[j + next] && is_prime[k + next * next])
                                bit_sum[i + 1] += sum[i][j][k];
                        }

                    for (int j = next; j < max_digit_sum; j++)
                        for (int k = next * next; k < max_squre_digit_sum; k++) {
                            left_sum[i + 1][j - next][k - next * next] += left_sum[i][j][k];
                        }
                }
            }
        }

        static long luckyNumbers(long a, long b) {
            return go(b) - go(a - 1);
        }

        static long go(long N) {
            if (N == 0)
                return 0;
            long ret = 0;
            boolean first = true;
            int pre_digit_sum = 0, pre_sdigit_sum = 0;
            for (int i = 19; i > 0; i--) {
                int bit = get_bit(N, i - 1);
                int least;
                if (bit != 0 && first) {
                    least = 1;
                    first = false;
                    for (int j = 1; j < i; j++)
                        ret += bit_sum[j];
                } else {
                    least = 0;
                }

                for (int nbit = least; nbit < bit; nbit++) {
                    int digit_sum = pre_digit_sum + nbit;
                    int sdigit_sum = pre_sdigit_sum + nbit * nbit;
                    ret += left_sum[i - 1][digit_sum][sdigit_sum];
                }
                pre_digit_sum += bit;
                pre_sdigit_sum += bit * bit;
            }
            if (is_prime[pre_digit_sum] && is_prime[pre_sdigit_sum])
                ret += 1;
            return ret;
        }

        static int get_bit(long n, int m) {
            for (int i = 0; i < m; i++)
                n /= 10;
            return (int) (n % 10);
        }
    }
    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int t = Integer.parseInt(scanner.nextLine().trim());

        for (int tItr = 0; tItr < t; tItr++) {
            String[] ab = scanner.nextLine().split(" ");

            long a = Long.parseLong(ab[0].trim());

            long b = Long.parseLong(ab[1].trim());

            long result = luckyNumbers(a, b);

            bufferedWriter.write(String.valueOf(result));
            bufferedWriter.newLine();
        }

        bufferedWriter.close();
    }
}

Lucky Numbers problem solution in C++.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<list>
#include<queue>
#include<set>
using namespace std;
typedef vector<int> VI;
typedef long long LL;
#define FOR(x, b, e) for(int x=b; x<=(e); ++x)
#define FORD(x, b, e) for(int x=b; x>=(e); --x)
#define REP(x, n) for(int x=0; x<(n); ++x)
#define VAR(v,n) __typeof(n) v=(n)
#define ALL(c) c.begin(),c.end()
#define SIZE(x) (int)x.size()
#define FOREACH(i,c) for(VAR(i,(c).begin());i!=(c).end();++i)
#define PB push_back
#define ST first
#define ND second
const int MILION=1000*1000;
const int MAX_SUMA=200;
const int MAX_KWADRAT=2000;
const int MAX_CYFR=20;

LL pot[MAX_CYFR];
bool pierwsza[MAX_SUMA][MAX_KWADRAT];
LL w[MAX_SUMA][MAX_KWADRAT][MAX_CYFR];


bool czyPierwsza(int k){
    if (k<2) return false;
    int i=2;
    while (i*i<=k){
        if (k%i==0) return false;
        i++;
    }
    return true;
}

void ustalPot(){
    pot[0]=1;
    FOR(i,1,MAX_CYFR-1){
        pot[i]=pot[i-1]*10;
    }
}

void ustalPierwsze(){
    FOR(i,2,MAX_SUMA-2){
        FOR(j,2,MAX_KWADRAT-2){
            if (czyPierwsza(i) && czyPierwsza(j)){
                pierwsza[i][j]=true;
            }
        }
    }
}

void ustalW(){
    REP(i,MAX_SUMA){
        REP(j,MAX_KWADRAT){
            if (pierwsza[i][j]){
                w[i][j][0]++;
            }
        }
    }
    FOR(p,1,MAX_CYFR-1){
        REP(i,MAX_SUMA-2){
            REP(j,MAX_KWADRAT-2){
                REP(k,10){
                    if (i+k<MAX_SUMA-2 && j+k*k < MAX_KWADRAT-2){
                        w[i][j][p]+=w[i+k][j+k*k][p-1];
                    }
                }
            }
        }
    }
}

