Question

2031. Count Subarrays With More Ones Than Zeros

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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 <= 105
• 0 <= 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 $nums$.

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

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;
  }
};

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;
}

Solution 2

from sortedcontainers import SortedList


class Solution:
  def subarraysWithMoreZerosThanOnes(self, nums: List[int]) -> int:
    sl = SortedList([0])
    mod = 10**9 + 7
    ans = s = 0
    for x in nums:
      s += x or -1
      ans += sl.bisect_left(s)
      ans %= mod
      sl.add(s)
    return ans