Count of Integers

Count of Integers - Problem

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: digit_sum(x) denotes the sum of the digits of x.

Input & Output

Example 1 — Basic Range

$ Input: num1 = "1", num2 = "12", min_sum = 1, max_sum = 8

› Output: 12

💡 Note: Numbers 1-12 have digit sums: 1(1), 2(2), 3(3), 4(4), 5(5), 6(6), 7(7), 8(8), 9(9), 10(1), 11(2), 12(3). All have digit sum ≤ 8 and ≥ 1, so count = 12.

Example 2 — Narrow Sum Range

$ Input: num1 = "1", num2 = "5", min_sum = 1, max_sum = 3

› Output: 3

💡 Note: Numbers 1,2,3 have digit sums 1,2,3 respectively (all valid). Numbers 4,5 have digit sums 4,5 (both > 3, invalid). Count = 3.

Example 3 — Single Number

$ Input: num1 = "15", num2 = "15", min_sum = 5, max_sum = 8

› Output: 1

💡 Note: Only number 15 in range. Digit sum = 1+5 = 6, which satisfies 5 ≤ 6 ≤ 8. Count = 1.

Constraints

1 ≤ num1 ≤ num2 < 10²²
1 ≤ min_sum ≤ max_sum ≤ 400

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to use digit DP to efficiently count valid numbers by building them position by position while tracking constraints. The optimal approach uses memoized recursion with states (position, digit_sum, tight_bound, started). Time: O(d × S × 2 × 2 × 10), Space: O(d × S × 2 × 2).

Common Approaches

✓ Brute Force Iteration

⏱️ Time: O(N * d) Space: O(1)

Convert string bounds to integers, then check every number from num1 to num2. For each number, calculate the sum of its digits and verify if it falls within the min_sum and max_sum range.

Digit DP with Memoization

⏱️ Time: O(d * S * 2 * 2 * 10) Space: O(d * S * 2 * 2)

Use digit DP technique where we build numbers digit by digit from left to right, keeping track of constraints (tight bounds, digit sum so far). Memoize states to avoid recomputation.

Brute Force Iteration — Algorithm Steps

Convert string bounds to integers
Iterate through each number in range
Calculate digit sum for each number
Count numbers with valid digit sums

Visualization

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Step-by-Step Walkthrough

Range

Convert strings to numbers, iterate from num1 to num2

Check

For each number, calculate sum of digits

Count

Count numbers with digit sum in valid range

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int digitSum(long long n) {
    int total = 0;
    while (n > 0) {
        total += n % 10;
        n /= 10;
    }
    return total;
}

int solution(char* num1, char* num2, int min_sum, int max_sum) {
    const int MOD = 1000000007;
    long long start = atoll(num1);
    long long end = atoll(num2);
    int count = 0;
    
    for (long long i = start; i <= end; i++) {
        int ds = digitSum(i);
        if (min_sum <= ds && ds <= max_sum) {
            count = (count + 1) % MOD;
        }
    }
    
    return count;
}

int main() {
    char num1[100], num2[100];
    int min_sum, max_sum;
    
    fgets(num1, sizeof(num1), stdin);
    num1[strcspn(num1, "\n")] = 0;
    
    fgets(num2, sizeof(num2), stdin);
    num2[strcspn(num2, "\n")] = 0;
    
    scanf("%d", &min_sum);
    scanf("%d", &max_sum);
    
    int result = solution(num1, num2, min_sum, max_sum);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(N * d)

N numbers in range, d digits per number to calculate sum

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Iteration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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