LeetCode #3747 — MEDIUM

Count Distinct Integers After Removing Zeros

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

The Problem

Problem Statement

You are given a positive integer n.

For every integer x from 1 to n, we write down the integer obtained by removing all zeros from the decimal representation of x.

Return an integer denoting the number of distinct integers written down.

Example 1:

Input: n = 10

Output: 9

Explanation:

The integers we wrote down are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1. There are 9 distinct integers (1, 2, 3, 4, 5, 6, 7, 8, 9).

Example 2:

Input: n = 3

Output: 3

Explanation:

The integers we wrote down are 1, 2, 3. There are 3 distinct integers (1, 2, 3).

Constraints:

1 <= n <= 10¹⁵

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer n. For every integer x from 1 to n, we write down the integer obtained by removing all zeros from the decimal representation of x. Return an integer denoting the number of distinct integers written down.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

Build integers less than or equal to <code>n</code> using only digits from 1 to 9
Count such integers using math or digit DP

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 #3747: Count Distinct Integers After Removing Zeros
class Solution {
    private char[] s;
    private Long[][][][] f;

    public long countDistinct(long n) {
        s = String.valueOf(n).toCharArray();
        f = new Long[s.length][2][2][2];
        return dfs(0, 0, 1, 1);
    }

    private long dfs(int i, int zero, int lead, int limit) {
        if (i == s.length) {
            return (zero == 0 && lead == 0) ? 1 : 0;
        }

        if (limit == 0 && f[i][zero][lead][limit] != null) {
            return f[i][zero][lead][limit];
        }

        int up = limit == 1 ? s[i] - '0' : 9;
        long ans = 0;
        for (int d = 0; d <= up; d++) {
            int nxtZero = zero == 1 || (d == 0 && lead == 0) ? 1 : 0;
            int nxtLead = (lead == 1 && d == 0) ? 1 : 0;
            int nxtLimit = (limit == 1 && d == up) ? 1 : 0;
            ans += dfs(i + 1, nxtZero, nxtLead, nxtLimit);
        }

        if (limit == 0) {
            f[i][zero][lead][limit] = ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3747: Count Distinct Integers After Removing Zeros
func countDistinct(n int64) int64 {
	s := []byte(fmt.Sprint(n))
	m := len(s)
	var f [20][2][2][2]int64
	for i := range f {
		for j := range f[i] {
			for k := range f[i][j] {
				for t := range f[i][j][k] {
					f[i][j][k][t] = -1
				}
			}
		}
	}

	var dfs func(i, zero, lead, limit int) int64
	dfs = func(i, zero, lead, limit int) int64 {
		if i == m {
			if zero == 0 && lead == 0 {
				return 1
			}
			return 0
		}

		if limit == 0 && f[i][zero][lead][limit] != -1 {
			return f[i][zero][lead][limit]
		}

		up := 9
		if limit == 1 {
			up = int(s[i] - '0')
		}

		var ans int64 = 0
		for d := 0; d <= up; d++ {
			nxtZero := zero
			if d == 0 && lead == 0 {
				nxtZero = 1
			}
			nxtLead := 0
			if lead == 1 && d == 0 {
				nxtLead = 1
			}
			nxtLimit := 0
			if limit == 1 && d == up {
				nxtLimit = 1
			}
			ans += dfs(i+1, nxtZero, nxtLead, nxtLimit)
		}

		if limit == 0 {
			f[i][zero][lead][limit] = ans
		}
		return ans
	}

	return dfs(0, 0, 1, 1)
}

# Accepted solution for LeetCode #3747: Count Distinct Integers After Removing Zeros
class Solution:
    def countDistinct(self, n: int) -> int:
        @cache
        def dfs(i: int, zero: bool, lead: bool, lim: bool) -> int:
            if i >= len(s):
                return 1 if (not zero and not lead) else 0
            up = int(s[i]) if lim else 9
            ans = 0
            for j in range(up + 1):
                nxt_zero = zero or (j == 0 and not lead)
                nxt_lead = lead and j == 0
                nxt_lim = lim and j == up
                ans += dfs(i + 1, nxt_zero, nxt_lead, nxt_lim)
            return ans

        s = str(n)
        return dfs(0, False, True, True)

// Accepted solution for LeetCode #3747: Count Distinct Integers After Removing Zeros
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3747: Count Distinct Integers After Removing Zeros
// class Solution {
//     private char[] s;
//     private Long[][][][] f;
// 
//     public long countDistinct(long n) {
//         s = String.valueOf(n).toCharArray();
//         f = new Long[s.length][2][2][2];
//         return dfs(0, 0, 1, 1);
//     }
// 
//     private long dfs(int i, int zero, int lead, int limit) {
//         if (i == s.length) {
//             return (zero == 0 && lead == 0) ? 1 : 0;
//         }
// 
//         if (limit == 0 && f[i][zero][lead][limit] != null) {
//             return f[i][zero][lead][limit];
//         }
// 
//         int up = limit == 1 ? s[i] - '0' : 9;
//         long ans = 0;
//         for (int d = 0; d <= up; d++) {
//             int nxtZero = zero == 1 || (d == 0 && lead == 0) ? 1 : 0;
//             int nxtLead = (lead == 1 && d == 0) ? 1 : 0;
//             int nxtLimit = (limit == 1 && d == up) ? 1 : 0;
//             ans += dfs(i + 1, nxtZero, nxtLead, nxtLimit);
//         }
// 
//         if (limit == 0) {
//             f[i][zero][lead][limit] = ans;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3747: Count Distinct Integers After Removing Zeros
function countDistinct(n: number): number {
    const s = n.toString();
    const m = s.length;

    const f: number[][][][] = Array.from({ length: m }, () =>
        Array.from({ length: 2 }, () => Array.from({ length: 2 }, () => Array(2).fill(-1))),
    );

    const dfs = (i: number, zero: number, lead: number, limit: number): number => {
        if (i === m) {
            return zero === 0 && lead === 0 ? 1 : 0;
        }

        if (limit === 0 && f[i][zero][lead][limit] !== -1) {
            return f[i][zero][lead][limit];
        }

        const up = limit === 1 ? parseInt(s[i]) : 9;
        let ans = 0;
        for (let d = 0; d <= up; d++) {
            const nxtZero = zero === 1 || (d === 0 && lead === 0) ? 1 : 0;
            const nxtLead = lead === 1 && d === 0 ? 1 : 0;
            const nxtLimit = limit === 1 && d === up ? 1 : 0;
            ans += dfs(i + 1, nxtZero, nxtLead, nxtLimit);
        }

        if (limit === 0) {
            f[i][zero][lead][limit] = ans;
        }
        return ans;
    };

    return dfs(0, 0, 1, 1);
}

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_10 n × D)

Space

O(log_10 n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.