LeetCode #1922 — MEDIUM

Count Good Numbers

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

A digit string is good if the digits (0-indexed) at even indices are even and the digits at odd indices are prime (2, 3, 5, or 7).

For example, "2582" is good because the digits (2 and 8) at even positions are even and the digits (5 and 2) at odd positions are prime. However, "3245" is not good because 3 is at an even index but is not even.

Given an integer n, return the total number of good digit strings of length n. Since the answer may be large, return it modulo 10⁹ + 7.

A digit string is a string consisting of digits 0 through 9 that may contain leading zeros.

Example 1:

Input: n = 1
Output: 5
Explanation: The good numbers of length 1 are "0", "2", "4", "6", "8".

Example 2:

Input: n = 4
Output: 400

Example 3:

Input: n = 50
Output: 564908303

Constraints:

1 <= n <= 10¹⁵

Step 01

Brute Force Baseline

Problem summary: A digit string is good if the digits (0-indexed) at even indices are even and the digits at odd indices are prime (2, 3, 5, or 7). For example, "2582" is good because the digits (2 and 8) at even positions are even and the digits (5 and 2) at odd positions are prime. However, "3245" is not good because 3 is at an even index but is not even. Given an integer n, return the total number of good digit strings of length n. Since the answer may be large, return it modulo 109 + 7. A digit string is a string consisting of digits 0 through 9 that may contain leading zeros.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

Is there a formula we can use to find the count of all the good numbers?
Exponentiation can be done very fast if we looked at the binary bits of n.

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 #1922: Count Good Numbers
class Solution {
    private final int mod = (int) 1e9 + 7;

    public int countGoodNumbers(long n) {
        return (int) (qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod);
    }

    private long qpow(long x, long n) {
        long res = 1;
        while (n != 0) {
            if ((n & 1) == 1) {
                res = res * x % mod;
            }
            x = x * x % mod;
            n >>= 1;
        }
        return res;
    }
}

// Accepted solution for LeetCode #1922: Count Good Numbers
const mod int64 = 1e9 + 7

func countGoodNumbers(n int64) int {
	return int(myPow(5, (n+1)>>1) * myPow(4, n>>1) % mod)
}

func myPow(x, n int64) int64 {
	var res int64 = 1
	for n != 0 {
		if (n & 1) == 1 {
			res = res * x % mod
		}
		x = x * x % mod
		n >>= 1
	}
	return res
}

# Accepted solution for LeetCode #1922: Count Good Numbers
class Solution:
    def countGoodNumbers(self, n: int) -> int:
        mod = 10**9 + 7
        return pow(5, (n + 1) >> 1, mod) * pow(4, n >> 1, mod) % mod

// Accepted solution for LeetCode #1922: Count Good Numbers
/**
 * [1922] Count Good Numbers
 *
 * A digit string is good if the digits (0-indexed) at even indices are even and the digits at odd indices are prime (2, 3, 5, or 7).
 *
 * 	For example, "2582" is good because the digits (2 and 8) at even positions are even and the digits (5 and 2) at odd positions are prime. However, "3245" is not good because 3 is at an even index but is not even.
 *
 * Given an integer n, return the total number of good digit strings of length n. Since the answer may be large, return it modulo 10^9 + 7.
 * A digit string is a string consisting of digits 0 through 9 that may contain leading zeros.
 *  
 * Example 1:
 *
 * Input: n = 1
 * Output: 5
 * Explanation: The good numbers of length 1 are "0", "2", "4", "6", "8".
 *
 * Example 2:
 *
 * Input: n = 4
 * Output: 400
 *
 * Example 3:
 *
 * Input: n = 50
 * Output: 564908303
 *
 *  
 * Constraints:
 *
 * 	1 <= n <= 10^15
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-good-numbers/
// discuss: https://leetcode.com/problems/count-good-numbers/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_good_numbers(n: i64) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1922_example_1() {
        let n = 1;

        let result = 5;

        assert_eq!(Solution::count_good_numbers(n), result);
    }

    #[test]
    #[ignore]
    fn test_1922_example_2() {
        let n = 4;

        let result = 400;

        assert_eq!(Solution::count_good_numbers(n), result);
    }

    #[test]
    #[ignore]
    fn test_1922_example_3() {
        let n = 50;

        let result = 564908303;

        assert_eq!(Solution::count_good_numbers(n), result);
    }
}

// Accepted solution for LeetCode #1922: Count Good Numbers
function countGoodNumbers(n: number): number {
    const mod = 1000000007n;
    const qpow = (x: bigint, n: bigint): bigint => {
        let res = 1n;
        while (n > 0n) {
            if (n & 1n) {
                res = (res * x) % mod;
            }
            x = (x * x) % mod;
            n >>= 1n;
        }
        return res;
    };
    const a = qpow(5n, BigInt(n + 1) / 2n);
    const b = qpow(4n, BigInt(n) / 2n);
    return Number((a * b) % mod);
}

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.