#3791Hard~60 min
Number of Balanced Integers in a Range
Time O((logn)^2) · Space O(logn) · Official statement on LeetCode
Solutions
// Time: O((logn)^2)
// Space: O(logn)
// dp
class Solution {
public:
long long countBalanced(long long low, long long high) {
const auto& count = [&](int64_t n) {
vector<int> digits;
for (; n; n /= 10) {
digits.emplace_back(n % 10);
}
ranges::reverse(digits);
const auto& shift = size(digits) / 2 * 9;
vector<vector<int64_t>> dp(size(digits) * 9 + 1, vector<int64_t>(2));
dp[shift][1] = 1;
for (int i = 0; i < size(digits); ++i) {
vector<vector<int64_t>> new_dp(size(digits) * 9 + 1, vector<int64_t>(2));
for (int curr = 0; curr < size(dp); ++curr) {
for (int tight = 0; tight <= 1; ++tight) {
if (dp[curr][tight] == 0) {
continue;
}
for (int d = 0, bound = tight ? digits[i] : 9; d <= bound; ++d) {
new_dp[(i & 1) ? curr - d : curr + d][tight && d == bound] += dp[curr][tight];
}
}
}
dp = move(new_dp);
}
return dp[shift][0];
};
return count(high + 1) - count(low);
}
};
// Time: O((logn)^2)
// Space: O((logn)^2)
// memoization
class Solution2 {
public:
long long countBalanced(long long low, long long high) {
const auto& count = [&](int64_t n) {
vector<int> digits;
for (; n; n /= 10) {
digits.emplace_back(n % 10);
}
ranges::reverse(digits);
vector<vector<int64_t>> memo(size(digits), vector<int64_t>(size(digits) * 9 + 1, -1));
const int shift = size(digits) / 2 * 9;
const auto memoization = [&](this auto&& memoization, int i, int curr, bool tight) -> int64_t {
if (i == size(digits)) {
return curr == shift;
}
if (!tight && memo[i][curr] != -1) {
return memo[i][curr];
}
const auto& bound = tight ? digits[i] : 9;
int64_t result = 0;
for (int d = 0; d <= bound; ++d) {
result += memoization(i + 1, (i & 1) ? curr - d : curr + d, tight && d == bound);
}
if (!tight) {
memo[i][curr] = result;
}
return result;
};
return memoization(0, shift, true);
};
return count(high) - count(low - 1);
}
};
// Time: O((logn)^2)
// Space: O((logn)^2)
// memoization
class Solution3 {
public:
long long countBalanced(long long low, long long high) {
const auto& count = [&](int64_t n) {
vector<int> digits;
for (; n; n /= 10) {
digits.emplace_back(n % 10);
}
ranges::reverse(digits);
vector<vector<vector<int64_t>>> memo(size(digits), vector<vector<int64_t>>(size(digits) * 9 + 1, vector<int64_t>(2, -1)));
const int shift = size(digits) / 2 * 9;
const auto memoization = [&](this auto&& memoization, int i, int curr, bool tight) -> int64_t {
if (i == size(digits)) {
return curr == shift;
}
if (memo[i][curr][tight] == -1) {
const auto& bound = tight ? digits[i] : 9;
int64_t result = 0;
for (int d = 0; d <= bound; ++d) {
result += memoization(i + 1, (i & 1) ? curr - d : curr + d, tight && d == bound);
}
memo[i][curr][tight] = result;
}
return memo[i][curr][tight];
};
return memoization(0, shift, true);
};
return count(high) - count(low - 1);
}
};
Beginner Explanation
What is Number of Balanced Integers in a Range?
Number of Balanced Integers in a Range (LeetCode #3791) is a Hard problem that primarily trains dynamic programming.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with dynamic programming.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DP, Memoization.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Number of Balanced Integers in a Range
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to dynamic programming.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O((logn)^2)) and space (O(logn)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O((logn)^2) time and O(logn) space.
Pattern focus: dynamic programming
Use the pattern as a checklist:
- dynamic programming — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O((logn)^2) |
| Space | O(logn) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Number of Balanced Integers in a Range
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to dynamic programming:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: dynamic programming.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Number of Balanced Integers in a Range in a second language (cpp, python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Conceptual diagram for Number of Balanced Integers in a Range: show input structure (dynamic programming), highlight the moving parts of the dynamic programming approach, and annotate each step with the maintained invariant and complexity.
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dynamic programming approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Number of Balanced Integers in a Range (#3791) — Hard. Pattern: dynamic programming. Complexity: O((logn)^2) time / O(logn) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Number of Balanced Integers in a Range?+
The reference solutions aim for O((logn)^2) time and O(logn) space. Always re-derive complexity from the code you write in the interview.
What pattern does Number of Balanced Integers in a Range use?+
It primarily maps to dynamic programming, within the broader topic of dynamic programming.
Is Number of Balanced Integers in a Range good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/number-of-balanced-integers-in-a-range/