LeetCode #3234 — MEDIUM

Count the Number of Substrings With Dominant Ones

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

You are given a binary string s.

Return the number of substrings with dominant ones.

A string has dominant ones if the number of ones in the string is greater than or equal to the square of the number of zeros in the string.

Example 1:

Input: s = "00011"

Output: 5

Explanation:

The substrings with dominant ones are shown in the table below.

i	j	s[i..j]	Number of Zeros	Number of Ones
3	3	1	0	1
4	4	1	0	1
2	3	01	1	1
3	4	11	0	2
2	4	011	1	2

Example 2:

Input: s = "101101"

Output: 16

Explanation:

The substrings with non-dominant ones are shown in the table below.

Since there are 21 substrings total and 5 of them have non-dominant ones, it follows that there are 16 substrings with dominant ones.

i	j	s[i..j]	Number of Zeros	Number of Ones
1	1	0	1	0
4	4	0	1	0
1	4	0110	2	2
0	4	10110	2	3
1	5	01101	2	3

Constraints:

1 <= s.length <= 4 * 10⁴
s consists only of characters '0' and '1'.

Step 01

Brute Force Baseline

Problem summary: You are given a binary string s. Return the number of substrings with dominant ones. A string has dominant ones if the number of ones in the string is greater than or equal to the square of the number of zeros in the string.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"00011"

Example 2

"101101"

Core Insight

What unlocks the optimal approach

Let us fix the starting index <code>l</code> of the substring and count the number of indices <code>r</code> such that <code>l <= r</code> and the substring <code>s[l..r]</code> has dominant ones.
A substring with dominant ones has at most <code>sqrt(n)</code> zeros.
We cannot iterate over every <code>r</code> and check if the <code>s[l..r]</code> has dominant ones. Instead, we iterate over the next <code>sqrt(n)</code> zeros to the left of <code>l</code> and count the number of substrings with dominant ones where the current zero is the rightmost zero of the substring.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3234: Count the Number of Substrings With Dominant Ones
class Solution {
    public int numberOfSubstrings(String s) {
        int n = s.length();
        int[] nxt = new int[n + 1];
        nxt[n] = n;
        for (int i = n - 1; i >= 0; --i) {
            nxt[i] = nxt[i + 1];
            if (s.charAt(i) == '0') {
                nxt[i] = i;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int cnt0 = s.charAt(i) == '0' ? 1 : 0;
            int j = i;
            while (j < n && 1L * cnt0 * cnt0 <= n) {
                int cnt1 = nxt[j + 1] - i - cnt0;
                if (cnt1 >= cnt0 * cnt0) {
                    ans += Math.min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
                }
                j = nxt[j + 1];
                ++cnt0;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3234: Count the Number of Substrings With Dominant Ones
func numberOfSubstrings(s string) int {
	n := len(s)
	nxt := make([]int, n+1)
	nxt[n] = n
	for i := n - 1; i >= 0; i-- {
		nxt[i] = nxt[i+1]
		if s[i] == '0' {
			nxt[i] = i
		}
	}
	ans := 0
	for i := 0; i < n; i++ {
		cnt0 := 0
		if s[i] == '0' {
			cnt0 = 1
		}
		j := i
		for j < n && int64(cnt0*cnt0) <= int64(n) {
			cnt1 := nxt[j+1] - i - cnt0
			if cnt1 >= cnt0*cnt0 {
				ans += min(nxt[j+1]-j, cnt1-cnt0*cnt0+1)
			}
			j = nxt[j+1]
			cnt0++
		}
	}
	return ans
}

# Accepted solution for LeetCode #3234: Count the Number of Substrings With Dominant Ones
class Solution:
    def numberOfSubstrings(self, s: str) -> int:
        n = len(s)
        nxt = [n] * (n + 1)
        for i in range(n - 1, -1, -1):
            nxt[i] = nxt[i + 1]
            if s[i] == "0":
                nxt[i] = i
        ans = 0
        for i in range(n):
            cnt0 = int(s[i] == "0")
            j = i
            while j < n and cnt0 * cnt0 <= n:
                cnt1 = (nxt[j + 1] - i) - cnt0
                if cnt1 >= cnt0 * cnt0:
                    ans += min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1)
                j = nxt[j + 1]
                cnt0 += 1
        return ans

// Accepted solution for LeetCode #3234: Count the Number of Substrings With Dominant Ones
impl Solution {
    pub fn number_of_substrings(s: String) -> i32 {
        let n = s.len();
        let mut nxt = vec![n; n + 1];

        for i in (0..n).rev() {
            nxt[i] = nxt[i + 1];
            if &s[i..i + 1] == "0" {
                nxt[i] = i;
            }
        }

        let mut ans = 0;
        for i in 0..n {
            let mut cnt0 = if &s[i..i + 1] == "0" { 1 } else { 0 };
            let mut j = i;
            while j < n && (cnt0 * cnt0) as i64 <= n as i64 {
                let cnt1 = nxt[j + 1] - i - cnt0;
                if cnt1 >= (cnt0 * cnt0) {
                    ans += std::cmp::min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
                }
                j = nxt[j + 1];
                cnt0 += 1;
            }
        }

        ans as i32
    }
}

// Accepted solution for LeetCode #3234: Count the Number of Substrings With Dominant Ones
function numberOfSubstrings(s: string): number {
    const n = s.length;
    const nxt: number[] = Array(n + 1).fill(0);
    nxt[n] = n;
    for (let i = n - 1; i >= 0; --i) {
        nxt[i] = nxt[i + 1];
        if (s[i] === '0') {
            nxt[i] = i;
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        let cnt0 = s[i] === '0' ? 1 : 0;
        let j = i;
        while (j < n && cnt0 * cnt0 <= n) {
            const cnt1 = nxt[j + 1] - i - cnt0;
            if (cnt1 >= cnt0 * cnt0) {
                ans += Math.min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
            }
            j = nxt[j + 1];
            ++cnt0;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × sqrtn)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.