Merge pull request #4264 from SanskarSinghiit/master2

Adds Leetcode 308 solution 🧩 and addresses changes requested

Author: ajay-dhangar

dsa-solutions/lc-solutions/0300-0399/0308-range-sum-query-2d-mutable.md

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+---
+id: range-sum-query-2d-mutable
+title: Range Sum Query 2D - Mutable
+sidebar_label: 0308-Range-Sum-Query-2D-Mutable
+tags: [Segment Tree, 2D Matrix, Hard]
+description: Given a 2D matrix, handle multiple queries of the following types - updating the value of a cell and calculating the sum of elements inside a rectangle defined by its upper left and lower right corners.
+
+---
+
+
+## Problem Statement
+
+Given a 2D matrix `matrix`, handle multiple queries of the following types:
+
+1. **Update** the value of a cell in `matrix`.
+2. **Calculate** the sum of the elements inside a rectangle defined by its upper left corner `(row1, col1)` and lower right corner `(row2, col2)`.
+
+Implement the `NumMatrix` class:
+
+- **NumMatrix(int[][] matrix)**: Initializes the object with the integer matrix `matrix`.
+- **void update(int row, int col, int val)**: Updates the value of `matrix[row][col]` to be `val`.
+- **int sumRegion(int row1, int col1, int row2, int col2)**: Returns the sum of the elements of `matrix` inside the rectangle defined by its upper left corner `(row1, col1)` and lower right corner `(row2, col2)`.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input:
+["NumMatrix", "sumRegion", "update", "sumRegion"]
+[[[[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]], [2, 1, 4, 3], [3, 2, 2], [2, 1, 4, 3]]
+Output:
+[null, 8, null, 10]
+```
+
+**Example 2:**
+
+```plaintext
+Input: m = 1, n = 1, positions = [[0,0]]
+Output: [1]
+```
+
+### Constraints
+- m == matrix.length
+- n == matrix[i].length
+- 1 <= m, n <= 200
+- -1000 <= matrix[i][j] <= 1000
+- 0 <= row < m
+- 0 <= col < n
+- -1000 <= val <= 1000
+- 0 <= row1 <= row2 < m
+- 0 <= col1 <= col2 < n
+- At most 5000 calls will be made to sumRegion and update.
+
+### Follow up
+Implement a solution with Binary Indexed Tree or Segment Tree.
+
+## Solution
+
+### Approach 
+
+We use Binary Indexed Tree (BIT) to handle the update and sum queries efficiently.
+
+#### Algorithm
+- Binary Indexed Tree: Maintain a BIT to manage and update the sums of elements efficiently.
+- Update Operation: Update the value in the BIT when a cell value changes.
+- Query Operation: Calculate the sum for a given rectangle using the BIT.
+
+#### Python
+
+```py
+# segment tree
+class Node:
+    def __init__(self):
+        self.l = 0
+        self.r = 0
+        self.v = 0
+
+class SegmentTree:
+    def __init__(self, nums):
+        n = len(nums)
+        self.nums = nums
+        self.tr = [Node() for _ in range(4 * n)]
+        self.build(1, 1, n)
+
+    def build(self, u, l, r):
+        self.tr[u].l = l
+        self.tr[u].r = r
+        if l == r:
+            self.tr[u].v = self.nums[l - 1]
+            return
+        mid = (l + r) >> 1
+        self.build(u << 1, l, mid)
+        self.build(u << 1 | 1, mid + 1, r)
+        self.pushup(u)
+
+    def modify(self, u, x, v):
+        if self.tr[u].l == x and self.tr[u].r == x:
+            self.tr[u].v = v
+            return
+        mid = (self.tr[u].l + self.tr[u].r) >> 1
+        if x <= mid:
+            self.modify(u << 1, x, v)
+        else:
+            self.modify(u << 1 | 1, x, v)
+        self.pushup(u)
+
+    def query(self, u, l, r):
+        if self.tr[u].l >= l and self.tr[u].r <= r:
+            return self.tr[u].v
+        mid = (self.tr[u].l + self.tr[u].r) >> 1
+        v = 0
+        if l <= mid:
+            v += self.query(u << 1, l, r)
+        if r > mid:
+            v += self.query(u << 1 | 1, l, r)
+        return v
+
+    def pushup(self, u):
+        self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v
+
+class NumMatrix:
+
+    def __init__(self, matrix: List[List[int]]):
+        self.trees = [SegmentTree(row) for row in matrix]
+
+    def update(self, row: int, col: int, val: int) -> None:
+        tree = self.trees[row]
+        tree.modify(1, col + 1, val)
+
+    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
+        return sum(self.trees[row].query(1, col1 + 1, col2 + 1) for row in range(row1, row2 + 1))
+
+
+# Your NumMatrix object will be instantiated and called as such:
+# obj = NumMatrix(matrix)
+# obj.update(row,col,val)
+# param_2 = obj.sumRegion(row1,col1,row2,col2)
+
+############
+'''
+It uses a binary indexed tree (BIT) or Fenwick tree to efficiently update and query sums of submatrices.
+The NumMatrix class constructor initializes the BIT and matrix data structure. 
+The update method updates the matrix and BIT with the difference in values. 
