feat: update solutions to lc problems: No.1248,1252 (#3144)

* No.1248.Count Number of Nice Subarrays
* No.1252.Cells with Odd Values in a Matrix

Author: yanglbme

solution/1200-1299/1248.Count Number of Nice Subarrays/README.md

@@ -155,13 +155,13 @@ func numberOfSubarrays(nums []int, k int) (ans int) {
 ```ts
 function numberOfSubarrays(nums: number[], k: number): number {
     const n = nums.length;
-    const cnt = new Array(n + 1).fill(0);
+    const cnt = Array(n + 1).fill(0);
     cnt[0] = 1;
     let ans = 0;
     let t = 0;
     for (const v of nums) {
         t += v & 1;
-        if (t - k >= 0) {
+        if (t >= k) {
             ans += cnt[t - k];
         }
         cnt[t] += 1;

solution/1200-1299/1248.Count Number of Nice Subarrays/README_EN.md

@@ -153,13 +153,13 @@ func numberOfSubarrays(nums []int, k int) (ans int) {
 ```ts
 function numberOfSubarrays(nums: number[], k: number): number {
     const n = nums.length;
-    const cnt = new Array(n + 1).fill(0);
+    const cnt = Array(n + 1).fill(0);
     cnt[0] = 1;
     let ans = 0;
     let t = 0;
     for (const v of nums) {
         t += v & 1;
-        if (t - k >= 0) {
+        if (t >= k) {
             ans += cnt[t - k];
         }
         cnt[t] += 1;

solution/1200-1299/1248.Count Number of Nice Subarrays/Solution.ts

@@ -1,12 +1,12 @@
 function numberOfSubarrays(nums: number[], k: number): number {
     const n = nums.length;
-    const cnt = new Array(n + 1).fill(0);
+    const cnt = Array(n + 1).fill(0);
     cnt[0] = 1;
     let ans = 0;
     let t = 0;
     for (const v of nums) {
         t += v & 1;
-        if (t - k >= 0) {
+        if (t >= k) {
             ans += cnt[t - k];
         }
         cnt[t] += 1;

solution/1200-1299/1252.Cells with Odd Values in a Matrix/README.md

@@ -80,11 +80,11 @@ tags:
 
 ### 方法一：模拟
 
-创建一个矩阵 $g$ 来存放操作的结果。对于 $indices$ 中的每一对 $(r_i, c_i)$，我们将矩阵第 $r_i$ 行的所有数加 $1$，第 $c_i$ 列的所有元素加 $1$。
+我们创建一个矩阵 $g$ 来存放操作的结果。对于 $\text{indices}$ 中的每一对 $(r_i, c_i)$，我们将矩阵第 $r_i$ 行的所有数加 $1$，第 $c_i$ 列的所有元素加 $1$。
 
 模拟结束后，遍历矩阵，统计奇数的个数。
 
-时间复杂度 $O(indices.length*(m+n)+mn)$，空间复杂度 $O(mn)$。
+时间复杂度 $O(k \times (m + n) + m \times n)$，空间复杂度 $O(m \times n)$。其中 $k$ 为 $\text{indices}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -137,12 +137,19 @@ public:
         vector<vector<int>> g(m, vector<int>(n));
         for (auto& e : indices) {
             int r = e[0], c = e[1];
-            for (int i = 0; i < m; ++i) ++g[i][c];
-            for (int j = 0; j < n; ++j) ++g[r][j];
+            for (int i = 0; i < m; ++i) {
+                ++g[i][c];
+            }
+            for (int j = 0; j < n; ++j) {
+                ++g[r][j];
+            }
         }
         int ans = 0;
-        for (auto& row : g)
-            for (int v : row) ans += v % 2;
+        for (auto& row : g) {
+            for (int v : row) {
+                ans += v % 2;
+            }
+        }
         return ans;
     }
 };
@@ -183,11 +190,11 @@ func oddCells(m int, n int, indices [][]int) int {
 
 ### 方法二：空间优化
 
-用行数组 $row$ 和列数组 $col$ 来记录每一行、每一列被增加的次数。对于 $indices$ 中的每一对 $(r_i, c_i)$，我们将 $row[r_i]$ 和 $col[c_i]$ 分别加 $1$。
+我们可以使用行数组 $\text{row}$ 和列数组 $\text{col}$ 来记录每一行、每一列被增加的次数。对于 $\text{indices}$ 中的每一对 $(r_i, c_i)$，我们将 $\text{row}[r_i]$ 和 $\text{col}[c_i]$ 分别加 $1$。
 
