Question

2719. Count of Integers

中文文档

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 $num$.

Similar problems:

• 2801. Count Stepping Numbers in Range

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

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;
  }
};

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;
}