What is the time complexity of Count the Number of Powerful Integers?

The optimal solution has a time complexity of O(log(finish)).

What is the space complexity of Count the Number of Powerful Integers?

The space complexity is O(log(finish)).

What data structures are used to solve Count the Number of Powerful 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 the Number of Powerful Integers?

Insight 1: The problem can be solved by calculating the number of powerful integers up to 'finish' and subtracting the number of powerful integers up to 'start - 1'. This reduces the problem to counting powerful integers up to a certain number. Insight 2: Dynamic Programming (DP) with memoization can be used to efficiently count the number of valid prefixes. The 'is_tight' flag is crucial to handle cases where the current digits must match the original number.

Which companies ask LeetCode 2999. Count the Number of Powerful Integers in interviews?

This problem is frequently asked at HashedIn, Sprinklr.

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Count the Number of Powerful Integers - LeetCode 2999 Solution

Count the Number of Powerful Integers - Complete Solution Guide

Count the Number of Powerful Integers is LeetCode problem 2999, 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 three integers start , finish , and limit . You are also given a 0-indexed string s representing a positive integer. A positive integer x is called powerful if it ends with s (in other words, s is a suffix of x ) and each digit in x is at most limit . Return the total number of powerful integers in the range [start..finish] . A string x is a suffix of a string y if and only if x is a substring of y that starts from some index ( including 0 ) in y and extends to the index y.length -

Detailed Explanation

The problem asks us to count the number of 'powerful integers' within a given range [start, finish]. A powerful integer must satisfy two conditions: it must end with a given suffix string 's', and each of its digits must be less than or equal to a given limit. The problem involves counting integers satisfying these constraints within the inclusive range [start, finish].

Solution Approach

The solution uses dynamic programming with memoization to count the number of integers whose digits are within the limit and that end with 's' in a given range. The approach consists of breaking down the problem into smaller subproblems of counting valid prefixes and storing results in a memoization table to avoid redundant calculations. It uses the principle of inclusion-exclusion by calculating the count up to 'finish' and subtracting the count up to 'start - 1'.

Step-by-Step Algorithm

Step 1: Calculate 's_val' which is the integer representation of the suffix 's'. Also determine the length of the suffix ('len_s') and compute the power of 10 equivalent to the length of 's' (power_of_10_s). This is used to isolate the prefix.
Step 2: Define a recursive function 'dp(num_str, index, is_tight)' to count valid prefixes. 'num_str' is the string representation of the prefix, 'index' represents the current digit being considered, and 'is_tight' is a boolean flag indicating whether the digits chosen so far are constrained by the upper limit (i.e., the prefix of the original number being analyzed).
Step 3: Implement the base case for the 'dp' function: if 'index' reaches the end of 'num_str', it means a valid prefix has been found, so return 1.
Step 4: Implement memoization within the 'dp' function. Check if the result for the current state (index, is_tight) is already cached in the 'memo'. If so, return the cached value.
Step 5: Calculate the 'upper_bound' for the current digit. If 'is_tight' is true, the 'upper_bound' is the digit at the current 'index' in 'num_str'. Otherwise, the 'upper_bound' is 'limit'.
Step 6: Iterate from 0 to 'min(limit, upper_bound)' to consider all possible digits. For each digit, recursively call 'dp' to count valid prefixes, setting 'new_tight' to true if the current digit equals 'upper_bound' (and 'is_tight' was already true) and false otherwise.
Step 7: Sum up the results from the recursive calls. Store the result in the 'memo' and return it.
Step 8: Implement the function 'count_valid_numbers(n_str)' to call the 'dp' function with initial values (index = 0, is_tight = true). Clear the 'memo' before calling 'dp'.
Step 9: Implement the function 'get_powerful_count_up_to(n)' to calculate the number of powerful integers up to 'n'. First, check if 'n < s_val'. If it is, return 0 because no powerful integers are possible.
Step 10: Calculate the 'upper_prefix' by subtracting 's_val' from 'n' and then dividing by 'power_of_10_s'. Convert 'upper_prefix' to a string.
Step 11: Call 'count_valid_numbers' with the 'upper_prefix' string and return the result.
Step 12: Calculate the number of powerful integers up to 'finish' and 'start - 1' using 'get_powerful_count_up_to'. Subtract the two results to get the number of powerful integers in the range [start, finish].

