1297. Maximum Number of Occurrences of a Substring

Description

Given a string s, return the maximum number of occurrences of any substring under the following rules:

The number of unique characters in the substring must be less than or equal to maxLetters.
The substring size must be between minSize and maxSize inclusive.

Example 1:

Input: s = "aababcaab", maxLetters = 2, minSize = 3, maxSize = 4
Output: 2
Explanation: Substring "aab" has 2 occurrences in the original string.
It satisfies the conditions, 2 unique letters and size 3 (between minSize and maxSize).

Example 2:

Input: s = "aaaa", maxLetters = 1, minSize = 3, maxSize = 3
Output: 2
Explanation: Substring "aaa" occur 2 times in the string. It can overlap.

Constraints:

1 <= s.length <= 10⁵
1 <= maxLetters <= 26
1 <= minSize <= maxSize <= min(26, s.length)
s consists of only lowercase English letters.

Solutions

class Solution {
    public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
        int ans = 0;
        Map<String, Integer> cnt = new HashMap<>();
        for (int i = 0; i < s.length() - minSize + 1; ++i) {
            String t = s.substring(i, i + minSize);
            Set<Character> ss = new HashSet<>();
            for (int j = 0; j < minSize; ++j) {
                ss.add(t.charAt(j));
            }
            if (ss.size() <= maxLetters) {
                cnt.put(t, cnt.getOrDefault(t, 0) + 1);
                ans = Math.max(ans, cnt.get(t));
            }
        }
        return ans;
    }
}

class Solution {
public:
    int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
        int ans = 0;
        unordered_map<string, int> cnt;
        for (int i = 0; i < s.size() - minSize + 1; ++i) {
            string t = s.substr(i, minSize);
            unordered_set<char> ss(t.begin(), t.end());
            if (ss.size() <= maxLetters) {
                ans = max(ans, ++cnt[t]);
            }
        }
        return ans;
    }
};

class Solution:
    def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
        ans = 0
        cnt = Counter()
        for i in range(len(s) - minSize + 1):
            t = s[i : i + minSize]
            ss = set(t)
            if len(ss) <= maxLetters:
                cnt[t] += 1
                ans = max(ans, cnt[t])
        return ans

func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
	cnt := map[string]int{}
	for i := 0; i < len(s)-minSize+1; i++ {
		t := s[i : i+minSize]
		ss := map[rune]bool{}
		for _, c := range t {
			ss[c] = true
		}
		if len(ss) <= maxLetters {
			cnt[t]++
			if ans < cnt[t] {
				ans = cnt[t]
			}
		}
	}
	return
}

function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
    const cnt = new Map<string, number>();
    let ans = 0;
    for (let i = 0; i < s.length - minSize + 1; ++i) {
        const t = s.slice(i, i + minSize);
        const ss = new Set(t.split(''));
        if (ss.size <= maxLetters) {
            cnt.set(t, (cnt.get(t) || 0) + 1);
            ans = Math.max(ans, cnt.get(t)!);
        }
    }
    return ans;
}

use std::collections::{HashMap, HashSet};

impl Solution {
    pub fn max_freq(s: String, max_letters: i32, min_size: i32, _max_size: i32) -> i32 {
        let n = s.len();
        let m = min_size as usize;
        let max_letters = max_letters as usize;
        let bytes = s.as_bytes();
        let mut cnt: HashMap<&[u8], i32> = HashMap::new();
        let mut ans = 0;

        for i in 0..=n - m {
            let t = &bytes[i..i + m];

            let mut set = HashSet::new();
            for &c in t {
                set.insert(c);
                if set.len() > max_letters {
                    break;
                }
            }
            if set.len() <= max_letters {
                let v = cnt.entry(t).or_insert(0);
                *v += 1;
                ans = ans.max(*v);
            }
        }

        ans
    }
}

class Hashing {
    private final long[] p;
    private final long[] h;
    private final long mod;

    public Hashing(String word) {
        this(word, 13331, 998244353);
    }

    public Hashing(String word, long base, int mod) {
        int n = word.length();
        p = new long[n + 1];
        h = new long[n + 1];
        p[0] = 1;
        this.mod = mod;
        for (int i = 1; i <= n; i++) {
            p[i] = p[i - 1] * base % mod;
            h[i] = (h[i - 1] * base + word.charAt(i - 1)) % mod;
        }
    }

