LeetCode #2719 — HARD

Count of Integers

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 10⁹ + 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 1,2,3,4,5,6,7,8,10,11, and 12. Thus, we return 11.

Example 2:

Input: num1 = "1", num2 = "5", min_sum = 1, max_sum = 5
Output: 5
Explanation: The 5 integers whose sum of digits lies between 1 and 5 are 1,2,3,4, and 5. Thus, we return 5.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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 109 + 7. Note that digit_sum(x) denotes the sum of the digits of x.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

"1"
"12"
1
8

Example 2

"1"
"5"
1
5

Core Insight

What unlocks the optimal approach

Let f(n, l, r) denotes the number of integers from 1 to n with the sum of digits between l and r.
The answer is f(num2, min_sum, max_sum) - f(num1-1, min_sum, max_sum).
You can calculate f(n, l, r) using digit dp.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2719: Count of Integers
import java.math.BigInteger;

class Solution {
    private final int mod = (int) 1e9 + 7;
    private Integer[][] f;
    private String num;
    private int min;
    private int max;

    public int count(String num1, String num2, int min_sum, int max_sum) {
        min = min_sum;
        max = max_sum;
        num = num2;
        f = new Integer[23][220];
        int a = dfs(0, 0, true);
        num = new BigInteger(num1).subtract(BigInteger.ONE).toString();
        f = new Integer[23][220];
        int b = dfs(0, 0, true);
        return (a - b + mod) % mod;
    }

    private int dfs(int pos, int s, boolean limit) {
        if (pos >= num.length()) {
            return s >= min && s <= max ? 1 : 0;
        }
        if (!limit && f[pos][s] != null) {
            return f[pos][s];
        }
        int ans = 0;
        int up = limit ? num.charAt(pos) - '0' : 9;
        for (int i = 0; i <= up; ++i) {
            ans = (ans + dfs(pos + 1, s + i, limit && i == up)) % mod;
        }
        if (!limit) {
            f[pos][s] = ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2719: Count of Integers
func count(num1 string, num2 string, min_sum int, max_sum int) int {
	const mod = 1e9 + 7
	f := [23][220]int{}
	for i := range f {
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	num := num2
	var dfs func(int, int, bool) int
	dfs = func(pos, s int, limit bool) int {
		if pos >= len(num) {
			if s >= min_sum && s <= max_sum {
				return 1
			}
			return 0
		}
		if !limit && f[pos][s] != -1 {
			return f[pos][s]
		}
		var ans int
		up := 9
		if limit {
			up = int(num[pos] - '0')
		}
		for i := 0; i <= up; i++ {
			ans = (ans + dfs(pos+1, s+i, limit && i == up)) % mod
		}
		if !limit {
			f[pos][s] = ans
		}
		return ans
	}
	a := dfs(0, 0, true)
	t := []byte(num1)
	for i := len(t) - 1; i >= 0; i-- {
		if t[i] != '0' {
			t[i]--
			break
		}
		t[i] = '9'
	}
	num = string(t)
	f = [23][220]int{}
	for i := range f {
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	b := dfs(0, 0, true)
	return (a - b + mod) % mod
}

# Accepted solution for LeetCode #2719: Count of Integers
class Solution:
    def count(self, num1: str, num2: str, min_sum: int, max_sum: int) -> int:
        @cache
        def dfs(pos: int, s: int, limit: bool) -> int:
            if pos >= len(num):
                return int(min_sum <= s <= max_sum)
            up = int(num[pos]) if limit else 9
            return (
                sum(dfs(pos + 1, s + i, limit and i == up) for i in range(up + 1)) % mod
            )

        mod = 10**9 + 7
        num = num2
        a = dfs(0, 0, True)
        dfs.cache_clear()
        num = str(int(num1) - 1)
        b = dfs(0, 0, True)
        return (a - b) % mod

// Accepted solution for LeetCode #2719: Count of Integers
/**
 * [2719] Count of Integers
 */
pub struct Solution {}

