LeetCode #3704 — HARD

Count No-Zero Pairs That Sum to N

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

Solve on LeetCode
The Problem

Problem Statement

A no-zero integer is a positive integer that does not contain the digit 0 in its decimal representation.

Given an integer n, count the number of pairs (a, b) where:

  • a and b are no-zero integers.
  • a + b = n

Return an integer denoting the number of such pairs.

Example 1:

Input: n = 2

Output: 1

Explanation:

The only pair is (1, 1).

Example 2:

Input: n = 3

Output: 2

Explanation:

The pairs are (1, 2) and (2, 1).

Example 3:

Input: n = 11

Output: 8

Explanation:

The pairs are (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), and (9, 2). Note that (1, 10) and (10, 1) do not satisfy the conditions because 10 contains 0 in its decimal representation.

Constraints:

  • 2 <= n <= 1015
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: A no-zero integer is a positive integer that does not contain the digit 0 in its decimal representation. Given an integer n, count the number of pairs (a, b) where: a and b are no-zero integers. a + b = n Return an integer denoting the number of such pairs.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming

Example 1

2

Example 2

3

Example 3

11
Step 02

Core Insight

What unlocks the optimal approach

  • Use digit DP over the decimal representation of <code>n</code>.
  • At each digit, track whether a carry is present and whether <code>a</code> or <code>b</code> has used a zero so far.
  • Transition by choosing digits for <code>a</code> and <code>b</code> that add up (with carry) to the current digit of <code>n</code>.
  • Subtract the cases where either number ends up being zero-containing when <code>n</code> itself is no-zero.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3704: Count No-Zero Pairs That Sum to N
class Solution {
    public long countNoZeroPairs(long n) {
        char[] cs = Long.toString(n).toCharArray();
        int m = cs.length;
        int[] digits = new int[m + 1];
        for (int i = 0; i < m; i++) {
            digits[i] = cs[m - 1 - i] - '0';
        }
        digits[m] = 0;

        long[][][] dp = new long[2][2][2];
        dp[0][1][1] = 1;

        for (int pos = 0; pos < m + 1; pos++) {
            long[][][] ndp = new long[2][2][2];
            int target = digits[pos];
            for (int carry = 0; carry <= 1; carry++) {
                for (int aliveA = 0; aliveA <= 1; aliveA++) {
                    for (int aliveB = 0; aliveB <= 1; aliveB++) {
                        long ways = dp[carry][aliveA][aliveB];
                        if (ways == 0) {
                            continue;
                        }
                        int[] aDigits;
                        int[] aNext;
                        if (aliveA == 1) {
                            if (pos == 0) {
                                aDigits = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9};
                                aNext = new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1};
                            } else {
                                aDigits = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 0};
                                aNext = new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 0};
                            }
                        } else {
                            aDigits = new int[] {0};
                            aNext = new int[] {0};
                        }

                        int[] bDigits;
                        int[] bNext;
                        if (aliveB == 1) {
                            if (pos == 0) {
                                bDigits = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9};
                                bNext = new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1};
                            } else {
                                bDigits = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 0};
                                bNext = new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 0};
                            }
                        } else {
                            bDigits = new int[] {0};
                            bNext = new int[] {0};
                        }

                        for (int ai = 0; ai < aDigits.length; ai++) {
                            int da = aDigits[ai];
                            int na = aNext[ai];
                            for (int bi = 0; bi < bDigits.length; bi++) {
                                int db = bDigits[bi];
                                int nb = bNext[bi];
                                int s = da + db + carry;
                                if (s % 10 != target) {
                                    continue;
                                }
                                int ncarry = s / 10;
                                ndp[ncarry][na][nb] += ways;
                            }
                        }
                    }
                }
            }
            dp = ndp;
        }

        return dp[0][0][0];
    }

    public long countPairs(long n) {
        return countNoZeroPairs(n);
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(L · 9^2)
Space
O(1)

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.

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