    public long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
}

class Solution {
    public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
        int n = s.length();
        Hashing hashing = new Hashing(s);
        int[] freq = new int[256];
        int k = 0;
        int ans = 0;
        Map<Long, Integer> cnt = new HashMap<>();

        for (int i = 1; i <= n; i++) {
            if (++freq[s.charAt(i - 1)] == 1) {
                k++;
            }

            if (i >= minSize) {
                if (k <= maxLetters) {
                    long x = hashing.query(i - minSize + 1, i);
                    ans = Math.max(ans, cnt.merge(x, 1, Integer::sum));
                }
                int j = i - minSize;
                if (--freq[s.charAt(j)] == 0) {
                    k--;
                }
            }
        }

        return ans;
    }
}

class Hashing {
public:
    vector<long long> p, h;
    long long mod;

    Hashing(const string& word)
        : Hashing(word, 13331, 998244353) {}

    Hashing(const string& word, long long base, long long mod)
        : mod(mod) {
        int n = word.size();
        p.assign(n + 1, 1);
        h.assign(n + 1, 0);
        for (int i = 1; i <= n; i++) {
            p[i] = p[i - 1] * base % mod;
            h[i] = (h[i - 1] * base + word[i - 1]) % mod;
        }
    }

    long long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
};

class Solution {
public:
    int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
        int n = s.size();
        Hashing hashing(s);
        vector<int> freq(256, 0);
        int k = 0;
        int ans = 0;
        unordered_map<long long, int> cnt;

        for (int i = 1; i <= n; i++) {
            if (++freq[s[i - 1]] == 1) {
                k++;
            }

            if (i >= minSize) {
                if (k <= maxLetters) {
                    long long x = hashing.query(i - minSize + 1, i);
                    ans = max(ans, ++cnt[x]);
                }
                int j = i - minSize;
                if (--freq[s[j]] == 0) {
                    k--;
                }
            }
        }

        return ans;
    }
};

class Hashing:
    __slots__ = ["mod", "h", "p"]

    def __init__(
        self, s: Union[str, List[str]], base: int = 13331, mod: int = 998244353
    ):
        self.mod = mod
        self.h = [0] * (len(s) + 1)
        self.p = [1] * (len(s) + 1)
        for i in range(1, len(s) + 1):
            self.h[i] = (self.h[i - 1] * base + ord(s[i - 1])) % mod
            self.p[i] = (self.p[i - 1] * base) % mod

    def query(self, l: int, r: int) -> int:
        return (self.h[r] - self.h[l - 1] * self.p[r - l + 1]) % self.mod


class Solution:
    def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
        freq = Counter()
        hashing = Hashing(s)
        cnt = Counter()
        ans = k = 0
        for i, c in enumerate(s, 1):
            freq[c] += 1
            if freq[c] == 1:
                k += 1
            if i >= minSize:
                if k <= maxLetters:
                    x = hashing.query(i - minSize + 1, i)
                    cnt[x] += 1
                    ans = max(ans, cnt[x])
                j = i - minSize
                freq[s[j]] -= 1
                if freq[s[j]] == 0:
                    k -= 1
        return ans

func maxFreq(s string, maxLetters int, minSize int, maxSize int) int {
	n := len(s)
	hashing := NewHashing(s)
	freq := make([]int, 256)
	k := 0
	ans := 0
	cnt := make(map[uint64]int)

	for i := 1; i <= n; i++ {
		c := s[i-1]
		freq[c]++
		if freq[c] == 1 {
			k++
		}

		if i >= minSize {
			if k <= maxLetters {
				x := hashing.Query(i-minSize+1, i)
				cnt[x]++
				if cnt[x] > ans {
					ans = cnt[x]
				}
			}
			j := i - minSize
			c2 := s[j]
			freq[c2]--
			if freq[c2] == 0 {
				k--
			}
		}
	}

	return ans
}

type Hashing struct {
	p, h []uint64
	mod  uint64
	base uint64
}

func NewHashing(word string) *Hashing {
	return NewHashingWithBase(word, 13331, 998244353)
}

