Problem Description

You need to find the minimum number of prime numbers that sum up to a given integer n.

You're given two integers:

• n: the target sum you need to achieve
• m: you can only use the first m prime numbers

The key rules are:

• You can only select from the first m prime numbers (e.g., if m = 3, you can only use 2, 3, and 5)
• You can use each prime number multiple times (it's a multiset, so repetition is allowed)
• The selected primes must sum to exactly n
• Return the minimum count of primes needed

If it's impossible to form the sum n using the first m primes, return -1.

For example:

• If n = 9 and m = 3, you can use primes {2, 3, 5}
• One way: 3 + 3 + 3 = 9 (using three primes)
• Another way: 2 + 2 + 5 = 9 (using three primes)
• The minimum number of primes needed is 3

The solution uses dynamic programming where f[i] represents the minimum number of primes needed to sum to i. Starting from f[0] = 0, for each prime p from the first m primes, we update all possible sums by considering adding that prime: f[i] = min(f[i], f[i - p] + 1).

Intuition

The problem asks for the minimum number of primes that sum to n. This is a classic optimization problem that resembles the coin change problem - we want to make change for n using the fewest "coins" (primes).

Think about building up to the target sum n incrementally. For any number i, the minimum primes needed to reach it depends on the minimum primes needed to reach smaller numbers. If we already know the minimum primes to reach i - p (where p is a prime we can use), then we can reach i by adding one more prime p.

This naturally leads to a dynamic programming approach:

• Start with the base case: we need 0 primes to sum to 0
• For each number from 1 to n, calculate the minimum primes needed
• For each position i, try adding each available prime p to a previous sum i - p

The recurrence relation becomes: f[i] = min(f[i], f[i - p] + 1) for each prime p we can use.

Why does this work? Because we're systematically considering all possible ways to form each sum, keeping only the minimum count. By building from smaller sums to larger ones, we ensure that when we calculate f[i], we already have the optimal solutions for all smaller values.

The preprocessing step generates the first 1000 primes upfront because we might need up to the m-th prime, and having them ready avoids recalculating during the main algorithm.

Solution Approach

The solution consists of two main parts: preprocessing prime numbers and applying dynamic programming.

Step 1: Preprocessing Prime Numbers

First, we generate the first 1000 prime numbers using a simple primality testing algorithm:

• Start with x = 2 and an empty list of primes
• For each number x, check if it's divisible by any previously found prime p where p * p ≤ x
• If no prime divides x, then x is prime and add it to the list
• Continue until we have 1000 primes

primes = []
x = 2
while len(primes) < 1000:
    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

Step 2: Dynamic Programming

We use a DP array f where f[i] represents the minimum number of primes needed to sum to i.

Initialization:

• f[0] = 0 (zero primes sum to 0)
• f[i] = inf for all other i from 1 to n (initially impossible)

State Transition: For each prime x from the first m primes and for each sum i from x to n:

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

This means: to reach sum i, we can either:

• Keep the current minimum f[i]
• Or use one prime x plus the minimum primes needed for i - x

Implementation Details:

def minNumberOfPrimes(self, n: int, m: int) -> int:
    f = [0] + [inf] * n  # f[0] = 0, others = inf
  
    for x in primes[:m]:  # Use only first m primes
        for i in range(x, n + 1):  # Update sums from x to n
            f[i] = min(f[i], f[i - x] + 1)
  
    return f[n] if f[n] < inf else -1

The outer loop iterates through each available prime, and the inner loop updates all possible sums that can include this prime. After processing all primes, f[n] contains the minimum number of primes needed to sum to n, or remains inf if it's impossible.

Time Complexity: O(m * n) where m is the number of available primes and n is the target sum. Space Complexity: O(n) for the DP array.

Example Walkthrough

Let's walk through a concrete example where n = 10 and m = 3, meaning we need to sum to 10 using only the first 3 prime numbers: {2, 3, 5}.

Initial Setup:

• First 3 primes: [2, 3, 5]
• DP array f of size 11 (indices 0 to 10)
• Initialize: f = [0, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf]

Processing Prime 2: Starting from index 2, we update each position by considering adding prime 2:

• f[2] = min(inf, f[0] + 1) = min(inf, 0 + 1) = 1 (one 2 makes 2)
• f[3] = min(inf, f[1] + 1) = min(inf, inf + 1) = inf (can't make 3 yet)
• f[4] = min(inf, f[2] + 1) = min(inf, 1 + 1) = 2 (two 2s make 4)
• f[5] = min(inf, f[3] + 1) = min(inf, inf + 1) = inf
• f[6] = min(inf, f[4] + 1) = min(inf, 2 + 1) = 3 (three 2s make 6)
• f[7] = min(inf, f[5] + 1) = min(inf, inf + 1) = inf
• f[8] = min(inf, f[6] + 1) = min(inf, 3 + 1) = 4 (four 2s make 8)
• f[9] = min(inf, f[7] + 1) = min(inf, inf + 1) = inf
• f[10] = min(inf, f[8] + 1) = min(inf, 4 + 1) = 5 (five 2s make 10)

After prime 2: f = [0, inf, 1, inf, 2, inf, 3, inf, 4, inf, 5]

Processing Prime 3: Starting from index 3, we update positions considering prime 3:

