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3610. Minimum Number of Primes to Sum to Target 🔒

Description

You are given two integers n and m.

You have to select a multiset of prime numbers from the first m prime numbers such that the sum of the selected primes is exactly n. You may use each prime number multiple times.

Return the minimum number of prime numbers needed to sum up to n, or -1 if it is not possible.

 

Example 1:

Input: n = 10, m = 2

Output: 4

Explanation:

The first 2 primes are [2, 3]. The sum 10 can be formed as 2 + 2 + 3 + 3, requiring 4 primes.

Example 2:

Input: n = 15, m = 5

Output: 3

Explanation:

The first 5 primes are [2, 3, 5, 7, 11]. The sum 15 can be formed as 5 + 5 + 5, requiring 3 primes.

Example 3:

Input: n = 7, m = 6

Output: 1

Explanation:

The first 6 primes are [2, 3, 5, 7, 11, 13]. The sum 7 can be formed directly by prime 7, requiring only 1 prime.

 

Constraints:

  • 1 <= n <= 1000
  • 1 <= m <= 1000

Solutions

Solution 1: Preprocessing + Dynamic Programming

We can first preprocess to obtain the first $1000$ prime numbers, and then use dynamic programming to solve the problem.

Define $f[i]$ as the minimum number of primes needed to sum up to $i$. Initially, set $f[0] = 0$ and all other $f[i] = \infty$. For each prime $p$, we can update $f[i]$ from $f[i - p]$ as follows:

\[f[i] = \min(f[i], f[i - p] + 1)\]

If $f[n]$ is still $\infty$, it means it is impossible to obtain $n$ as the sum of the first $m$ primes, so return -1; otherwise, return $f[n]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(n + M)$, where $M$ is the number of preprocessed primes (here it is $1000$).

  • class Solution {
    static List<Integer> primes = new ArrayList<>();
    static {
        int x = 2;
        int M = 1000;
        while (primes.size() < M) {
            boolean is_prime = true;
            for (int p : primes) {
                if (p * p > x) {
                    break;
                }
                if (x % p == 0) {
                    is_prime = false;
                    break;
                }
            }
            if (is_prime) {
                primes.add(x);
            }
            x++;
        }
    }

    public int minNumberOfPrimes(int n, int m) {
        int[] f = new int[n + 1];
        final int inf = 1 << 30;
        Arrays.fill(f, inf);
        f[0] = 0;
        for (int x : primes.subList(0, m)) {
            for (int i = x; i <= n; i++) {
                f[i] = Math.min(f[i], f[i - x] + 1);
            }
        }
        return f[n] < inf ? f[n] : -1;
    }
}


  • class Solution {
public:
    int minNumberOfPrimes(int n, int m) {
        static vector<int> primes;
        if (primes.empty()) {
            int x = 2;
            int M = 1000;
            while ((int) primes.size() < M) {
                bool is_prime = true;
                for (int p : primes) {
                    if (p * p > x) break;
                    if (x % p == 0) {
                        is_prime = false;
                        break;
                    }
                }
                if (is_prime) primes.push_back(x);
                x++;
            }
        }

        vector<int> f(n + 1, INT_MAX);
        f[0] = 0;
        for (int x : vector<int>(primes.begin(), primes.begin() + m)) {
            for (int i = x; i <= n; ++i) {
                if (f[i - x] != INT_MAX) {
                    f[i] = min(f[i], f[i - x] + 1);
                }
            }
        }
        return f[n] < INT_MAX ? f[n] : -1;
    }
};


  • primes = []
x = 2
M = 1000
while len(primes) < M:
    is_prime = True
    for p in primes:
        if p * p > x:
            break
        if x % p == 0:
            is_prime = False
            break
    if is_prime:
        primes.append(x)
    x += 1


class Solution:
    def minNumberOfPrimes(self, n: int, m: int) -> int:
        min = lambda x, y: x if x < y else y
        f = [0] + [inf] * n
        for x in primes[:m]:
            for i in range(x, n + 1):
                f[i] = min(f[i], f[i - x] + 1)
        return f[n] if f[n] < inf else -1


  • var primes []int

func init() {
	x := 2
	M := 1000
	for len(primes) < M {
		is_prime := true
		for _, p := range primes {
			if p*p > x {
				break
			}
			if x%p == 0 {
				is_prime = false
				break
			}
		}
		if is_prime {
			primes = append(primes, x)
		}
		x++
	}
}

func minNumberOfPrimes(n int, m int) int {
	const inf = int(1e9)
	f := make([]int, n+1)
	for i := 1; i <= n; i++ {
		f[i] = inf
	}
	f[0] = 0

	for _, x := range primes[:m] {
		for i := x; i <= n; i++ {
			if f[i-x] < inf && f[i-x]+1 < f[i] {
				f[i] = f[i-x] + 1
			}
		}
	}

	if f[n] < inf {
		return f[n]
	}
	return -1
}


  • const primes: number[] = [];
let x = 2;
const M = 1000;
while (primes.length < M) {
    let is_prime = true;
    for (const p of primes) {
        if (p * p > x) break;
        if (x % p === 0) {
            is_prime = false;
            break;
        }
    }
    if (is_prime) primes.push(x);
    x++;
}

function minNumberOfPrimes(n: number, m: number): number {
    const inf = 1e9;
    const f: number[] = Array(n + 1).fill(inf);
    f[0] = 0;

    for (const x of primes.slice(0, m)) {
        for (let i = x; i <= n; i++) {
            if (f[i - x] < inf) {
                f[i] = Math.min(f[i], f[i - x] + 1);
            }
        }
    }

    return f[n] < inf ? f[n] : -1;
}

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