-操作结束后，可以算出 $(i, j)$ 位置的计数为 $row[i]+col[j]$。遍历矩阵，统计奇数的个数。
+操作结束后，可以算出 $(i, j)$ 位置的计数为 $\text{row}[i]+\text{col}[j]$。遍历矩阵，统计奇数的个数。
 
-时间复杂度 $O(indices.length+mn)$，空间复杂度 $O(m+n)$。
+时间复杂度 $O(k + m \times n)$，空间复杂度 $O(m + n)$。其中 $k$ 为 $\text{indices}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -241,8 +248,11 @@ public:
             col[c]++;
         }
         int ans = 0;
-        for (int i : row)
-            for (int j : col) ans += (i + j) % 2;
+        for (int i : row) {
+            for (int j : col) {
+                ans += (i + j) % 2;
+            }
+        }
         return ans;
     }
 };
@@ -277,11 +287,11 @@ func oddCells(m int, n int, indices [][]int) int {
 
 ### 方法三：数学优化
 
-我们注意到，只有当 $row[i]$ 和 $col[j]$ 中恰好为“一奇一偶”时，矩阵 $(i, j)$ 位置的数才会是奇数。
+我们注意到，只有当 $\text{row}[i]$ 和 $\text{col}[j]$ 中恰好为“一奇一偶”时，矩阵 $(i, j)$ 位置的数才会是奇数。
 
-我们统计 $row$ 中的奇数个数，记为 $cnt1$；$col$ 中的奇数个数，记为 $cnt2$。那么最终得到的奇数个数为 $cnt1*(n-cnt2)+cnt2*(m-cnt1)$。
+我们统计 $\text{row}$ 中的奇数个数，记为 $\text{cnt1}$；而 $\text{col}$ 中的奇数个数，记为 $\text{cnt2}$。那么最终得到的奇数个数为 $\text{cnt1} \times (n - \text{cnt2}) + \text{cnt2} \times (m - \text{cnt1})$。
 
-时间复杂度 $O(indices.length+m+n)$，空间复杂度 $O(m+n)$。
+时间复杂度 $O(k + m + n)$，空间复杂度 $O(m + n)$。其中 $k$ 为 $\text{indices}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -338,8 +348,12 @@ public:
             col[c]++;
         }
         int cnt1 = 0, cnt2 = 0;
-        for (int v : row) cnt1 += v % 2;
-        for (int v : col) cnt2 += v % 2;
+        for (int v : row) {
+            cnt1 += v % 2;
+        }
+        for (int v : col) {
+            cnt2 += v % 2;
+        }
         return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
     }
 };

solution/1200-1299/1252.Cells with Odd Values in a Matrix/README_EN.md

@@ -71,11 +71,11 @@ The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.
 
 ### Solution 1: Simulation
 
-We create a matrix $g$ to store the results of the operations. For each pair $(r_i, c_i)$ in $indices$, we add $1$ to all elements in the $r_i$th row and the $c_i$th column of the matrix.
+We create a matrix $g$ to store the result of operations. For each pair $(r_i, c_i)$ in $\text{indices}$, we add $1$ to all numbers in the $r_i$-th row of the matrix and add $1$ to all elements in the $c_i$-th column.
 
-After the simulation, we traverse the matrix and count the number of odd numbers.
+After the simulation ends, we traverse the matrix and count the number of odd numbers.
 
-The time complexity is $O(\text{indices.length} \times (m+n) + mn)$, and the space complexity is $O(mn)$.
+The time complexity is $O(k \times (m + n) + m \times n)$, and the space complexity is $O(m \times n)$. Here, $k$ is the length of $\text{indices}$.
 
 <!-- tabs:start -->
 
@@ -128,12 +128,19 @@ public:
         vector<vector<int>> g(m, vector<int>(n));
         for (auto& e : indices) {
             int r = e[0], c = e[1];
-            for (int i = 0; i < m; ++i) ++g[i][c];
-            for (int j = 0; j < n; ++j) ++g[r][j];
+            for (int i = 0; i < m; ++i) {
+                ++g[i][c];
+            }
+            for (int j = 0; j < n; ++j) {
+                ++g[r][j];
+            }
         }
         int ans = 0;
-        for (auto& row : g)
-            for (int v : row) ans += v % 2;
+        for (auto& row : g) {
+            for (int v : row) {
+                ans += v % 2;
+            }
+        }
         return ans;
     }
 };
@@ -174,11 +181,11 @@ func oddCells(m int, n int, indices [][]int) int {
 
 ### Solution 2: Space Optimization
 
-We use row array $row$ and column array $col$ to record the number of times each row and column are increased. For each pair $(r_i, c_i)$ in $indices$, we add $1$ to $row[r_i]$ and $col[c_i]$ respectively.
+We can use a row array $\text{row}$ and a column array $\text{col}$ to record the number of times each row and column is incremented. For each pair $(r_i, c_i)$ in $\text{indices}$, we add $1$ to $\text{row}[r_i]$ and $\text{col}[c_i]$ respectively.
 
