What is the time complexity of Count of Integers?

The optimal solution has a time complexity of O(n * max_sum).

What is the space complexity of Count of Integers?

The space complexity is O(n * max_sum).

What data structures are used to solve Count of Integers?

This problem uses Math, String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count of Integers?

Insight 1: The problem can be solved by calculating the number of 'good' integers less than or equal to num2 and subtracting the number of 'good' integers less than num1. This reduces the problem to a counting problem up to a certain bound. Insight 2: Dynamic programming with memoization is crucial to avoid redundant calculations, especially when dealing with large numbers. The DP state captures the current index in the number, the current digit sum, and whether we are still 'tight' (i.e., whether we have used the maximum possible digit at each position).

Which companies ask LeetCode 2719. Count of Integers in interviews?

This problem is frequently asked at Cisco, Morgan Stanley.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Count of Integers - LeetCode 2719 Solution

Q: How to solve Count of Integers?

The provided solution uses dynamic programming (with memoization) to efficiently count the number of integers less than or equal to a given number `S` that have a digit sum between `min_sum` and `max_sum`. It leverages the principle of inclusion-exclusion: Count(num1 <= x <= num2, min_sum <= digit_sum(x) <= max_sum) = Count(x <= num2, min_sum <= digit_sum(x) <= max_sum) - Count(x < num1, min_sum <= digit_sum(x) <= max_sum). The `count_le` function calculates the number of integers less than or equal to a given string `S` with digit sum in the range [min_sum, max_sum]. The `dp` function is the core of the dynamic programming solution. It recursively explores the possible digits at each position of the number, keeping track of the current digit sum and a flag `is_tight` to ensure the generated number does not exceed the input number.

Count of Integers - Complete Solution Guide

Count of Integers is LeetCode problem 2719, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two numeric strings num1 and num2 and two integers max_sum and min_sum . We denote an integer x to be good if: num1 <= x <= num2 min_sum <= digit_sum(x) <= max_sum . Return the number of good integers . Since the answer may be large, return it modulo 10 9 + 7 . Note that digit_sum(x) denotes the sum of the digits of x . Example 1: Input: num1 = "1", num2 = "12", min_sum = 1, max_sum = 8 Output: 11 Explanation: There are 11 integers whose sum of digits lies between 1 and 8 are 1,2,3

Detailed Explanation

The problem requires us to count the number of 'good' integers within a given range [num1, num2] (inclusive) such that the sum of their digits falls within another given range [min_sum, max_sum] (inclusive). Essentially, we need to iterate through integers between num1 and num2, calculate the digit sum of each integer, and check if it falls within the range [min_sum, max_sum]. If it does, we increment our count. Since the numbers can be very large (up to 10^22), a brute-force approach is infeasible. We have to use dynamic programming to efficiently solve the problem, because a direct iteration through all numbers in the range [num1, num2] would be too slow.

Solution Approach

The provided solution uses dynamic programming (with memoization) to efficiently count the number of integers less than or equal to a given number `S` that have a digit sum between `min_sum` and `max_sum`. It leverages the principle of inclusion-exclusion: Count(num1 <= x <= num2, min_sum <= digit_sum(x) <= max_sum) = Count(x <= num2, min_sum <= digit_sum(x) <= max_sum) - Count(x < num1, min_sum <= digit_sum(x) <= max_sum). The `count_le` function calculates the number of integers less than or equal to a given string `S` with digit sum in the range [min_sum, max_sum]. The `dp` function is the core of the dynamic programming solution. It recursively explores the possible digits at each position of the number, keeping track of the current digit sum and a flag `is_tight` to ensure the generated number does not exceed the input number.

