Leetcode #2031: Count Subarrays With More Ones Than Zeros
In this guide, we solve Leetcode #2031 Count Subarrays With More Ones Than Zeros in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Binary Indexed Tree, Segment Tree, Array, Hash Table, Binary Search, Divide and Conquer, Ordered Set, Merge Sort
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [0,1,1,0,1]
Output: 9
Explanation:
The subarrays of size 1 that have more ones than zeros are: [1], [1], [1]
The subarrays of size 2 that have more ones than zeros are: [1,1]
The subarrays of size 3 that have more ones than zeros are: [0,1,1], [1,1,0], [1,0,1]
The subarrays of size 4 that have more ones than zeros are: [1,1,0,1]
The subarrays of size 5 that have more ones than zeros are: [0,1,1,0,1]
Python Solution
class BinaryIndexedTree:
__slots__ = ["n", "c"]
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] += v
x += x & -x
def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= x & -x
return s
class Solution:
def subarraysWithMoreZerosThanOnes(self, nums: List[int]) -> int:
n = len(nums)
base = n + 1
tree = BinaryIndexedTree(n + base)
tree.update(base, 1)
mod = 10**9 + 7
ans = s = 0
for x in nums:
s += x or -1
ans += tree.query(s - 1 + base)
ans %= mod
tree.update(s + base, 1)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.