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2031. Count Subarrays With More Ones Than Zeros
Description
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.
Example 1:
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]
Example 2:
Input: nums = [0] Output: 0 Explanation: No subarrays have more ones than zeros.
Example 3:
Input: nums = [1] Output: 1 Explanation: The subarrays of size 1 that have more ones than zeros are: [1]
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1
Solutions
Solution 1: Prefix Sum + Binary Indexed Tree
The problem requires us to count the number of subarrays where the count of $1$ is greater than the count of $0$. If we treat $0$ in the array as $-1$, then the problem becomes counting the number of subarrays where the sum of elements is greater than $0$.
To calculate the sum of elements in a subarray, we can use the prefix sum. To count the number of subarrays where the sum of elements is greater than $0$, we can use a binary indexed tree to maintain the occurrence count of each prefix sum. Initially, the occurrence count of the prefix sum $0$ is $1$.
Next, we traverse the array $nums$, use variable $s$ to record the current prefix sum, and use variable $ans$ to record the answer. For each position $i$, we update the prefix sum $s$, then query the occurrence count of the prefix sum in the range $[0, s)$ in the binary indexed tree, add it to $ans$, and then update the occurrence count of $s$ in the binary indexed tree.
Finally, return $ans$.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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class BinaryIndexedTree { private int n; private int[] c; public BinaryIndexedTree(int n) { this.n = n; c = new int[n + 1]; } public void update(int x, int v) { for (; x <= n; x += x & -x) { c[x] += v; } } public int query(int x) { int s = 0; for (; x > 0; x -= x & -x) { s += c[x]; } return s; } } class Solution { public int subarraysWithMoreZerosThanOnes(int[] nums) { int n = nums.length; int base = n + 1; BinaryIndexedTree tree = new BinaryIndexedTree(n + base); tree.update(base, 1); final int mod = (int) 1e9 + 7; int ans = 0, s = 0; for (int x : nums) { s += x == 0 ? -1 : 1; ans += tree.query(s - 1 + base); ans %= mod; tree.update(s + base, 1); } return ans; } } -
class BinaryIndexedTree { private: int n; vector<int> c; public: BinaryIndexedTree(int n) : n(n) , c(n + 1, 0) {} void update(int x, int v) { for (; x <= n; x += x & -x) { c[x] += v; } } int query(int x) { int s = 0; for (; x > 0; x -= x & -x) { s += c[x]; } return s; } }; class Solution { public: int subarraysWithMoreZerosThanOnes(vector<int>& nums) { int n = nums.size(); int base = n + 1; BinaryIndexedTree tree(n + base); tree.update(base, 1); const int mod = 1e9 + 7; int ans = 0, s = 0; for (int x : nums) { s += (x == 0) ? -1 : 1; ans += tree.query(s - 1 + base); ans %= mod; tree.update(s + base, 1); } return ans; } }; -
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 -
type BinaryIndexedTree struct { n int c []int } func newBinaryIndexedTree(n int) *BinaryIndexedTree { return &BinaryIndexedTree{n: n, c: make([]int, n+1)} } func (bit *BinaryIndexedTree) update(x, v int) { for ; x <= bit.n; x += x & -x { bit.c[x] += v } } func (bit *BinaryIndexedTree) query(x int) (s int) { for ; x > 0; x -= x & -x { s += bit.c[x] } return } func subarraysWithMoreZerosThanOnes(nums []int) (ans int) { n := len(nums) base := n + 1 tree := newBinaryIndexedTree(n + base) tree.update(base, 1) const mod = int(1e9) + 7 s := 0 for _, x := range nums { if x == 0 { s-- } else { s++ } ans += tree.query(s - 1 + base) ans %= mod tree.update(s+base, 1) } return } -
class BinaryIndexedTree { private n: number; private c: number[]; constructor(n: number) { this.n = n; this.c = Array(n + 1).fill(0); } update(x: number, v: number): void { for (; x <= this.n; x += x & -x) { this.c[x] += v; } } query(x: number): number { let s = 0; for (; x > 0; x -= x & -x) { s += this.c[x]; } return s; } } function subarraysWithMoreZerosThanOnes(nums: number[]): number { const n: number = nums.length; const base: number = n + 1; const tree: BinaryIndexedTree = new BinaryIndexedTree(n + base); tree.update(base, 1); const mod: number = 1e9 + 7; let ans: number = 0; let s: number = 0; for (const x of nums) { s += x || -1; ans += tree.query(s - 1 + base); ans %= mod; tree.update(s + base, 1); } return ans; }