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1513. Number of Substrings With Only 1s

SourceWeekly Contest 197 Q2DifficultyMediumRating1351

Description

Given a binary string s, return the number of substrings with all characters 1's. Since the answer may be too large, return it modulo 109 + 7.

 

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.

Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.

Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.

Solutions

Solution 1: Traversal and Counting

Thinking

Count substrings that contain only ones. \(n\le 10^5\), so we cannot list all \(O(n^2)\) substrings. Runs of ones separated by zeros are independent.

A run of length \(k\) contributes \(k(k+1)/2\) substrings, which is the same as maintaining the current run length \(cur\): each extra \(1\) increments \(cur\) and adds \(cur\) to the answer; a \(0\) resets it. No explicit segmentation is required.

We traverse the string \(s\), using a variable \(\textit{cur}\) to record the current count of consecutive 1s, and a variable \(\textit{ans}\) to record the answer. When we traverse to character \(s[i]\), if \(s[i] = 0\), then set \(\textit{cur}\) to 0; otherwise, increment \(\textit{cur}\) by 1, then add \(\textit{cur}\) to \(\textit{ans}\), and take modulo \(10^9 + 7\).

After the traversal is complete, return \(\textit{ans}\).

The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).

Similar problems:

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class Solution:
    def numSub(self, s: str) -> int:
        mod = 10**9 + 7
        ans = cur = 0
        for c in s:
            if c == "0":
                cur = 0
            else:
                cur += 1
                ans = (ans + cur) % mod
        return ans

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class Solution {
    public int numSub(String s) {
        final int mod = 1_000_000_007;
        int ans = 0, cur = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) == '0') {
                cur = 0;
            } else {
                cur++;
                ans = (ans + cur) % mod;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int numSub(string s) {
        const int mod = 1e9 + 7;
        int ans = 0, cur = 0;
        for (char c : s) {
            if (c == '0') {
                cur = 0;
            } else {
                cur++;
                ans = (ans + cur) % mod;
            }
        }
        return ans;
    }
};

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func numSub(s string) (ans int) {
    const mod int = 1e9 + 7
    cur := 0
    for _, c := range s {
        if c == '0' {
            cur = 0
        } else {
            cur++
            ans = (ans + cur) % mod
        }
    }
    return
}

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function numSub(s: string): number {
    const mod = 1_000_000_007;
    let [ans, cur] = [0, 0];
    for (const c of s) {
        if (c === '0') {
            cur = 0;
        } else {
            cur++;
            ans = (ans + cur) % mod;
        }
    }
    return ans;
}

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impl Solution {
    pub fn num_sub(s: String) -> i32 {
        const MOD: i32 = 1_000_000_007;
        let mut ans: i32 = 0;
        let mut cur: i32 = 0;
        for c in s.chars() {
            if c == '0' {
                cur = 0;
            } else {
                cur += 1;
                ans = (ans + cur) % MOD;
            }
        }
        ans
    }
}

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