• f[3] = min(inf, f[0] + 1) = min(inf, 0 + 1) = 1 (one 3 makes 3)
• f[4] = min(2, f[1] + 1) = min(2, inf + 1) = 2 (stays as 2)
• f[5] = min(inf, f[2] + 1) = min(inf, 1 + 1) = 2 (2 + 3 = 5)
• f[6] = min(3, f[3] + 1) = min(3, 1 + 1) = 2 (3 + 3 = 6)
• f[7] = min(inf, f[4] + 1) = min(inf, 2 + 1) = 3 (2 + 2 + 3 = 7)
• f[8] = min(4, f[5] + 1) = min(4, 2 + 1) = 3 (2 + 3 + 3 = 8)
• f[9] = min(inf, f[6] + 1) = min(inf, 2 + 1) = 3 (3 + 3 + 3 = 9)
• f[10] = min(5, f[7] + 1) = min(5, 3 + 1) = 4 (2 + 2 + 3 + 3 = 10)

After prime 3: f = [0, inf, 1, 1, 2, 2, 2, 3, 3, 3, 4]

Processing Prime 5: Starting from index 5, we update positions considering prime 5:

• f[5] = min(2, f[0] + 1) = min(2, 0 + 1) = 1 (one 5 makes 5)
• f[6] = min(2, f[1] + 1) = min(2, inf + 1) = 2 (stays as 2)
• f[7] = min(3, f[2] + 1) = min(3, 1 + 1) = 2 (2 + 5 = 7)
• f[8] = min(3, f[3] + 1) = min(3, 1 + 1) = 2 (3 + 5 = 8)
• f[9] = min(3, f[4] + 1) = min(3, 2 + 1) = 3 (stays as 3)
• f[10] = min(4, f[5] + 1) = min(4, 1 + 1) = 2 (5 + 5 = 10)

Final DP array: f = [0, inf, 1, 1, 2, 1, 2, 2, 2, 3, 2]

Result: f[10] = 2, meaning the minimum number of primes needed to sum to 10 is 2 (which can be achieved with 5 + 5).

Solution Implementation

Time and Space Complexity

The code consists of two parts: prime number generation and the dynamic programming solution.

Prime Generation Part:

• Time: O(M × √M) where M = 1000. For each number checked, we iterate through existing primes up to its square root.
• Space: O(M) to store the primes list.

Dynamic Programming Part:

• Time: O(m × n). The outer loop iterates through the first m primes, and for each prime x, the inner loop iterates from x to n+1, which is approximately n iterations. Therefore, the total time complexity is O(m × n).
• Space: O(n) for the DP array f of size n+1.

Overall Complexity:

• Time Complexity: O(M × √M + m × n). Since M is a constant (1000) and the prime generation happens only once, the dominant term is O(m × n).
• Space Complexity: O(n + M), where n is for the DP array and M is for storing the preprocessed primes.

Common Pitfalls

1. Incorrect DP Array Update Order

A common mistake is updating the DP array in the wrong direction or using a 1D array incorrectly, which can lead to using the same prime multiple times unintentionally in a single iteration.

Incorrect approach:

# WRONG: This might count the same prime multiple times incorrectly
for prime in primes[:m]:
    for i in range(n, prime - 1, -1):  # Backward iteration
        f[i] = min(f[i], f[i - prime] + 1)

This backward iteration pattern is used for 0/1 knapsack problems where each item can only be used once. Since we can use each prime multiple times (unbounded knapsack), we need forward iteration.

Correct approach:

# CORRECT: Forward iteration allows using each prime multiple times
for prime in primes[:m]:
    for i in range(prime, n + 1):  # Forward iteration
        f[i] = min(f[i], f[i - prime] + 1)

2. Prime Generation Overflow or Inefficiency

When generating primes, using an inefficient algorithm or not optimizing the primality check can cause performance issues.

Inefficient approach:

# INEFFICIENT: Checking divisibility by all numbers
for divisor in range(2, x):
    if x % divisor == 0:
        is_prime = False

Optimized approach:

# EFFICIENT: Only check previously found primes up to sqrt(x)
for p in primes:
    if p * p > x:
        break
    if x % p == 0:
        is_prime = False
        break

3. Not Handling Edge Cases

Failing to handle special cases like n = 0 or m = 0 can cause errors.

Solution:

def minNumberOfPrimes(self, n: int, m: int) -> int:
    # Handle edge cases
    if n == 0:
        return 0
    if m == 0:
        return -1
  
    # Rest of the DP solution...

4. Integer Overflow with Large Values

Using a sentinel value that's too small instead of infinity can cause incorrect comparisons.

Problematic approach:

# WRONG: Using a fixed large number that might not be large enough
f = [0] + [10000] * n  # What if the answer is legitimately > 10000?

Correct approach:

# CORRECT: Using float('inf') ensures proper comparison
f = [0] + [float('inf')] * n

5. Off-by-One Errors in Array Indexing

Not properly handling array bounds when initializing or accessing the DP array.

Common mistake:

# WRONG: Missing the last element
f = [float('inf')] * n  # Should be n+1 elements
for i in range(prime, n):  # Should go up to n inclusive

Correct approach:

# CORRECT: Proper array size and iteration bounds
f = [0] + [float('inf')] * n  # n+1 elements total
for i in range(prime, n + 1):  # Include n in the range