feat: add solutions to lc problem: No.2031

No.2031.Count Subarrays With More Ones Than Zeros

Author: yanglbme

solution/2000-2099/2031.Count Subarrays With More Ones Than Zeros/README.md

@@ -53,22 +53,234 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：树状数组**
+
+树状数组。
+
+树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：
+
+1. **单点更新** `update(x, delta)`： 把序列 x 位置的数加上一个值 delta；
+1. **前缀和查询** `query(x)`：查询序列 `[1,...x]` 区间的区间和，即位置 x 的前缀和。
+
+这两个操作的时间复杂度均为 `O(log n)`。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class BinaryIndexedTree:
+    def __init__(self, n):
+        n += int(1e5 + 1)
+        self.n = n
+        self.c = [0] * (n + 1)
+
+    @staticmethod
+    def lowbit(x):
+        x += int(1e5 + 1)
+        return x & -x
+
+    def update(self, x, delta):
+        x += int(1e5 + 1)
+        while x <= self.n:
+            self.c[x] += delta
+            x += BinaryIndexedTree.lowbit(x)
+
+    def query(self, x):
+        x += int(1e5 + 1)
+        s = 0
+        while x > 0:
+            s += self.c[x]
+            x -= BinaryIndexedTree.lowbit(x)
+        return s
 
+
+class Solution:
+    def subarraysWithMoreZerosThanOnes(self, nums: List[int]) -> int:
+        n = len(nums)
+        s = [0]
+        for v in nums:
+            s.append(s[-1] + (v or -1))
+        tree = BinaryIndexedTree(n + 1)
+        MOD = int(1e9 + 7)
+        ans = 0
+        for v in s:
+            ans = (ans + tree.query(v - 1)) % MOD
+            tree.update(v, 1)
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class BinaryIndexedTree {
+    private int n;
+    private int[] c;
+
+    public BinaryIndexedTree(int n) {
+        n += (int) 1e5 + 1;
+        this.n = n;
+        c = new int[n + 1];
+    }
+
+    public void update(int x, int delta) {
+        x += (int) 1e5 + 1;
+        while (x <= n) {
+            c[x] += delta;
+            x += lowbit(x);
+        }
+    }
+
+    public int query(int x) {
+        x += (int) 1e5 + 1;
+        int s = 0;
+        while (x > 0) {
+            s += c[x];
+            x -= lowbit(x);
+        }
+        return s;
+    }
+
+    public static int lowbit(int x) {
+        x += (int) 1e5 + 1;
+        return x & -x;
+    }
+}
+
+class Solution {
+    private static final int MOD = (int) 1e9 + 7;
+
+    public int subarraysWithMoreZerosThanOnes(int[] nums) {
+        int n = nums.length;
+        int[] s = new int[n + 1];
+        for (int i = 0; i < n; ++i) {
+            s[i + 1] = s[i] + (nums[i] == 1 ? 1 : -1);
+        }
+        BinaryIndexedTree tree = new BinaryIndexedTree(n + 1);
+        int ans = 0;
+        for (int v : s) {
+            ans = (ans + tree.query(v - 1)) % MOD;
+            tree.update(v, 1);
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class BinaryIndexedTree {
+public:
+    int n;
+    vector<int> c;
+
+    BinaryIndexedTree(int _n): n(_n + 1e5 + 1), c(_n + 1 + 1e5 + 1){}
+
+    void update(int x, int delta) {
+        x += 1e5 + 1;
+        while (x <= n)
+        {
+            c[x] += delta;
+            x += lowbit(x);
+        }
+    }
+
+    int query(int x) {
+        x += 1e5 + 1;
+        int s = 0;
+        while (x > 0)
+        {
+            s += c[x];
+            x -= lowbit(x);
+        }
+        return s;
+    }
+
+    int lowbit(int x) {
+        x += 1e5 + 1;
+        return x & -x;
+    }
+};
+
+class Solution {
+public:
+    int subarraysWithMoreZerosThanOnes(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> s(n + 1);
+        for (int i = 0; i < n; ++i) s[i + 1] = s[i] + (nums[i] == 1 ? 1 : -1);
+        BinaryIndexedTree* tree = new BinaryIndexedTree(n + 1);
+        int ans = 0;
+        const int MOD = 1e9 + 7;
+        for (int v : s)
+        {
+            ans = (ans + tree->query(v - 1)) % MOD;
+            tree->update(v, 1);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+type BinaryIndexedTree struct {
+	n int
+	c []int
+}
+
+func newBinaryIndexedTree(n int) *BinaryIndexedTree {
+	n += 1e5 + 1
+	c := make([]int, n+1)
+	return &BinaryIndexedTree{n, c}
+}
+
+func (this *BinaryIndexedTree) lowbit(x int) int {
+	x += 1e5 + 1
+	return x & -x
+}
+
+func (this *BinaryIndexedTree) update(x, delta int) {
+	x += 1e5 + 1
+	for x <= this.n {
+		this.c[x] += delta
+		x += this.lowbit(x)
+	}
+}
+
+func (this *BinaryIndexedTree) query(x int) int {
+	s := 0
+	x += 1e5 + 1
+	for x > 0 {
+		s += this.c[x]
+		x -= this.lowbit(x)
+	}
+	return s
+}
 
+func subarraysWithMoreZerosThanOnes(nums []int) int {
+	n := len(nums)
+	s := make([]int, n+1)
+	for i, v := range nums {
+		if v == 0 {
+			v = -1
+		}
+		s[i+1] = s[i] + v
+	}
+	tree := newBinaryIndexedTree(n + 1)
+	ans := 0
+	mod := int(1e9 + 7)
+	for _, v := range s {
+		ans = (ans + tree.query(v-1)) % mod
+		tree.update(v, 1)
+	}
+	return ans
+}
 ```
 
 ### **...**