2999. Count the Number of Powerful Integers

Description

You are given three integers start, finish, and limit. You are also given a 0-indexed string s representing a positive integer.

A positive integer x is called powerful if it ends with s (in other words, s is a suffix of x) and each digit in x is at most limit.

Return the total number of powerful integers in the range [start..finish].

A string x is a suffix of a string y if and only if x is a substring of y that starts from some index (including 0) in y and extends to the index y.length - 1. For example, 25 is a suffix of 5125 whereas 512 is not.

Example 1:

Input: start = 1, finish = 6000, limit = 4, s = "124"
Output: 5
Explanation: The powerful integers in the range [1..6000] are 124, 1124, 2124, 3124, and, 4124. All these integers have each digit <= 4, and "124" as a suffix. Note that 5124 is not a powerful integer because the first digit is 5 which is greater than 4.
It can be shown that there are only 5 powerful integers in this range.

Example 2:

Input: start = 15, finish = 215, limit = 6, s = "10"
Output: 2
Explanation: The powerful integers in the range [15..215] are 110 and 210. All these integers have each digit <= 6, and "10" as a suffix.
It can be shown that there are only 2 powerful integers in this range.

Example 3:

Input: start = 1000, finish = 2000, limit = 4, s = "3000"
Output: 0
Explanation: All integers in the range [1000..2000] are smaller than 3000, hence "3000" cannot be a suffix of any integer in this range.

Constraints:

1 <= start <= finish <= 10¹⁵
1 <= limit <= 9
1 <= s.length <= floor(log₁₀(finish)) + 1
s only consists of numeric digits which are at most limit.
s does not have leading zeros.

Solutions

class Solution {
    private String s;
    private String t;
    private Long[] f;
    private int limit;

    public long numberOfPowerfulInt(long start, long finish, int limit, String s) {
        this.s = s;
        this.limit = limit;
        t = String.valueOf(start - 1);
        f = new Long[20];
        long a = dfs(0, true);
        t = String.valueOf(finish);
        f = new Long[20];
        long b = dfs(0, true);
        return b - a;
    }

    private long dfs(int pos, boolean lim) {
        if (t.length() < s.length()) {
            return 0;
        }
        if (!lim && f[pos] != null) {
            return f[pos];
        }
        if (t.length() - pos == s.length()) {
            return lim ? (s.compareTo(t.substring(pos)) <= 0 ? 1 : 0) : 1;
        }
        int up = lim ? t.charAt(pos) - '0' : 9;
        up = Math.min(up, limit);
        long ans = 0;
        for (int i = 0; i <= up; ++i) {
            ans += dfs(pos + 1, lim && i == (t.charAt(pos) - '0'));
        }
        if (!lim) {
            f[pos] = ans;
        }
        return ans;
    }
}

class Solution {
public:
    long long numberOfPowerfulInt(long long start, long long finish, int limit, string s) {
        string t = to_string(start - 1);
        long long f[20];
        memset(f, -1, sizeof(f));

        function<long long(int, bool)> dfs = [&](int pos, bool lim) -> long long {
            if (t.size() < s.size()) {
                return 0;
            }
            if (!lim && f[pos] != -1) {
                return f[pos];
            }
            if (t.size() - pos == s.size()) {
                return lim ? s <= t.substr(pos) : 1;
            }
            long long ans = 0;
            int up = min(lim ? t[pos] - '0' : 9, limit);
            for (int i = 0; i <= up; ++i) {
                ans += dfs(pos + 1, lim && i == (t[pos] - '0'));
            }
            if (!lim) {
                f[pos] = ans;
            }
            return ans;
        };

        long long a = dfs(0, true);
        t = to_string(finish);
        memset(f, -1, sizeof(f));
        long long b = dfs(0, true);
        return b - a;
    }
};

class Solution:
    def numberOfPowerfulInt(self, start: int, finish: int, limit: int, s: str) -> int:
        @cache
        def dfs(pos: int, lim: int):
            if len(t) < n:
                return 0
            if len(t) - pos == n:
                return int(s <= t[pos:]) if lim else 1
            up = min(int(t[pos]) if lim else 9, limit)
            ans = 0
            for i in range(up + 1):
                ans += dfs(pos + 1, lim and i == int(t[pos]))
            return ans

        n = len(s)
        t = str(start - 1)
        a = dfs(0, True)
        dfs.cache_clear()
        t = str(finish)
        b = dfs(0, True)
        return b - a

func numberOfPowerfulInt(start, finish int64, limit int, s string) int64 {
	t := strconv.FormatInt(start-1, 10)
	f := make([]int64, 20)
	for i := range f {
		f[i] = -1
	}