-After the operation, we can calculate that the count at position $(i, j)$ is $row[i] + col[j]$. We traverse the matrix and count the number of odd numbers.
+After the operations are completed, the count at position $(i, j)$ can be calculated as $\text{row}[i] + \text{col}[j]$. We traverse the matrix and count the number of odd numbers.
 
-The time complexity is $O(\text{indices.length} + mn)$, and the space complexity is $O(m+n)$.
+The time complexity is $O(k + m \times n)$, and the space complexity is $O(m + n)$. Here, $k$ is the length of $\text{indices}$.
 
 <!-- tabs:start -->
 
@@ -232,8 +239,11 @@ public:
             col[c]++;
         }
         int ans = 0;
-        for (int i : row)
-            for (int j : col) ans += (i + j) % 2;
+        for (int i : row) {
+            for (int j : col) {
+                ans += (i + j) % 2;
+            }
+        }
         return ans;
     }
 };
@@ -268,11 +278,11 @@ func oddCells(m int, n int, indices [][]int) int {
 
 ### Solution 3: Mathematical Optimization
 
-We notice that only when exactly one of $row[i]$ and $col[j]$ is odd, the number at position $(i, j)$ in the matrix will be odd.
+We notice that a number at position $(i, j)$ in the matrix will be odd only when exactly one of $\text{row}[i]$ and $\text{col}[j]$ is odd and the other is even.
 
-We count the number of odd numbers in $row$, denoted as $cnt1$; the number of odd numbers in $col$, denoted as $cnt2$. Then the final number of odd numbers is $cnt1 \times (n-cnt2) + cnt2 \times (m-cnt1)$.
+We count the number of odd numbers in $\text{row}$, denoted as $\text{cnt1}$, and the number of odd numbers in $\text{col}$, denoted as $\text{cnt2}$. Therefore, the final count of odd numbers is $\text{cnt1} \times (n - \text{cnt2}) + \text{cnt2} \times (m - \text{cnt1})$.
 
-The time complexity is $O(\text{indices.length} + m + n)$, and the space complexity is $O(m+n)$.
+The time complexity is $O(k + m + n)$, and the space complexity is $O(m + n)$. Here, $k$ is the length of $\text{indices}$.
 
 <!-- tabs:start -->
 
@@ -329,8 +339,12 @@ public:
             col[c]++;
         }
         int cnt1 = 0, cnt2 = 0;
-        for (int v : row) cnt1 += v % 2;
-        for (int v : col) cnt2 += v % 2;
+        for (int v : row) {
+            cnt1 += v % 2;
+        }
+        for (int v : col) {
+            cnt2 += v % 2;
+        }
         return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
     }
 };

solution/1200-1299/1252.Cells with Odd Values in a Matrix/Solution.cpp

@@ -4,12 +4,19 @@ class Solution {
         vector<vector<int>> g(m, vector<int>(n));
         for (auto& e : indices) {
             int r = e[0], c = e[1];
-            for (int i = 0; i < m; ++i) ++g[i][c];
-            for (int j = 0; j < n; ++j) ++g[r][j];
+            for (int i = 0; i < m; ++i) {
+                ++g[i][c];
+            }
+            for (int j = 0; j < n; ++j) {
+                ++g[r][j];
+            }
         }
         int ans = 0;
-        for (auto& row : g)
-            for (int v : row) ans += v % 2;
+        for (auto& row : g) {
+            for (int v : row) {
+                ans += v % 2;
+            }
+        }
         return ans;
     }
 };

solution/1200-1299/1252.Cells with Odd Values in a Matrix/Solution2.cpp

@@ -9,8 +9,11 @@ class Solution {
             col[c]++;
         }
         int ans = 0;
-        for (int i : row)
-            for (int j : col) ans += (i + j) % 2;
+        for (int i : row) {
+            for (int j : col) {
+                ans += (i + j) % 2;
+            }
+        }
         return ans;
     }
 };

solution/1200-1299/1252.Cells with Odd Values in a Matrix/Solution3.cpp

@@ -9,8 +9,12 @@ class Solution {
             col[c]++;
         }
         int cnt1 = 0, cnt2 = 0;
-        for (int v : row) cnt1 += v % 2;
-        for (int v : col) cnt2 += v % 2;
+        for (int v : row) {
+            cnt1 += v % 2;
+        }
+        for (int v : col) {
+            cnt2 += v % 2;
+        }
         return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
     }
 };