LeetCode 2719: Count of Integers Solution

Master LeetCode problem 2719 (Count of Integers), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2719. Count of Integers

Hard

Math String Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can iterate through all the integers between num1 and num2, calculate the sum of digits for each integer, and count the integers that satisfy the conditions min_sum <= digit_sum(x) <= max_sum. We can optimize the process by calculating the range of possible digit sums based on the input constraints and then count the valid integers within this range. Finally, we return the count modulo 10^9 + 7.

Algorithm:

Define a helper function digitSum to calculate the sum of digits for a given integer.
Calculate the range of possible digit sums based on the constraints.
Iterate through all integers between num1 and num2, count valid integers that satisfy the conditions.
Return the count modulo 10^9 + 7.

Time Complexity:

The time complexity of this algorithm is O(N), where N is the difference between num2 and num1.

Space Complexity:

The space complexity is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int countOfIntegers(String num1, String num2, int min_sum, int max_sum) {
        long count = 0;
        int mod = 1000000007;
        
        int minDigits = num1.length();
        int maxDigits = num2.length();
        
        for (int digits = minDigits; digits <= maxDigits; digits++) {
            long minVal = Math.max(Long.parseLong(num1), (long) Math.pow(10, digits - 1));
            long maxVal = Math.min(Long.parseLong(num2), (long) (Math.pow(10, digits) - 1));
            
            for (long i = minVal; i <= maxVal; i++) {
                long sum = digitSum(i);
                if (sum >= min_sum && sum <= max_sum) {
                    count++;
                }
            }
        }
        
        return (int) (count % mod);
    }
    
    private long digitSum(long num) {
        long sum = 0;
        while (num > 0) {
            sum += num % 10;
            num /= 10;
        }
        return sum;
    }
}

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