Key Insights

Insight 1: The problem can be solved by calculating the number of powerful integers up to 'finish' and subtracting the number of powerful integers up to 'start - 1'. This reduces the problem to counting powerful integers up to a certain number.
Insight 2: Dynamic Programming (DP) with memoization can be used to efficiently count the number of valid prefixes. The 'is_tight' flag is crucial to handle cases where the current digits must match the original number.
Insight 3: Extracting the prefix from the number (n - s_val) // power_of_10_s allows the problem to be solved without having to construct all possible integers. This is a key simplification.

Complexity Analysis

Time Complexity: O(log(finish))

Space Complexity: O(log(finish))

Topics

This problem involves: Math, String, Dynamic Programming.

Companies

Asked at: HashedIn, Sprinklr.

Count the Number of Powerful Integers | LeetCode 2999

Hard

Math String Dynamic Programming

Time: O(log(finish))Space: O(log(finish))

Solution Code

Available in:

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

class Solution {
    public long numberOfPowerfulInt(long start, long finish, int limit, String s) {
        long s_val = Long.parseLong(s);
        int len_s = s.length();
        long power_of_10_s = (long) Math.pow(10, len_s);
        
        Map<String, Long> memo = new HashMap<>();

        class DP {
            long solve(String num_str, int index, boolean is_tight) {
                if (index == num_str.length()) {
                    return 1;
                }
                
                String state = index + "," + is_tight;
                if (memo.containsKey(state)) {
                    return memo.get(state);
                }
                
                long res = 0;
                int upper_bound = Character.getNumericValue(num_str.charAt(index));
                if (!is_tight) {
                  upper_bound = 9;
                }
                
                for (int digit = 0; digit <= Math.min(limit, upper_bound); digit++) {
                    boolean new_tight = is_tight && (digit == upper_bound);
                    res += solve(num_str, index + 1, new_tight);
                }
                
                memo.put(state, res);
                return res;
            }
        }

        DP dpSolver = new DP();

        long count_valid_numbers(String n_str) {
            memo.clear();
            return dpSolver.solve(n_str, 0, true);
        }

        long get_powerful_count_up_to(long n) {
            if (n < s_val) {
                return 0;
            }
            
            long upper_prefix = (n - s_val) / power_of_10_s;
            
            return count_valid_numbers(String.valueOf(upper_prefix));
        }

        long count_finish = get_powerful_count_up_to(finish);
        long count_start_minus_1 = get_powerful_count_up_to(start - 1);
        
        return count_finish - count_start_minus_1;
    }
}

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Problem Info

Description

Problem 2999: Count the Number of Powerful Integers - Detailed explanation and solution approaches.

