LeetCode #3272 — HARD

Find the Count of Good Integers

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given two positive integers n and k.

An integer x is called k-palindromic if:

x is a palindrome.
x is divisible by k.

An integer is called good if its digits can be rearranged to form a k-palindromic integer. For example, for k = 2, 2020 can be rearranged to form the k-palindromic integer 2002, whereas 1010 cannot be rearranged to form a k-palindromic integer.

Return the count of good integers containing n digits.

Note that any integer must not have leading zeros, neither before nor after rearrangement. For example, 1010 cannot be rearranged to form 101.

Example 1:

Input: n = 3, k = 5

Output: 27

Explanation:

Some of the good integers are:

551 because it can be rearranged to form 515.
525 because it is already k-palindromic.

Example 2:

Input: n = 1, k = 4

Output: 2

Explanation:

The two good integers are 4 and 8.

Example 3:

Input: n = 5, k = 6

Output: 2468

Constraints:

1 <= n <= 10
1 <= k <= 9

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers n and k. An integer x is called k-palindromic if: x is a palindrome. x is divisible by k. An integer is called good if its digits can be rearranged to form a k-palindromic integer. For example, for k = 2, 2020 can be rearranged to form the k-palindromic integer 2002, whereas 1010 cannot be rearranged to form a k-palindromic integer. Return the count of good integers containing n digits. Note that any integer must not have leading zeros, neither before nor after rearrangement. For example, 1010 cannot be rearranged to form 101.

Baseline thinking

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

Pattern signal: Hash Map · Math

Example 1

3
5

Example 2

1
4

Example 3

5
6

Core Insight

What unlocks the optimal approach

How to generate all K-palindromic strings of length <code>n</code>? Do we need to go through all <code>n</code> digits?
Use permutations to calculate the number of possible rearrangements.

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

Largest constraint values

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 #3272: Find the Count of Good Integers
class Solution {
    public long countGoodIntegers(int n, int k) {
        long[] fac = new long[n + 1];
        fac[0] = 1;
        for (int i = 1; i <= n; i++) {
            fac[i] = fac[i - 1] * i;
        }

        long ans = 0;
        Set<String> vis = new HashSet<>();
        int base = (int) Math.pow(10, (n - 1) / 2);

        for (int i = base; i < base * 10; i++) {
            String s = String.valueOf(i);
            StringBuilder sb = new StringBuilder(s).reverse();
            s += sb.substring(n % 2);
            if (Long.parseLong(s) % k != 0) {
                continue;
            }

            char[] arr = s.toCharArray();
            Arrays.sort(arr);
            String t = new String(arr);
            if (vis.contains(t)) {
                continue;
            }
            vis.add(t);
            int[] cnt = new int[10];
            for (char c : arr) {
                cnt[c - '0']++;
            }

            long res = (n - cnt[0]) * fac[n - 1];
            for (int x : cnt) {
                res /= fac[x];
            }
            ans += res;
        }

        return ans;
    }
}

// Accepted solution for LeetCode #3272: Find the Count of Good Integers
func factorial(n int) []int64 {
	fac := make([]int64, n+1)
	fac[0] = 1
	for i := 1; i <= n; i++ {
		fac[i] = fac[i-1] * int64(i)
	}
	return fac
}

func countGoodIntegers(n int, k int) (ans int64) {
	fac := factorial(n)
	vis := make(map[string]bool)
	base := int(math.Pow(10, float64((n-1)/2)))

	for i := base; i < base*10; i++ {
		s := strconv.Itoa(i)
		rev := reverseString(s)
		s += rev[n%2:]
		num, _ := strconv.ParseInt(s, 10, 64)
		if num%int64(k) != 0 {
			continue
		}
		bs := []byte(s)
		slices.Sort(bs)
		t := string(bs)

		if vis[t] {
			continue
		}
		vis[t] = true
		cnt := make([]int, 10)
		for _, c := range t {
			cnt[c-'0']++
		}
		res := (int64(n) - int64(cnt[0])) * fac[n-1]
		for _, x := range cnt {
			res /= fac[x]
		}
		ans += res
	}

	return
}

func reverseString(s string) string {
	t := []byte(s)
	for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
		t[i], t[j] = t[j], t[i]
	}
	return string(t)
}

# Accepted solution for LeetCode #3272: Find the Count of Good Integers
class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        fac = [factorial(i) for i in range(n + 1)]
        ans = 0
        vis = set()
        base = 10 ** ((n - 1) // 2)
        for i in range(base, base * 10):
            s = str(i)
            s += s[::-1][n % 2 :]
            if int(s) % k:
                continue
            t = "".join(sorted(s))
            if t in vis:
                continue
            vis.add(t)
            cnt = Counter(t)
            res = (n - cnt["0"]) * fac[n - 1]
            for x in cnt.values():
                res //= fac[x]
            ans += res
        return ans

// Accepted solution for LeetCode #3272: Find the Count of Good Integers
impl Solution {
    pub fn count_good_integers(n: i32, k: i32) -> i64 {
        use std::collections::HashSet;
        let n = n as usize;
        let k = k as i64;
        let mut fac = vec![1_i64; n + 1];
        for i in 1..=n {
            fac[i] = fac[i - 1] * i as i64;
        }

        let mut ans = 0;
        let mut vis = HashSet::new();
        let base = 10_i64.pow(((n - 1) / 2) as u32);

        for i in base..base * 10 {
            let s = i.to_string();
            let rev: String = s.chars().rev().collect();
            let full_s = if n % 2 == 0 {
                format!("{}{}", s, rev)
            } else {
                format!("{}{}", s, &rev[1..])
            };

            let num: i64 = full_s.parse().unwrap();
            if num % k != 0 {
                continue;
            }

            let mut arr: Vec<char> = full_s.chars().collect();
            arr.sort_unstable();
            let t: String = arr.iter().collect();
            if vis.contains(&t) {
                continue;
            }
            vis.insert(t);

            let mut cnt = vec![0; 10];
            for c in arr {
                cnt[c as usize - '0' as usize] += 1;
            }

            let mut res = (n - cnt[0]) as i64 * fac[n - 1];
            for &x in &cnt {
                if x > 0 {
                    res /= fac[x];
                }
            }
            ans += res;
        }

        ans
    }
}

// Accepted solution for LeetCode #3272: Find the Count of Good Integers
function countGoodIntegers(n: number, k: number): number {
    const fac = factorial(n);
    let ans = 0;
    const vis = new Set<string>();
    const base = Math.pow(10, Math.floor((n - 1) / 2));

    for (let i = base; i < base * 10; i++) {
        let s = `${i}`;
        const rev = reverseString(s);
        if (n % 2 === 1) {
            s += rev.substring(1);
        } else {
            s += rev;
        }

        if (+s % k !== 0) {
            continue;
        }

        const bs = Array.from(s).sort();
        const t = bs.join('');

        if (vis.has(t)) {
            continue;
        }

        vis.add(t);

        const cnt = Array(10).fill(0);
        for (const c of t) {
            cnt[+c]++;
        }

        let res = (n - cnt[0]) * fac[n - 1];
        for (const x of cnt) {
            res /= fac[x];
        }
        ans += res;
    }

    return ans;
}

function factorial(n: number): number[] {
    const fac = Array(n + 1).fill(1);
    for (let i = 1; i <= n; i++) {
        fac[i] = fac[i - 1] * i;
    }
    return fac;
}

function reverseString(s: string): string {
    return s.split('').reverse().join('');
}

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(10^m × n × log n)

Space

O(10^m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.