# 2719. Count of Integers

## Description

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

## Solutions

Solution 1: Digit DP

The problem is actually asking for the number of integers in the range $[num1,..num2]$ whose digit sum is in the range $[min_sum,..max_sum]$. For this kind of range $[l,..r]$ problem, we can consider transforming it into finding the answers for $[1,..r]$ and $[1,..l-1]$, and then subtracting the latter from the former.

For the answer to $[1,..r]$, we can use digit DP to solve it. We design a function $dfs(pos, s, limit)$, which represents the number of schemes when we are currently processing the $pos$th digit, the digit sum is $s$, and whether the current number has an upper limit $limit$. Here, $pos$ is enumerated from high to low.

For $dfs(pos, s, limit)$, we can enumerate the value of the current digit $i$, and then recursively calculate $dfs(pos+1, s+i, limit \bigcap i==up)$, where $up$ represents the upper limit of the current digit. If $limit$ is true, then $up$ is the upper limit of the current digit, otherwise $up$ is $9$. If $pos$ is greater than or equal to the length of $num$, then we can judge whether $s$ is in the range $[min_sum,..max_sum]$. If it is, return $1$, otherwise return $0$.

The time complexity is $O(10 \times n \times max_sum)$, and the space complexity is $O(n \times max_sum)$. Here, $n$ represents the length of $num$.

• 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;
}
}

• class Solution {
public:
int count(string num1, string num2, int min_sum, int max_sum) {
const int mod = 1e9 + 7;
int f[23][220];
memset(f, -1, sizeof(f));
string num = num2;

function<int(int, int, bool)> dfs = [&](int pos, int s, bool limit) -> int {
if (pos >= num.size()) {
return s >= min_sum && s <= max_sum ? 1 : 0;
}
if (!limit && f[pos][s] != -1) {
return f[pos][s];
}
int up = limit ? num[pos] - '0' : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
ans += dfs(pos + 1, s + i, limit && i == up);
ans %= mod;
}
if (!limit) {
f[pos][s] = ans;
}
return ans;
};

int a = dfs(0, 0, true);
for (int i = num1.size() - 1; ~i; --i) {
if (num1[i] == '0') {
num1[i] = '9';
} else {
num1[i] -= 1;
break;
}
}
num = num1;
memset(f, -1, sizeof(f));
int b = dfs(0, 0, true);
return (a - b + mod) % mod;
}
};

• 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


• 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
}

• 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;
}