feat: add solutions to lc problems: No.2716~2719

solution/2700-2799/2716.Minimize String Length/README_EN.md

@@ -50,6 +50,12 @@
 
 ## Solutions
 
+**Solution 1: Hash Table**
+
+The problem can actually be transformed into finding the number of different characters in the string. Therefore, we only need to count the number of different characters in the string.
+
+The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, the character set is lowercase English letters, so $C=26$.
+
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 ### **Python3**

solution/2700-2799/2717.Semi-Ordered Permutation/README_EN.md

@@ -59,6 +59,14 @@ It can be proved that there is no sequence of less than three operations that ma
 
 ## Solutions
 
+**Solution 1: Find the Positions of 1 and n**
+
+We can first find the indices $i$ and $j$ of $1$ and $n$, respectively. Then, based on the relative positions of $i$ and $j$, we can determine the number of swaps required.
+
+If $i < j$, the number of swaps required is $i + n - j - 1$. If $i > j$, the number of swaps required is $i + n - j - 2$.
+
+The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
+
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 ### **Python3**

solution/2700-2799/2718.Sum of Matrix After Queries/README_EN.md

@@ -46,6 +46,19 @@
 
 ## Solutions
 
+**Solution 1: Hash Table**
+
+Since the value of each row and column depends on the last modification, we can traverse all queries in reverse order and use hash tables $row$ and $col$ to record which rows and columns have been modified.
+
+For each query $(t, i, v)$:
+
+-   If $t = 0$, we check whether the $i$th row has been modified. If not, we add $v \times (n - |col|)$ to the answer, where $|col|$ represents the size of $col$, and then add $i$ to $row$.
+-   If $t = 1$, we check whether the $i$th column has been modified. If not, we add $v \times (n - |row|)$ to the answer, where $|row|$ represents the size of $row$, and then add $i$ to $col$.
+
+Finally, return the answer.
+
+The time complexity is $O(m)$, and the space complexity is $O(n)$. Here, $m$ represents the number of queries.
+
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 ### **Python3**

solution/2700-2799/2719.Count of Integers/README.md

@@ -50,7 +50,7 @@
 
 **方法一：数位 DP**
 
-题目实际上求的是区间 $[num1,..num2]$ 中数位和在 $[min\_sum,..max\_sum]$ 的数的个数。对于这种区间 $[l,..r]$ 的问题，我们可以转化为求 $[1,..r]$ 和 $[1,..l-1]$ 的答案，然后相减即可。
+题目实际上求的是区间 $[num1,..num2]$ 中，数位和在 $[min\_sum,..max\_sum]$ 的数的个数。对于这种区间 $[l,..r]$ 的问题，我们可以考虑转化为求 $[1,..r]$ 和 $[1,..l-1]$ 的答案，然后相减即可。
 
 对于 $[1,..r]$ 的答案，我们可以使用数位 DP 来求解。我们设计一个函数 $dfs(pos, s, limit)$ 表示当前处理到第 $pos$ 位，数位和为 $s$，当前数是否有上界限制 $limit$ 的方案数。其中 $pos$ 从高到低枚举。
 
@@ -74,19 +74,19 @@ class Solution:
         @cache
         def dfs(pos: int, s: int, limit: bool) -> int:
             if pos >= len(num):
-                return 1 if min_sum <= s <= max_sum else 0
+                return int(min_sum <= s <= max_sum)
             up = int(num[pos]) if limit else 9
             return (
                 sum(dfs(pos + 1, s + i, limit and i == up) for i in range(up + 1)) % mod
             )
 
