Count Subarrays With More Ones Than Zeros

Count Subarrays With More Ones Than Zeros - Problem

Array Hash Table Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set Medium

You are given a binary array nums containing only the integers 0 and 1.

Return the number of subarrays in nums that have more 1's than 0's.

Since the answer may be very large, return it modulo 10⁹ + 7.

A subarray is a contiguous sequence of elements within an array.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,0,1,1,0]

› Output: 9

💡 Note: Subarrays with more 1s than 0s: [1] at indices (0,0), (2,2), (3,3); [1,0,1] at (0,2); [0,1] at (1,2); [1,1] at (2,3); [1,0,1,1] at (0,3); [0,1,1] at (1,3); [1,1,0] at (2,4). Total = 9 subarrays.

Example 2 — All Ones

$ Input: nums = [1,1,1]

› Output: 6

💡 Note: All possible subarrays have more 1s than 0s: [1] appears 3 times, [1,1] appears 2 times, [1,1,1] appears 1 time. Total = 3+2+1 = 6 subarrays.

Example 3 — Single Element

$ Input: nums = [0]

› Output: 0

💡 Note: The only subarray [0] has 0 ones and 1 zero, so 0 ≤ 1, which means it doesn't have more ones than zeros.

Constraints

1 ≤ nums.length ≤ 10⁵
nums[i] is either 0 or 1

Visualization

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Understanding the Visualization

Input

Binary array [1,0,1,1,0]

Process

Find all subarrays where count(1s) > count(0s)

Output

Total count of valid subarrays = 9

Key Takeaway

🎯 Key Insight: Transform counting problem into prefix sum problem by treating 0s as -1s

Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is to transform the problem by treating 0s as -1s, making it a prefix sum problem where we count subarrays with positive sum. The optimized O(n²) approach uses running counts to avoid recomputation. Advanced solutions can achieve better complexity using data structures like Binary Indexed Trees. Time: O(n²), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Prefix Sum with Hash Map	O(n²)	O(n)	Transform problem using prefix differences and count using hash map
Optimized - Prefix Sum Approach	O(n²)	O(1)	Use running counts to avoid recounting elements in each subarray
Brute Force - Check All Subarrays	O(n³)	O(1)	Generate all possible subarrays and count 1s vs 0s in each

Prefix Sum with Hash Map — Algorithm Steps

Convert 0s to -1s in the array
Calculate prefix sums
For each position, count previous smaller prefix sums
Use hash map or ordered data structure for counting

Visualization

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Step-by-Step Walkthrough

Transform

Convert 0s to -1s

Prefix Sum

Calculate running sum from each position

Count Positive

Count subarrays where sum > 0

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    const int MOD = 1000000007;
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int prefixSum = 0;
        for (int j = i; j < numsSize; j++) {
            prefixSum += nums[j] == 1 ? 1 : -1;
            if (prefixSum > 0) count++;
        }
    }
    
    return count % MOD;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* ptr = line + 1;
    while (*ptr != ']') {
        nums[size++] = strtol(ptr, &ptr, 10);
        if (*ptr == ',') ptr++;
    }
    
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each position, we might need to check all previous positions in worst case

⚠ Quadratic Growth

Space Complexity

O(n)

Hash map stores up to n prefix sum values

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Prefix Sum with Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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