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Programmingoneonone
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HackerRank Counting the Ways problem solution

YASH PAL, 31 July 202430 November 2025

In this HackerRank Counting, the Ways problem solution Little Walter likes playing with his toy scales. He has N types of weights. The Ith weight type has weight Ai. There are infinitely many weights of each type.

Recently, Walter defined a function, F(X), denoting the number of different ways to combine several weights so their total weight is equal to X. Ways are considered to be different if there is a type that has a different number of weights used in these two ways.

HackerRank Counting the Ways problem solution

Problem solution in Python.

MOD = 10**9 + 7

def lcm(lst):
    ans = 1
    for x in lst:
        ans = ans*x//gcd(ans, x)
    return ans

def gcd(a,b):
    if a<b:
        a, b = b, a
    while b > 0:
        a, b = b, a%b
    return a

def getsoltable(a, m, MOD=MOD):
    soltable = [1] + [0] * (len(a)*m-1)
    for x in a:
        oldsoltable = soltable
        soltable = list(soltable)
        for i in range(x, len(soltable)):
            soltable[i] = (oldsoltable[i] + soltable[i - x]) % MOD
    return soltable

def countsols(const, soltable, lcm):
    offset = const % lcm
    pts = soltable[offset::lcm]
    assert len(pts) == len(a)
    coef = polycoef(pts)
    return polyval(coef, const//lcm)

def polycoef(pts):
    coef = []
    for x, y in enumerate(pts):
        fact = descpower = 1
        for i, c in enumerate(coef):
            y -= descpower*c//fact
            descpower *= x - i
            fact *= i + 1
        coef.append(y)
    return coef
        
def polyval(coef, x):
    ans = 0
    fact = descpower = 1
    for i, c in enumerate(coef):
        ans += c * descpower * pow(fact, MOD-2, MOD)
        descpower = descpower * (x - i) % MOD
        fact *= i + 1
    return ans % MOD
        
n = int(input())
a = [1] + [int(fld) for fld in input().strip().split()]
L, R = [int(fld ) for fld in input().strip().split()]
m = lcm(a)
soltable = getsoltable(a, m)
print((countsols(R, soltable, m) - countsols(L-1, soltable, m)) % MOD)

Problem solution in Java.

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

public class Solution {

    static int MOD = 1000000007;
    
    static int N;
    static int[] arr;
    static List<Integer> listS;
    static int[] pSum;
    static int[] frec;
    static int sizePS;
    static int maxS = 100010;
    static int[][] memo = new int[60][maxS];
    
    static void init() {
        listS = new ArrayList<Integer>();
        for (int s = 1; s < (1 << N); s++){ 
            int x = 0;
            for (int t = 0; t < N; t++) {
                if ((s & (1 << t)) > 0) {
                    x += arr[t];
                }
            }
            listS.add(x);            
        }
        Collections.sort(listS);
        sizePS = listS.size();
        pSum = new int[sizePS];
        frec = new int[sizePS];
        int last = listS.get(0);
        pSum[0] = last;
        frec[0] = 1;
        int index = 0;
        for(int k = 1;k < sizePS;k++) {
            int next = listS.get(k);
            if(next == last) {
                frec[index]++;
            } else {
                last = next;
                pSum[++index] = next;
                frec[index] = 1;
            }
        }
        sizePS = index+1;
    }

    static int countWays(long l, long r) {
        init();
        int aR = recCalc(r, 0);
        memo = new int[60][maxS];
        return (aR + MOD - recCalc(l-1, 0)) % MOD; 
    }

    private static int recCalc(long l, int pos) {
        if(l == 0) {
            return 0;
        }
        int ans = 0;
        if((ans = memo[pos][(int) (l%maxS)]) >= 1) {
            return ans-1;
        }
        
        ans = recCalc(l/2, pos+1);
        for(int k = 0;k < sizePS;k++) {
            int a = pSum[k];
            if(l >= a) {
                int pa = 1+recCalc((l-a)/2, pos+1);
                ans = (int) ((ans + (long)frec[k] * pa) % MOD);
            }else {
                break;
            }
        }
        memo[pos][(int) (l%maxS)] = ans+1;
        return ans;
    }

    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")));
        N = scanner.nextInt();
        scanner.skip("(rn|[nru2028u2029u0085])*");
        arr = new int[N];
        String[] arrItems = scanner.nextLine().split(" ");
        scanner.skip("(rn|[nru2028u2029u0085])*");
        for (int arrItr = 0; arrItr < N; arrItr++) {
            int arrItem = Integer.parseInt(arrItems[arrItr]);
            arr[arrItr] = arrItem;
        }
        String[] lr = scanner.nextLine().split(" ");
        scanner.skip("(rn|[nru2028u2029u0085])*");
        long l = Long.parseLong(lr[0]);
        long r = Long.parseLong(lr[1]);
        int result = countWays(l, r);
        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();
        bufferedWriter.close();
        scanner.close();
    }
}

Problem solution in C++.

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

const long long MOD = 1e9 + 7;

int n;
int a[10];
long long L, R;

const int N = 202000;
int dp0[N];
int dp1[N];

inline void add(int &x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
}

long long solve(long long v) {
    bitset<62> s(v);
    memset(dp0, 0, sizeof(dp0));
    dp0[0] = 1;

    for (int k = 0; k < 62; k++) {
        for (int i = 0; i < n; i++) {
            for (int j = N - a[i] - 1; j >= 0; j--) {
                add(dp0[j + a[i]], dp0[j]);
            }
        }
        if (s[k]) {
            add(dp0[0], dp0[1]);
            for (int i = 1; i < N - 1; i++) {
                dp0[i] = dp0[i + 1];
            }
        }
        memset(dp1, 0, sizeof(dp1));
        for (int i = 0; i < N; i++) {
            add(dp1[(i + 1) / 2], dp0[i]); 
        }
        swap(dp0, dp1);
    }

    return dp0[0];
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    cin >> L >> R;

    int ans = solve(R) - solve(L - 1);
    if (ans < 0) ans += MOD;
    cout << ans << endl;
}

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