Topics

Math String Dynamic Programming

Companies

HashedInSprinklr

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2999

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

12 steps

1
Step 1: Calculate 's_val' which is the integer representation of the suffix 's'. Also determine the length of the suffix ('len_s') and compute the power of 10 equivalent to the length of 's' (power_of_10_s). This is used to isolate the prefix.
2
Step 2: Define a recursive function 'dp(num_str, index, is_tight)' to count valid prefixes. 'num_str' is the string representation of the prefix, 'index' represents the current digit being considered, and 'is_tight' is a boolean flag indicating whether the digits chosen so far are constrained by the upper limit (i.e., the prefix of the original number being analyzed).
3
Step 3: Implement the base case for the 'dp' function: if 'index' reaches the end of 'num_str', it means a valid prefix has been found, so return 1.
4
Step 4: Implement memoization within the 'dp' function. Check if the result for the current state (index, is_tight) is already cached in the 'memo'. If so, return the cached value.
5
Step 5: Calculate the 'upper_bound' for the current digit. If 'is_tight' is true, the 'upper_bound' is the digit at the current 'index' in 'num_str'. Otherwise, the 'upper_bound' is 'limit'.
6
Step 6: Iterate from 0 to 'min(limit, upper_bound)' to consider all possible digits. For each digit, recursively call 'dp' to count valid prefixes, setting 'new_tight' to true if the current digit equals 'upper_bound' (and 'is_tight' was already true) and false otherwise.
7
Step 7: Sum up the results from the recursive calls. Store the result in the 'memo' and return it.
8
Step 8: Implement the function 'count_valid_numbers(n_str)' to call the 'dp' function with initial values (index = 0, is_tight = true). Clear the 'memo' before calling 'dp'.
9
Step 9: Implement the function 'get_powerful_count_up_to(n)' to calculate the number of powerful integers up to 'n'. First, check if 'n < s_val'. If it is, return 0 because no powerful integers are possible.
10
Step 10: Calculate the 'upper_prefix' by subtracting 's_val' from 'n' and then dividing by 'power_of_10_s'. Convert 'upper_prefix' to a string.
11
Step 11: Call 'count_valid_numbers' with the 'upper_prefix' string and return the result.
12
Step 12: Calculate the number of powerful integers up to 'finish' and 'start - 1' using 'get_powerful_count_up_to'. Subtract the two results to get the number of powerful integers in the range [start, finish].

💡

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(log(finish))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(log(finish))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is determined by the 'dp' function, which explores all valid digit combinations for the prefix. The depth of the recursion is limited by the number of digits in the prefix, which is proportional to log(finish). Memoization ensures that each state is computed only once. The space complexity is dominated by the memoization table, which stores results for different states of the DP. The number of states is determined by the length of the prefix and the 'is_tight' flag.

📖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

Iterate through each number from 'start' to 'finish', check if it ends with 's', and if all digits are less than or equal to 'limit'.

Time: O((finish - start + 1) * len(str(i))), where i is each number from start to finish. This is approximately O(finish * log(finish)) in the worst case.Space: O(1)

Trade-offs: Brute force is simple to implement but inefficient for large ranges of [start, finish]. Dynamic programming is significantly more efficient but more complex to implement.

✨

Key Insights

1Insight 1: The problem can be solved by calculating the number of powerful integers up to 'finish' and subtracting the number of powerful integers up to 'start - 1'. This reduces the problem to counting powerful integers up to a certain number.
2Insight 2: Dynamic Programming (DP) with memoization can be used to efficiently count the number of valid prefixes. The 'is_tight' flag is crucial to handle cases where the current digits must match the original number.
3Insight 3: Extracting the prefix from the number (n - s_val) // power_of_10_s allows the problem to be solved without having to construct all possible integers. This is a key simplification.

📚

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: 'start' is 1. The logic for subtracting the result of 'start - 1' still works because get_powerful_count_up_to(0) returns 0.
Edge case 2: 'finish' is smaller than 's_val'. The logic handles this correctly by returning 0 for get_powerful_count_up_to(finish), as no powerful integers can be formed.
Edge case 3: When 's' has a length close to that of 'finish', the calculated prefix might be very small or even zero. The DP handles such cases appropriately.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the 'is_tight' constraint properly. This can lead to overcounting, especially when the digit at a particular index is equal to the limit.

✗Common mistake 2: Not clearing the memoization table between calls to 'count_valid_numbers'. This can lead to incorrect results when calculating the counts up to 'finish' and 'start - 1'.

✗Common mistake 3: Integer overflow when performing calculations with large numbers, especially when computing `(n - s_val) / power_of_10_s`.

✅

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: Counting valid identifiers or keys in a database that must adhere to specific naming conventions (e.g., ending with a certain suffix or having digits below a certain value).

🏦Real-world scenario 2: Generating test cases that follow a certain pattern (suffix) and have digit constraints. This can be used to create realistic and diverse test data.

🎮Real-world scenario 3: Analyzing network traffic where packets need to be filtered based on specific patterns (suffix) and data limits (digit constraint).

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: September 28, 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 the Number of Powerful 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.

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