Step-by-Step Algorithm

Step 1: Define a helper function `count_le(S, max_s)` that counts the number of integers less than or equal to the string `S` whose digit sum is at most `max_s`.
Step 2: Inside `count_le`, use a recursive function `dp(index, current_sum, is_tight)` to implement the dynamic programming solution. `index` represents the current digit position being considered, `current_sum` is the sum of the digits seen so far, and `is_tight` is a boolean flag indicating whether the generated number is still equal to the corresponding prefix of `S`.
Step 3: The base cases for the `dp` function are: if `current_sum > max_s`, return 0 because the current path is invalid. If `index == n` (reached the end of the number), return 1 because a valid number has been formed.
Step 4: In the recursive step, determine the maximum digit `limit` that can be placed at the current position. If `is_tight` is true, `limit` is the digit at the current position in `S`; otherwise, `limit` is 9.
Step 5: Iterate through all possible digits from 0 to `limit`. For each digit, update `current_sum` and `is_tight` and recursively call `dp` to explore the remaining digits.
Step 6: Memoize the result of each `dp` call to avoid redundant calculations.
Step 7: In the main `count` function, calculate `count_le(num2, max_sum) - count_le(num2, min_sum - 1)` to get the count of integers less than or equal to num2 with a digit sum between min_sum and max_sum.
Step 8: Similarly, calculate `count_le(num1_minus_1, max_sum) - count_le(num1_minus_1, min_sum - 1)` where num1_minus_1 = num1 - 1 to get the count of integers less than or equal to num1 - 1 with a digit sum between min_sum and max_sum.
Step 9: Return the difference of the two counts, applying the modulo operator to handle potential negative results and large values.

Key Insights

Insight 1: The problem can be solved by calculating the number of 'good' integers less than or equal to num2 and subtracting the number of 'good' integers less than num1. This reduces the problem to a counting problem up to a certain bound.
Insight 2: Dynamic programming with memoization is crucial to avoid redundant calculations, especially when dealing with large numbers. The DP state captures the current index in the number, the current digit sum, and whether we are still 'tight' (i.e., whether we have used the maximum possible digit at each position).
Insight 3: The constraint `min_sum <= max_sum <= 400` is vital. It limits the possible range of `current_sum` in the DP, making the memoization table size manageable.

Complexity Analysis

Time Complexity: O(n * max_sum)

Space Complexity: O(n * max_sum)

Topics

This problem involves: Math, String, Dynamic Programming.

Companies

Asked at: Cisco, Morgan Stanley.

Count of Integers | LeetCode 2719

Hard

Math String Dynamic Programming

Time: O(n * max_sum)Space: O(n * max_sum)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    private final int MOD = (int) 1e9 + 7;

    public int count(String num1, String num2, int min_sum, int max_sum) {
        long ans_num2 = (count_le(num2, max_sum) - count_le(num2, min_sum - 1) + MOD) % MOD;
        
        String num1_minus_1 = String.valueOf(Long.parseLong(num1) - 1);
        long ans_num1_minus_1 = (count_le(num1_minus_1, max_sum) - count_le(num1_minus_1, min_sum - 1) + MOD) % MOD;

        return (int)((ans_num2 - ans_num1_minus_1 + MOD) % MOD);
    }

    private long count_le(String S, int max_s) {
        int n = S.length();
        Map<String, Long> memo = new HashMap<>();

        if (max_s < 0) {
            return 0;
        }

        return dp(0, 0, true, S, max_s, memo);
    }

    private long dp(int index, int current_sum, boolean is_tight, String S, int max_s, Map<String, Long> memo) {
        if (current_sum > max_s) {
            return 0;
        }
        if (index == S.length()) {
            return 1;
        }

        String state = index + "," + current_sum + "," + is_tight;
        if (memo.containsKey(state)) {
            return memo.get(state);
        }

        int limit = is_tight ? S.charAt(index) - '0' : 9;
        long ans = 0;
        for (int digit = 0; digit <= limit; digit++) {
            boolean new_tight = is_tight && (digit == limit);
            ans = (ans + dp(index + 1, current_sum + digit, new_tight, S, max_s, memo)) % MOD;
        }

        memo.put(state, ans);
        return ans;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2719: Count of Integers - Detailed explanation and solution approaches.