// submission codes start here

struct DigitDP {
    num1: Vec<u32>,
    num2: Vec<u32>,
    min_sum: u32,
    max_sum: u32,

    dp_vec: Vec<Vec<i32>>,
}

impl DigitDP {
    pub fn new(num1: Vec<u32>, num2: Vec<u32>, min_sum: u32, max_sum: u32) -> DigitDP {
        let mut dp = Vec::with_capacity(num2.len());

        for _ in 0..num2.len() {
            let length = num2.len() * 9 + 1;
            let mut array = Vec::with_capacity(length);
            array.resize(length, -1);

            dp.push(array)
        }

        DigitDP {
            num1,
            num2,
            min_sum,
            max_sum,
            dp_vec: dp,
        }
    }

    pub fn dp(&mut self, sum: u32, l_limit: bool, r_limit: bool, index: usize) -> i32 {
        if index >= self.num2.len() {
            return if sum >= self.min_sum && sum <= self.max_sum {
                1
            } else {
                0
            };
        }

        if !l_limit && !r_limit && self.dp_vec[index][sum as usize] != -1 {
            return self.dp_vec[index][sum as usize];
        }

        let low = if l_limit { self.num1[index] } else { 0 };

        let high = if r_limit { self.num2[index] } else { 9 };

        let mut result = 0;
        for i in low..=high {
            result += DigitDP::dp(
                self,
                sum + i,
                i == self.num1[index] && l_limit,
                i == self.num2[index] && r_limit,
                index + 1,
            );

            if result > 1000000007 {
                result = result % 1000000007
            }
        }

        if !l_limit && !r_limit {
            self.dp_vec[index][sum as usize] = result;
        }

        result
    }
}

impl Solution {
    pub fn count(num1: String, num2: String, min_sum: i32, max_sum: i32) -> i32 {
        let mut num1 = Solution::str_to_vec(num1);
        let num2 = Solution::str_to_vec(num2);

        // 将两个数组的长度对齐，在num1前面填0
        for _ in 0..(num2.len() - num1.len()) {
            num1.insert(0, 0);
        }

        let mut digit_dp = DigitDP::new(num1, num2, min_sum as u32, max_sum as u32);

        digit_dp.dp(0, true, true, 0)
    }

    fn str_to_vec(num: String) -> Vec<u32> {
        let mut num_vec = Vec::new();

        for c in num.chars() {
            match c.to_digit(10) {
                None => {}
                Some(i) => num_vec.push(i),
            }
        }

        num_vec
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2719() {
        assert_eq!(
            Solution::count(String::from("1"), String::from("12"), 1, 8),
            11
        );

        assert_eq!(
            Solution::count(String::from("1"), String::from("5"), 1, 5),
            5
        );
    }
}

// Accepted solution for LeetCode #2719: Count of Integers
function count(num1: string, num2: string, min_sum: number, max_sum: number): number {
    const mod = 1e9 + 7;
    const f: number[][] = Array.from({ length: 23 }, () => Array(220).fill(-1));
    let num = num2;
    const dfs = (pos: number, s: number, limit: boolean): number => {
        if (pos >= num.length) {
            return s >= min_sum && s <= max_sum ? 1 : 0;
        }
        if (!limit && f[pos][s] !== -1) {
            return f[pos][s];
        }
        let ans = 0;
        const up = limit ? +num[pos] : 9;
        for (let i = 0; i <= up; i++) {
            ans = (ans + dfs(pos + 1, s + i, limit && i === up)) % mod;
        }
        if (!limit) {
            f[pos][s] = ans;
        }
        return ans;
    };
    const a = dfs(0, 0, true);
    num = (BigInt(num1) - 1n).toString();
    f.forEach(v => v.fill(-1));
    const b = dfs(0, 0, true);
    return (a - b + mod) % mod;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(10 × n × max\_sum)

Space

O(n × max\_sum)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.