Count the Number of Substrings With Dominant Ones - Problem

String Enumeration Medium

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.

In mathematical terms, for a substring with ones count of 1s and zeros count of 0s: ones ≥ zeros²

Input & Output

Example 1 — Basic Case

$ Input: s = "00011"

› Output: 5

💡 Note: Valid substrings: "1" (ones=1 ≥ zeros²=0), "11" (ones=2 ≥ zeros²=0), "1" (ones=1 ≥ zeros²=0), "01" (ones=1 ≥ zeros²=1), "011" (ones=2 ≥ zeros²=1)

Example 2 — All Ones

$ Input: s = "101101"

› Output: 15

💡 Note: Most substrings work since zeros are limited: single 1s, pairs like "10", "01", and longer valid combinations

Example 3 — Many Zeros

$ Input: s = "1000"

› Output: 2

💡 Note: Valid substrings: "1" at start (ones=1 ≥ zeros²=0), and "10" (ones=1 ≥ zeros²=1). Other substrings have too many zeros: "100" needs 1≥4 (fail), "1000" needs 1≥9 (fail)

Constraints

1 ≤ s.length ≤ 4 × 10⁴
s consists only of characters '0' and '1'

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is that when zeros become large, the condition ones ≥ zeros² becomes very restrictive, allowing for early termination optimizations. The brute force approach checks all O(n²) substrings in O(n³) time, while the optimized version uses mathematical constraints to prune impossible cases. Best approach uses early termination: Time O(n²), Space O(1).

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(1)

For each starting position, extend to all possible ending positions. Count ones and zeros in each substring and check if ones ≥ zeros².

Optimized - Group by Zero Count

⏱️ Time: O(n²) Space: O(n)

Since the condition depends on zeros², we can group substrings by their zero count. For k zeros, we need at least k² ones. This allows us to process more efficiently.

Brute Force - Check All Substrings — Algorithm Steps

Iterate through all starting positions i
For each i, iterate through all ending positions j ≥ i
Count ones and zeros in substring s[i:j+1]
Check if ones ≥ zeros² and increment counter

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible substrings from each starting position

Count 1s and 0s

For each substring, count ones and zeros

Check Condition

Verify if ones ≥ zeros² and increment counter

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int solution(char* s) {
    int n = strlen(s);
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        int ones = 0;
        int zeros = 0;
        for (int j = i; j < n; j++) {
            if (s[j] == '1') {
                ones++;
            } else {
                zeros++;
            }
            
            if (ones >= zeros * zeros) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char s[100005];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: O(n²) substrings × O(n) counting for each

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track counts and result

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Count the Number of Substrings With Dominant Ones - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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