How do you solve Count No-Zero Pairs That Sum to N on LeetCode?

Count No-Zero Pairs That Sum to N (LeetCode #3704) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Pairs must consist of digits 1-9 only.

What is the time complexity of Count No-Zero Pairs That Sum to N?

The optimal solution for Count No-Zero Pairs That Sum to N runs in O(n) time and uses O(n) space. The digit DP approach efficiently counts pairs without generating them, leading to linear time complexity.

What is the brute force solution for Count No-Zero Pairs That Sum to N?

The brute force approach for Count No-Zero Pairs That Sum to N is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all pairs of integers (a, b) that sum to n. Count valid pairs where both a and b are no-zero integers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count No-Zero Pairs That Sum to N?

Count No-Zero Pairs That Sum to N uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Digit DP, Backtracking.

What are the interview tips for Count No-Zero Pairs That Sum to N?

Explain your thought process clearly. Consider edge cases like small values of n. Practice digit DP problems for familiarity.

What are common mistakes when solving Count No-Zero Pairs That Sum to N?

Forgetting to check for zeros in pairs. Not handling carry correctly.

#3704

Count No-Zero Pairs That Sum to N

Hard

MathDynamic ProgrammingDigit DPBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use digit dynamic programming to count valid pairs without generating them. Track digits and carry to ensure no zeros are used.

⚙️

Algorithm

3 steps

1Step 1: Convert n to its string representation to analyze each digit.
2Step 2: Use a recursive function with memoization to count valid pairs based on current digit position and carry.
3Step 3: For each digit, calculate possible values for a and b, ensuring neither contains zero.

solution.py12 lines

1def count_no_zero_pairs(n):
2    s = str(n)
3    dp = {}  
4    def dfs(pos, carry, used_zero):
5        if pos == len(s): return 1 if not used_zero else 0
6        total = 0
7        for a in range(1, 10):
8            b = int(s[pos]) - a - carry
9            if 1 <= b < 10:
10                total += dfs(pos + 1, 1 if b < 0 else 0, used_zero or (a == 0 or b == 0))
11        return total
12    return dfs(0, 0, False)

ℹ

Complexity note: The digit DP approach efficiently counts pairs without generating them, leading to linear time complexity.

1Pairs must consist of digits 1-9 only.
2Carry affects the next digit's calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.