3261. Count Substrings That Satisfy K-Constraint II

Description

You are given a binary string s and an integer k.

You are also given a 2D integer array queries, where queries[i] = [li, ri].

A binary string satisfies the k-constraint if either of the following conditions holds:

• The number of 0's in the string is at most k.
• The number of 1's in the string is at most k.

Return an integer array answer, where answer[i] is the number of substrings of s[li..ri] that satisfy the k-constraint.

 

Example 1:

Input: s = "0001111", k = 2, queries = [[0,6]]

Output: [26]

Explanation:

For the query [0, 6], all substrings of s[0..6] = "0001111" satisfy the k-constraint except for the substrings s[0..5] = "000111" and s[0..6] = "0001111".

Example 2:

Input: s = "010101", k = 1, queries = [[0,5],[1,4],[2,3]]

Output: [15,9,3]

Explanation:

The substrings of s with a length greater than 3 do not satisfy the k-constraint.

 

Constraints:

• 1 <= s.length <= 105
• s[i] is either '0' or '1'.
• 1 <= k <= s.length
• 1 <= queries.length <= 105
• queries[i] == [li, ri]
• 0 <= li <= ri < s.length
• All queries are distinct.

Solutions

Solution 1: Sliding Window + Prefix Sum

We use two variables $\textit{cnt0}$ and $\textit{cnt1}$ to record the number of $0$s and $1$s in the current window, respectively. Pointers $i$ and $j$ mark the left and right boundaries of the window. We use an array $d$ to record the first position to the right of each position $i$ that does not satisfy the $k$ constraint, initially setting $d[i] = n$. Additionally, we use a prefix sum array $\textit{pre}[i]$ of length $n + 1$ to record the number of substrings that satisfy the $k$ constraint with the right boundary at position $i$.

When we move the window to the right, if the number of $0$s and $1$s in the window both exceed $k$, we update $d[i]$ to $j$, indicating that the first position to the right of $i$ that does not satisfy the $k$ constraint is $j$. Then we move $i$ one position to the right until the number of $0$s and $1$s in the window are both less than or equal to $k$. At this point, we can calculate the number of substrings that satisfy the $k$ constraint with the right boundary at $j$, which is $j - i + 1$, and update this in the prefix sum array.

Finally, for each query $[l, r]$, we first find the first position $p$ to the right of $l$ that does not satisfy the $k$ constraint, which is $p = \min(r + 1, d[l])$. All substrings in the range $[l, p - 1]$ satisfy the $k$ constraint, and the number of such substrings is $(1 + p - l) \times (p - l) / 2$. Then, we calculate the number of substrings that satisfy the $k$ constraint with the right boundary in the range $[p, r]$, which is $\textit{pre}[r + 1] - \textit{pre}[p]$. Finally, we add the two results together.

The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the string $s$ and the query array $\textit{queries}$, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def countKConstraintSubstrings(
        self, s: str, k: int, queries: List[List[int]]
    ) -> List[int]:
        cnt = [0, 0]
        i, n = 0, len(s)
        d = [n] * n
        pre = [0] * (n + 1)
        for j, x in enumerate(map(int, s)):
            cnt[x] += 1
            while cnt[0] > k and cnt[1] > k:
                d[i] = j
                cnt[int(s[i])] -= 1
                i += 1
            pre[j + 1] = pre[j] + j - i + 1
        ans = []
        for l, r in queries:
            p = min(r + 1, d[l])
            a = (1 + p - l) * (p - l) // 2
            b = pre[r + 1] - pre[p]
            ans.append(a + b)
        return ans

class Solution {
    public long[] countKConstraintSubstrings(String s, int k, int[][] queries) {
        int[] cnt = new int[2];
        int n = s.length();
        int[] d = new int[n];
        Arrays.fill(d, n);
        long[] pre = new long[n + 1];
        for (int i = 0, j = 0; j < n; ++j) {
            cnt[s.charAt(j) - '0']++;
            while (cnt[0] > k && cnt[1] > k) {
                d[i] = j;
                cnt[s.charAt(i++) - '0']--;
            }
            pre[j + 1] = pre[j] + j - i + 1;
        }
        int m = queries.length;
        long[] ans = new long[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            int p = Math.min(r + 1, d[l]);
            long a = (1L + p - l) * (p - l) / 2;
            long b = pre[r + 1] - pre[p];
            ans[i] = a + b;
        }
        return ans;
    }
}

class Solution {
public:
    vector<long long> countKConstraintSubstrings(string s, int k, vector<vector<int>>& queries) {
        int cnt[2]{};
        int n = s.size();
        vector<int> d(n, n);
        long long pre[n + 1];
        pre[0] = 0;
        for (int i = 0, j = 0; j < n; ++j) {
            cnt[s[j] - '0']++;
            while (cnt[0] > k && cnt[1] > k) {
                d[i] = j;
                cnt[s[i++] - '0']--;
            }
            pre[j + 1] = pre[j] + j - i + 1;
        }
        vector<long long> ans;
        for (const auto& q : queries) {
            int l = q[0], r = q[1];
            int p = min(r + 1, d[l]);
            long long a = (1LL + p - l) * (p - l) / 2;
            long long b = pre[r + 1] - pre[p];
            ans.push_back(a + b);
        }
        return ans;
    }
};

func countKConstraintSubstrings(s string, k int, queries [][]int) (ans []int64) {
    cnt := [2]int{}
    n := len(s)
    d := make([]int, n)
    for i := range d {
        d[i] = n
    }
    pre := make([]int, n+1)
    for i, j := 0, 0; j < n; j++ {
        cnt[s[j]-'0']++
        for cnt[0] > k && cnt[1] > k {
            d[i] = j
            cnt[s[i]-'0']--
            i++
        }
        pre[j+1] = pre[j] + j - i + 1
    }
    for _, q := range queries {
        l, r := q[0], q[1]
        p := min(r+1, d[l])
        a := (1 + p - l) * (p - l) / 2
        b := pre[r+1] - pre[p]
        ans = append(ans, int64(a+b))
    }
    return
}

function countKConstraintSubstrings(s: string, k: number, queries: number[][]): number[] {
    const cnt: [number, number] = [0, 0];
    const n = s.length;
    const d: number[] = Array(n).fill(n);
    const pre: number[] = Array(n + 1).fill(0);
    for (let i = 0, j = 0; j < n; ++j) {
        cnt[+s[j]]++;
        while (Math.min(cnt[0], cnt[1]) > k) {
            d[i] = j;
            cnt[+s[i++]]--;
        }
        pre[j + 1] = pre[j] + j - i + 1;
    }
    const ans: number[] = [];
    for (const [l, r] of queries) {
        const p = Math.min(r + 1, d[l]);
        const a = ((1 + p - l) * (p - l)) / 2;
        const b = pre[r + 1] - pre[p];
        ans.push(a + b);
    }
    return ans;
}

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