func NewHashingWithBase(word string, base uint64, mod uint64) *Hashing {
	n := len(word)
	p := make([]uint64, n+1)
	h := make([]uint64, n+1)
	p[0] = 1
	for i := 1; i <= n; i++ {
		p[i] = p[i-1] * base % mod
		h[i] = (h[i-1]*base + uint64(word[i-1])) % mod
	}
	return &Hashing{p, h, mod, base}
}

func (hs *Hashing) Query(l, r int) uint64 {
	return (hs.h[r] + hs.mod - hs.h[l-1]*hs.p[r-l+1]%hs.mod) % hs.mod
}

class Hashing {
    private p: bigint[];
    private h: bigint[];
    private mod: bigint;

    constructor(s: string, base: bigint = 13331n, mod: bigint = 998244353n) {
        const n = s.length;
        this.mod = mod;
        this.p = new Array<bigint>(n + 1).fill(1n);
        this.h = new Array<bigint>(n + 1).fill(0n);
        for (let i = 1; i <= n; i++) {
            this.p[i] = (this.p[i - 1] * base) % mod;
            this.h[i] = (this.h[i - 1] * base + BigInt(s.charCodeAt(i - 1))) % mod;
        }
    }

    query(l: number, r: number): bigint {
        return (this.h[r] - ((this.h[l - 1] * this.p[r - l + 1]) % this.mod) + this.mod) % this.mod;
    }
}

function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
    const n = s.length;
    const hashing = new Hashing(s);
    const freq = new Array<number>(256).fill(0);
    let k = 0;
    let ans = 0;
    const cnt = new Map<bigint, number>();

    for (let i = 1; i <= n; i++) {
        const c = s.charCodeAt(i - 1);
        if (++freq[c] === 1) {
            k++;
        }

        if (i >= minSize) {
            if (k <= maxLetters) {
                const x = hashing.query(i - minSize + 1, i);
                const v = (cnt.get(x) || 0) + 1;
                cnt.set(x, v);
                ans = Math.max(ans, v);
            }
            const j = i - minSize;
            const c2 = s.charCodeAt(j);
            if (--freq[c2] === 0) {
                k--;
            }
        }
    }

    return ans;
}

impl Solution {
    pub fn max_freq(s: String, max_letters: i32, min_size: i32, _max_size: i32) -> i32 {
        let n = s.len();
        let bytes = s.as_bytes();
        let hashing = Hashing::new(bytes.to_vec());
        let mut freq = [0i32; 256];
        let mut k = 0;
        let mut ans = 0;
        let mut cnt: std::collections::HashMap<u64, i32> = std::collections::HashMap::new();

        for i in 1..=n {
            let c = bytes[i - 1] as usize;
            freq[c] += 1;
            if freq[c] == 1 {
                k += 1;
            }

            if i as i32 >= min_size {
                if k <= max_letters {
                    let x = hashing.query(i - min_size as usize, i - 1);
                    let v = cnt.entry(x).and_modify(|v| *v += 1).or_insert(1);
                    ans = ans.max(*v);
                }
                let j = i - min_size as usize;
                let c2 = bytes[j] as usize;
                freq[c2] -= 1;
                if freq[c2] == 0 {
                    k -= 1;
                }
            }
        }

        ans
    }
}

struct Hashing {
    p: Vec<u64>,
    h: Vec<u64>,
    base: u64,
    modv: u64,
}

impl Hashing {
    fn new(s: Vec<u8>) -> Self {
        Self::with_params(s, 13331, 998244353)
    }

    fn with_params(s: Vec<u8>, base: u64, modv: u64) -> Self {
        let n = s.len();
        let mut p = vec![0u64; n + 1];
        let mut h = vec![0u64; n + 1];
        p[0] = 1;
        for i in 1..=n {
            p[i] = p[i - 1].wrapping_mul(base) % modv;
            h[i] = (h[i - 1].wrapping_mul(base) + s[i - 1] as u64) % modv;
        }
        Self { p, h, base, modv }
    }

    fn query(&self, l: usize, r: usize) -> u64 {
        let mut res = self.h[r + 1] + self.modv - (self.h[l] * self.p[r - l + 1] % self.modv);
        res %= self.modv;
        res
    }
}

1297 - Maximum Number of Occurrences of a Substring

1297. Maximum Number of Occurrences of a Substring

Description

Solutions

All Problems

All Solutions

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1297. Maximum Number of Occurrences of a Substring

Description

Solutions

All Problems

All Solutions