LeetCode 1513. Number of Substrings With Only 1s Solution in Java, C++, Python & More | Explanation + Code

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

We traverse the string s, using a variable cur to record the current count of consecutive 1s, and a variable ans to record the answer. When we traverse to character s[i], if s[i] = 0, then set cur to 0; otherwise, increment cur by 1, then add cur to ans, and take modulo 10⁹ + 7.

After the traversal is complete, return ans.

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

Similar problems:

• 413. Arithmetic Slices
• 2348. Number of Zero-Filled Subarrays

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)