LL daj(int c, int k, LL T){
    if (T<1){
        return pierwsza[c][k];
    }
    LL wyn=0;
    int wyk=0;
    int cyf=1;
    while (pot[wyk]<=T) wyk++;
    wyk--;
    while (cyf*pot[wyk]<=T) cyf++;
    cyf--;
    wyn+=daj(c+cyf,k+cyf*cyf,T-cyf*pot[wyk]);
    if (wyk>-1){
        FOR(i,0,cyf-1){
            wyn+=w[c+i][k+i*i][wyk];
        }
    }
    return wyn;
}


int main(){
    ustalPot();
    ustalPierwsze();
    ustalW();
    int t;
    LL a,b;
    scanf("%d",&t);
    REP(i,t){
        scanf("%lld%lld",&a,&b);
        printf("%lldn",daj(0,0,b)-daj(0,0,a-1));
    }
    return 0;
}

Problem solution in C.

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
#include <stdio.h>
#include <string.h>


int isprime[10000];
unsigned long long dp[19][2*82][2*730];
unsigned long long ans[19][10][164][1460];

int start_sum_sqr[19][163];
int end_sum_sqr[19][163];

void prime(){
	memset(isprime,0,10000*4);
	isprime[0]=1;
	isprime[1]=1;	
	for (int i  = 2; i < 10000; i += 1){
		if(isprime[i]==0){
			for (int j  = i*i; j < 10000; j += i){
				isprime[j]=1;
			}	
		}
	}
}





void inc(int i,int j,int k, int val){
	if(i<19 && j<164 && k< 1460)
		dp[i][j][k] += val;
}


unsigned long long setDP(){

	memset(dp, 0, sizeof(dp[0][0][0])*19*82*730);
	
	dp[0][0][0]=1;
    for (int i = 0; i < 18; ++i) {
        for (int j = 0; j <= 9 * i; ++j) {
            for (int k = 0; k <= 9 * 9 * i; ++k) {
                for (int l = 0; l < 10; ++l) {
                    dp[i + 1][j + l][k + l*l] += dp[i][j][k];
                }
            }
        }
    }

}

unsigned long long solve(unsigned long long num){
	if(num<=0)return 0;
	int digs[20],len=0;
	while(num){
		digs[len++]=(num%10); 
		num=num/10;
	}
	
	int sum=0,sqr_sum=0;
	unsigned long long ret=0;
	for (int i  = len-1; i >= 0; i -= 1){
		int dig = digs[i];	
		
		for (int j  = 0; j < dig; j += 1){
			if(ans[i][j][sum][sqr_sum]!=0){
				ret+=ans[i][j][sum][sqr_sum];
				continue;
			}
			unsigned long long  x=0;
			for (int k  = 0; k <= 9*i; k += 1){
				if(isprime[k+sum+j]) continue;
				for (int l  = start_sum_sqr[i][k]; l <= end_sum_sqr[i][k]; l += 1){
					if(isprime[l+ j*j + sqr_sum]==0) x+=dp[i][k][l];
				}
			}
			ans[i][j][sum][sqr_sum] = x;
			ret+=x;
		}
		sum+=dig;
		sqr_sum+=(dig*dig);
	}
	if(isprime[sum]==0 && isprime[sqr_sum]==0) ret++;
	return ret;
}

void set_sqrs(){
	for (int i = 0; i < 19; i += 1){
		for (int j = 0; j < 163; j += 1){
			for (int k = 0; k < 1460; k += 1){
				if(dp[i][j][k]!=0){
					start_sum_sqr[i][j]=k;
					break;
				}
			}
			for (int k = 1459; k>=0; k -= 1){
				if(dp[i][j][k]!=0){
					end_sum_sqr[i][j]=k;
					break;
				}
			}
		}
	}
}

int main(){
	prime();
	setDP();	
	set_sqrs();
	unsigned long long tc,a,b; 
	scanf("%lld",&tc);
	while(tc--){
		scanf("%lld %lld", &a, &b);
		if(b == 1000000000000000000ll)b--;
		printf("%lldn", solve(b) - solve(a-1));
	}
	return 0;
}

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