Topics

Math String Dynamic Programming

Companies

CiscoMorgan Stanley

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2719

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Define a helper function `count_le(S, max_s)` that counts the number of integers less than or equal to the string `S` whose digit sum is at most `max_s`.
2
Step 2: Inside `count_le`, use a recursive function `dp(index, current_sum, is_tight)` to implement the dynamic programming solution. `index` represents the current digit position being considered, `current_sum` is the sum of the digits seen so far, and `is_tight` is a boolean flag indicating whether the generated number is still equal to the corresponding prefix of `S`.
3
Step 3: The base cases for the `dp` function are: if `current_sum > max_s`, return 0 because the current path is invalid. If `index == n` (reached the end of the number), return 1 because a valid number has been formed.
4
Step 4: In the recursive step, determine the maximum digit `limit` that can be placed at the current position. If `is_tight` is true, `limit` is the digit at the current position in `S`; otherwise, `limit` is 9.
5
Step 5: Iterate through all possible digits from 0 to `limit`. For each digit, update `current_sum` and `is_tight` and recursively call `dp` to explore the remaining digits.
6
Step 6: Memoize the result of each `dp` call to avoid redundant calculations.
7
Step 7: In the main `count` function, calculate `count_le(num2, max_sum) - count_le(num2, min_sum - 1)` to get the count of integers less than or equal to num2 with a digit sum between min_sum and max_sum.
8
Step 8: Similarly, calculate `count_le(num1_minus_1, max_sum) - count_le(num1_minus_1, min_sum - 1)` where num1_minus_1 = num1 - 1 to get the count of integers less than or equal to num1 - 1 with a digit sum between min_sum and max_sum.
9
Step 9: Return the difference of the two counts, applying the modulo operator to handle potential negative results and large values.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * max_sum)

💾

Space Complexity

O(n * max_sum)

📝

Why?

The time complexity is determined by the number of unique states that the DP function explores. Each state (index, current_sum, is_tight) is calculated at most once due to memoization. The `count_le` function calculates count for number `num2` and `num1 - 1`. Thus, calculating the length of `num2` to be n, there are n * max_sum possible states. The space complexity is determined by the memory required to store the memoization table to avoid recalculating DP states.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-force approach

Iterate through all numbers from `num1` to `num2`, calculate the digit sum of each number, and check if it lies within the range `[min_sum, max_sum]`. However, this approach is not feasible for large numbers (up to 10^22) due to its high time complexity.

Time: O((num2 - num1 + 1) * log(num2)), where log(num2) is the number of digits in num2 (the time to calculate digit sum).Space: O(1)

Trade-offs: Brute force is simple to understand, but extremely inefficient. It's only suitable for very small input ranges. The dynamic programming approach is much more efficient for larger ranges.

✨

Key Insights

1Insight 1: The problem can be solved by calculating the number of 'good' integers less than or equal to num2 and subtracting the number of 'good' integers less than num1. This reduces the problem to a counting problem up to a certain bound.
2Insight 2: Dynamic programming with memoization is crucial to avoid redundant calculations, especially when dealing with large numbers. The DP state captures the current index in the number, the current digit sum, and whether we are still 'tight' (i.e., whether we have used the maximum possible digit at each position).
3Insight 3: The constraint `min_sum <= max_sum <= 400` is vital. It limits the possible range of `current_sum` in the DP, making the memoization table size manageable.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `num1` is '1' and `num2` is a large number. The dynamic programming approach efficiently handles this case without iterating through all the numbers.
Edge case 2: `min_sum` is greater than `max_sum`. In this case, there are no valid numbers, and the solution correctly returns 0.
Edge case 3: `num1` and `num2` are the same. The solution correctly computes the count by reducing to `count_le` on the given numbers.
Edge case 4: Leading zeros in `num1 - 1`. The solution removes leading zeros to ensure correctness.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the modulo operation. Since the answer can be very large, it's crucial to apply the modulo operator after each addition to prevent integer overflow.

✗Common mistake 2: Not handling the `is_tight` flag correctly. The `is_tight` flag is essential to ensure that the generated numbers do not exceed the input number `S`.

✗Common mistake 3: Integer overflow when calculating `num1 - 1` for very large `num1`. The code needs to handle the case where num1 is 10000...., using string manipulation.

✗Common mistake 4: Incorrect indexing in memoization table in the C solution leading to incorrect results or memory access errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This type of counting problem arises in cryptography, where one might need to count the number of keys or messages that satisfy certain conditions on their digit sum or other numerical properties.

🏦Real-world scenario 2: In number theory and combinatorics, similar counting problems appear when analyzing the distribution of numbers with certain properties.

🎮Real-world scenario 3: Database query optimization. Suppose we need to find records in a database within a numerical ID range, satisfying some constraints based on the sum of digits of some ID property. This algorithm helps estimate the number of such records, to optimize the execution plan.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count of Integers"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?