Count of Integers
Time O(m * n) · Space O(m + n) · Official statement on LeetCode
Solutions
// Time: O(n * m), m = max_sum
// Space: O(m)
// combinatorics, dp
class Solution {
public:
int count(string num1, string num2, int min_sum, int max_sum) {
static const int MOD = 1e9 + 7;
const auto& check = [&](const auto& x) {
const auto& total = accumulate(cbegin(x), cend(x), 0, [](const auto& accu, const auto& c) {
return accu + (c - '0');
});
return min_sum <= total && total <= max_sum;
};
const auto& f = [&](const auto& x) {
vector<vector<int>> dp(2, vector<int>(max_sum + 1));
dp[0][0] = dp[1][0] = 1;
for (int i = size(x) - 1; i >= 0; --i) {
vector<vector<int>> new_dp(2, vector<int>(max_sum + 1));
for (int t = 0; t < 2; ++t) {
for (int total = 0; total <= max_sum; ++total) {
for (int d = 0; d <= min((t == 1 ? x[i] - '0' : 9), total); ++d) {
new_dp[t][total] = (new_dp[t][total] + dp[static_cast<int>(t && d == x[i] - '0')][total - d]) % MOD;
}
}
}
dp = move(new_dp);
}
int result = 0;
for (int total = min_sum; total <= max_sum; ++total) {
result = (result + dp[1][total]) % MOD;
}
return result;
};
return (((f(num2) - f(num1)) % MOD + MOD) % MOD + check(num1)) % MOD;
}
};
// Time: O(n * m), m = max_sum
// Space: O(m)
// combinatorics, dp
class Solution2 {
public:
int count(string num1, string num2, int min_sum, int max_sum) {
static const int MOD = 1e9 + 7;
const auto& dec = [](auto& x) {
for (int i = size(x) - 1; i >= 0; --i) {
if (x[i] != '0') {
--x[i];
break;
}
x[i] = '9';
}
return x;
};
const auto& f = [&](const auto& x) {
vector<vector<int>> dp(2, vector<int>(max_sum + 1));
dp[0][0] = dp[1][0] = 1;
for (int i = size(x) - 1; i >= 0; --i) {
vector<vector<int>> new_dp(2, vector<int>(max_sum + 1));
for (int t = 0; t < 2; ++t) {
for (int total = 0; total <= max_sum; ++total) {
for (int d = 0; d <= min((t == 1 ? x[i] - '0' : 9), total); ++d) {
new_dp[t][total] = (new_dp[t][total] + dp[static_cast<int>(t && d == x[i] - '0')][total - d]) % MOD;
}
}
}
dp = move(new_dp);
}
int result = 0;
for (int total = min_sum; total <= max_sum; ++total) {
result = (result + dp[1][total]) % MOD;
}
return result;
};
return ((f(num2) - f(dec(num1)) % MOD) + MOD) % MOD;
}
};
Beginner Explanation
What is Count of Integers?
Count of Integers (LeetCode #2719) 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: Combinatorics, DP.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Count of Integers
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(m * n)) and space (O(m + n)) 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(m * n) time and O(m + n) 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(m * n) |
| Space | O(m + n) |
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 Count of Integers
- 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 Count of Integers 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
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
Count of Integers (#2719) — Hard. Pattern: dynamic programming. Complexity: O(m * n) time / O(m + n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Count of Integers?+
The reference solutions aim for O(m * n) time and O(m + n) space. Always re-derive complexity from the code you write in the interview.
What pattern does Count of Integers use?+
It primarily maps to dynamic programming, within the broader topic of dynamic programming.
Is Count of Integers 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/count-of-integers/