+The sumRegion method computes the sum of a submatrix using prefix sums computed with the BIT. 
+The sum method computes a prefix sum in the BIT.
+
+
+"Fenwick tree" vs "Segment tree"
+https://stackoverflow.com/questions/64190332/fenwick-tree-vs-segment-tree
+
+'''
+
+class NumMatrix:
+    def __init__(self, matrix: List[List[int]]):
+        if not matrix or not matrix[0]:
+            self.m, self.n = 0, 0
+            return
+
+        self.m, self.n = len(matrix), len(matrix[0])
+        self.bit = [[0] * (self.n + 1) for _ in range(self.m + 1)]
+        self.matrix = [[0] * self.n for _ in range(self.m)]
+        
+        for i in range(self.m):
+            for j in range(self.n):
+                self.update(i, j, matrix[i][j])
+
+    def update(self, row: int, col: int, val: int) -> None:
+        diff = val - self.matrix[row][col]
+        self.matrix[row][col] = val
+        i = row + 1
+        while i <= self.m:
+            j = col + 1
+            while j <= self.n:
+                self.bit[i][j] += diff
+                j += j & -j
+            i += i & -i
+
+    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
+        return self.sum(row2 + 1, col2 + 1) - self.sum(row2 + 1, col1) - self.sum(row1, col2 + 1) + self.sum(row1, col1)
+
+    def sum(self, row: int, col: int) -> int:
+        res = 0
+        i = row
+        while i > 0:
+            j = col
+            while j > 0:
+                res += self.bit[i][j]
+                j -= j & -j
+            i -= i & -i
+        return res
+
+```
+
+#### Java
+
+```java
+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 delta) {
+        while (x <= n) {
+            c[x] += delta;
+            x += lowbit(x);
+        }
+    }
+
+    public int query(int x) {
+        int s = 0;
+        while (x > 0) {
+            s += c[x];
+            x -= lowbit(x);
+        }
+        return s;
+    }
+
+    public static int lowbit(int x) {
+        return x & -x;
+    }
+}
+
+class NumMatrix {
+    private BinaryIndexedTree[] trees;
+
+    public NumMatrix(int[][] matrix) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+        trees = new BinaryIndexedTree[m];
+        for (int i = 0; i < m; ++i) {
+            BinaryIndexedTree tree = new BinaryIndexedTree(n);
+            for (int j = 0; j < n; ++j) {
+                tree.update(j + 1, matrix[i][j]);
+            }
+            trees[i] = tree;
+        }
+    }
+
+    public void update(int row, int col, int val) {
+        BinaryIndexedTree tree = trees[row];
+        int prev = tree.query(col + 1) - tree.query(col);
+        tree.update(col + 1, val - prev);
+    }
+
+    public int sumRegion(int row1, int col1, int row2, int col2) {
+        int s = 0;
+        for (int i = row1; i <= row2; ++i) {
+            BinaryIndexedTree tree = trees[i];
+            s += tree.query(col2 + 1) - tree.query(col1);
+        }
+        return s;
+    }
+}
+
+```
+
+#### C++
+
+```cpp
+class BinaryIndexedTree {
+public:
+    int n;
+    vector<int> c;
+
+    BinaryIndexedTree(int _n)
+        : n(_n)
+        , c(_n + 1) {}
+
+    void update(int x, int delta) {
+        while (x <= n) {
+            c[x] += delta;
+            x += lowbit(x);
+        }
+    }
+
+    int query(int x) {
+        int s = 0;
+        while (x > 0) {
+            s += c[x];
+            x -= lowbit(x);
+        }
+        return s;
+    }
+
+    int lowbit(int x) {
+        return x & -x;
+    }
+};
+
+class NumMatrix {
+public:
+    vector<BinaryIndexedTree*> trees;
+
+    NumMatrix(vector<vector<int>>& matrix) {
+        int m = matrix.size();
+        int n = matrix[0].size();
+        trees.resize(m);
+        for (int i = 0; i < m; ++i) {
+            BinaryIndexedTree* tree = new BinaryIndexedTree(n);
+            for (int j = 0; j < n; ++j) {
+                tree->update(j + 1, matrix[i][j]);
+            }
+            trees[i] = tree;
+        }
+    }
+
+    void update(int row, int col, int val) {
+        BinaryIndexedTree* tree = trees[row];
+        int prev = tree->query(col + 1) - tree->query(col);
+        tree->update(col + 1, val - prev);
+    }
+
+    int sumRegion(int row1, int col1, int row2, int col2) {
+        int s = 0;
+        for (int i = row1; i <= row2; ++i) {
+            BinaryIndexedTree* tree = trees[i];
+            s += tree->query(col2 + 1) - tree->query(col1);
+        }
+        return s;
+    }
+};
+
+```
+
+### Complexity Analysis
+
+- **Time Complexity**: $O(\log m \cdot \log n)$ for both update and sumRegion operations, where $m$ and $n$ are the dimensions of the matrix.
+- **Space Complexity**: $O(m \cdot n)$ for storing the BITs and matrix.