	var dfs func(int, bool) int64
	dfs = func(pos int, lim bool) int64 {
		if len(t) < len(s) {
			return 0
		}
		if !lim && f[pos] != -1 {
			return f[pos]
		}
		if len(t)-pos == len(s) {
			if lim {
				if s <= t[pos:] {
					return 1
				}
				return 0
			}
			return 1
		}

		ans := int64(0)
		up := 9
		if lim {
			up = int(t[pos] - '0')
		}
		up = min(up, limit)
		for i := 0; i <= up; i++ {
			ans += dfs(pos+1, lim && i == int(t[pos]-'0'))
		}
		if !lim {
			f[pos] = ans
		}
		return ans
	}

	a := dfs(0, true)
	t = strconv.FormatInt(finish, 10)
	for i := range f {
		f[i] = -1
	}
	b := dfs(0, true)
	return b - a
}

function numberOfPowerfulInt(start: number, finish: number, limit: number, s: string): number {
    let t: string = (start - 1).toString();
    let f: number[] = Array(20).fill(-1);

    const dfs = (pos: number, lim: boolean): number => {
        if (t.length < s.length) {
            return 0;
        }
        if (!lim && f[pos] !== -1) {
            return f[pos];
        }
        if (t.length - pos === s.length) {
            if (lim) {
                return s <= t.substring(pos) ? 1 : 0;
            }
            return 1;
        }

        let ans: number = 0;
        const up: number = Math.min(lim ? +t[pos] : 9, limit);
        for (let i = 0; i <= up; i++) {
            ans += dfs(pos + 1, lim && i === +t[pos]);
        }

        if (!lim) {
            f[pos] = ans;
        }
        return ans;
    };

    const a: number = dfs(0, true);
    t = finish.toString();
    f = Array(20).fill(-1);
    const b: number = dfs(0, true);

    return b - a;
}

public class Solution {
    private string s;
    private string t;
    private long?[] f;
    private int limit;

    public long NumberOfPowerfulInt(long start, long finish, int limit, string s) {
        this.s = s;
        this.limit = limit;
        t = (start - 1).ToString();
        f = new long?[20];
        long a = Dfs(0, true);
        t = finish.ToString();
        f = new long?[20];
        long b = Dfs(0, true);
        return b - a;
    }

    private long Dfs(int pos, bool lim) {
        if (t.Length < s.Length) {
            return 0;
        }
        if (!lim && f[pos].HasValue) {
            return f[pos].Value;
        }
        if (t.Length - pos == s.Length) {
            return lim ? (string.Compare(s, t.Substring(pos)) <= 0 ? 1 : 0) : 1;
        }
        int up = lim ? t[pos] - '0' : 9;
        up = Math.Min(up, limit);
        long ans = 0;
        for (int i = 0; i <= up; ++i) {
            ans += Dfs(pos + 1, lim && i == (t[pos] - '0'));
        }
        if (!lim) {
            f[pos] = ans;
        }
        return ans;
    }
}

impl Solution {
    pub fn number_of_powerful_int(start: i64, finish: i64, limit: i32, s: String) -> i64 {
        fn count(x: i64, limit: i32, s: &str) -> i64 {
            let t = x.to_string();
            if t.len() < s.len() {
                return 0;
            }

            let t_bytes: Vec<u8> = t.bytes().collect();
            let mut f = [-1_i64; 20];

            fn dfs(pos: usize, lim: bool, t: &[u8], s: &str, limit: i32, f: &mut [i64; 20]) -> i64 {
                if t.len() < s.len() {
                    return 0;
                }

                if !lim && f[pos] != -1 {
                    return f[pos];
                }

                if t.len() - pos == s.len() {
                    if lim {
                        let suffix = &t[pos..];
                        let suffix_str = String::from_utf8_lossy(suffix);
                        return if suffix_str.as_ref() >= s { 1 } else { 0 };
                    } else {
                        return 1;
                    }
                }

                let mut ans = 0;
                let up = if lim {
                    (t[pos] - b'0').min(limit as u8)
                } else {
                    limit as u8
                };

                for i in 0..=up {
                    let next_lim = lim && i == t[pos] - b'0';
                    ans += dfs(pos + 1, next_lim, t, s, limit, f);
                }

                if !lim {
                    f[pos] = ans;
                }

                ans
            }

            dfs(0, true, &t_bytes, s, limit, &mut f)
        }

        let a = count(start - 1, limit, &s);
        let b = count(finish, limit, &s);
        b - a
    }
}

2999 - Count the Number of Powerful Integers

2999. Count the Number of Powerful Integers

Description

Solutions

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2999. Count the Number of Powerful Integers

Description

Solutions

All Problems

All Solutions