LeetCode #1175 — EASY

Prime Arrangements

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)

(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)

Since the answer may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: n = 5
Output: 12
Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.

Example 2:

Input: n = 100
Output: 682289015

Constraints:

1 <= n <= 100

Step 01

Brute Force Baseline

Problem summary: Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.) (Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.) Since the answer may be large, return the answer modulo 10^9 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

Solve the problem for prime numbers and composite numbers separately.
Multiply the number of permutations of prime numbers over prime indices with the number of permutations of composite numbers over composite indices.
The number of permutations equals the factorial.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1175: Prime Arrangements
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int numPrimeArrangements(int n) {
        int cnt = count(n);
        long ans = f(cnt) * f(n - cnt);
        return (int) (ans % MOD);
    }

    private long f(int n) {
        long ans = 1;
        for (int i = 2; i <= n; ++i) {
            ans = (ans * i) % MOD;
        }
        return ans;
    }

    private int count(int n) {
        int cnt = 0;
        boolean[] primes = new boolean[n + 1];
        Arrays.fill(primes, true);
        for (int i = 2; i <= n; ++i) {
            if (primes[i]) {
                ++cnt;
                for (int j = i + i; j <= n; j += i) {
                    primes[j] = false;
                }
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #1175: Prime Arrangements
func numPrimeArrangements(n int) int {
	count := func(n int) int {
		cnt := 0
		primes := make([]bool, n+1)
		for i := range primes {
			primes[i] = true
		}
		for i := 2; i <= n; i++ {
			if primes[i] {
				cnt++
				for j := i + i; j <= n; j += i {
					primes[j] = false
				}
			}
		}
		return cnt
	}

	mod := int(1e9) + 7
	f := func(n int) int {
		ans := 1
		for i := 2; i <= n; i++ {
			ans = (ans * i) % mod
		}
		return ans
	}

	cnt := count(n)
	ans := f(cnt) * f(n-cnt)
	return ans % mod
}

# Accepted solution for LeetCode #1175: Prime Arrangements
class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        def count(n):
            cnt = 0
            primes = [True] * (n + 1)
            for i in range(2, n + 1):
                if primes[i]:
                    cnt += 1
                    for j in range(i + i, n + 1, i):
                        primes[j] = False
            return cnt

        cnt = count(n)
        ans = factorial(cnt) * factorial(n - cnt)
        return ans % (10**9 + 7)

// Accepted solution for LeetCode #1175: Prime Arrangements
struct Solution;

impl Solution {
    fn number_of_primes(n: usize) -> i32 {
        let mut a: Vec<bool> = vec![true; n + 1];
        a[0] = false;
        a[1] = false;
        let mut i: usize = 2;
        while i * i <= n {
            if a[i] {
                let mut j: usize = 2;
                while i * j <= n {
                    a[i * j] = false;
                    j += 1;
                }
            }
            i += 1;
        }
        let mut res = 0;
        for k in 0..=n {
            if a[k] {
                res += 1;
            }
        }
        res
    }

    fn num_prime_arrangements(n: i32) -> i32 {
        let primes = Self::number_of_primes(n as usize);
        let mut product = 1i64;
        for i in 1..=primes {
            product *= i as i64;
            product %= 1_000_000_007;
        }
        for i in 1..=(n - primes) {
            product *= i as i64;
            product %= 1_000_000_007;
        }
        product as i32
    }
}

#[test]
fn test() {
    assert_eq!(Solution::num_prime_arrangements(5), 12);
    assert_eq!(Solution::num_prime_arrangements(100), 682_289_015);
}

// Accepted solution for LeetCode #1175: Prime Arrangements
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1175: Prime Arrangements
// class Solution {
//     private static final int MOD = (int) 1e9 + 7;
// 
//     public int numPrimeArrangements(int n) {
//         int cnt = count(n);
//         long ans = f(cnt) * f(n - cnt);
//         return (int) (ans % MOD);
//     }
// 
//     private long f(int n) {
//         long ans = 1;
//         for (int i = 2; i <= n; ++i) {
//             ans = (ans * i) % MOD;
//         }
//         return ans;
//     }
// 
//     private int count(int n) {
//         int cnt = 0;
//         boolean[] primes = new boolean[n + 1];
//         Arrays.fill(primes, true);
//         for (int i = 2; i <= n; ++i) {
//             if (primes[i]) {
//                 ++cnt;
//                 for (int j = i + i; j <= n; j += i) {
//                     primes[j] = false;
//                 }
//             }
//         }
//         return cnt;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.