         mod = 10**9 + 7
         num = num2
-        ans = dfs(0, 0, True)
+        a = dfs(0, 0, True)
         dfs.cache_clear()
         num = str(int(num1) - 1)
-        ans -= dfs(0, 0, True)
-        return ans % mod
+        b = dfs(0, 0, True)
+        return (a - b) % mod
 ```
 
 ### **Java**
@@ -108,11 +108,11 @@ class Solution {
         max = max_sum;
         num = num2;
         f = new Integer[23][220];
-        int ans = dfs(0, 0, true);
+        int a = dfs(0, 0, true);
         num = new BigInteger(num1).subtract(BigInteger.ONE).toString();
         f = new Integer[23][220];
-        ans = (ans - dfs(0, 0, true) + mod) % mod;
-        return ans;
+        int b = dfs(0, 0, true);
+        return (a - b + mod) % mod;
     }
 
     private int dfs(int pos, int s, boolean limit) {
@@ -165,8 +165,8 @@ public:
             return ans;
         };
 
-        int ans = dfs(0, 0, true);
-        for (int i = num1.size() - 1; i >= 0; --i) {
+        int a = dfs(0, 0, true);
+        for (int i = num1.size() - 1; ~i; --i) {
             if (num1[i] == '0') {
                 num1[i] = '9';
             } else {
@@ -176,8 +176,8 @@ public:
         }
         num = num1;
         memset(f, -1, sizeof(f));
-        ans -= dfs(0, 0, true);
-        return (ans + mod) % mod;
+        int b = dfs(0, 0, true);
+        return (a - b + mod) % mod;
     }
 };
 ```
@@ -218,7 +218,7 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 		}
 		return ans
 	}
-	ans := dfs(0, 0, true)
+	a := dfs(0, 0, true)
 	t := []byte(num1)
 	for i := len(t) - 1; i >= 0; i-- {
 		if t[i] != '0' {
@@ -234,8 +234,8 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 			f[i][j] = -1
 		}
 	}
-	ans -= dfs(0, 0, true)
-	return (ans%mod + mod) % mod
+	b := dfs(0, 0, true)
+	return (a - b + mod) % mod
 }
 ```
 
@@ -244,9 +244,7 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 ```ts
 function count(num1: string, num2: string, min_sum: number, max_sum: number): number {
     const mod = 1e9 + 7;
-    let f: number[][] = Array(23)
-        .fill(0)
-        .map(() => Array(220).fill(-1));
+    const f: number[][] = Array.from({ length: 23 }, () => Array(220).fill(-1));
     let num = num2;
     const dfs = (pos: number, s: number, limit: boolean): number => {
         if (pos >= num.length) {
@@ -265,13 +263,11 @@ function count(num1: string, num2: string, min_sum: number, max_sum: number): nu
         }
         return ans;
     };
-    let ans = dfs(0, 0, true);
+    const a = dfs(0, 0, true);
     num = (BigInt(num1) - 1n).toString();
-    f = Array(23)
-        .fill(0)
-        .map(() => Array(220).fill(-1));
-    ans = (ans - dfs(0, 0, true) + mod) % mod;
-    return ans;
+    f.forEach(v => v.fill(-1));
+    const b = dfs(0, 0, true);
+    return (a - b + mod) % mod;
 }
 ```
 

solution/2700-2799/2719.Count of Integers/README_EN.md

@@ -42,6 +42,20 @@
 
 ## Solutions
 
+**Solution 1: Digit DP**
+
+The problem is actually asking for the number of integers in the range $[num1,..num2]$ whose digit sum is in the range $[min\_sum,..max\_sum]$. For this kind of range $[l,..r]$ problem, we can consider transforming it into finding the answers for $[1,..r]$ and $[1,..l-1]$, and then subtracting the latter from the former.
+
+For the answer to $[1,..r]$, we can use digit DP to solve it. We design a function $dfs(pos, s, limit)$, which represents the number of schemes when we are currently processing the $pos$th digit, the digit sum is $s$, and whether the current number has an upper limit $limit$. Here, $pos$ is enumerated from high to low.
+
+For $dfs(pos, s, limit)$, we can enumerate the value of the current digit $i$, and then recursively calculate $dfs(pos+1, s+i, limit \bigcap  i==up)$, where $up$ represents the upper limit of the current digit. If $limit$ is true, then $up$ is the upper limit of the current digit, otherwise $up$ is $9$. If $pos$ is greater than or equal to the length of $num$, then we can judge whether $s$ is in the range $[min\_sum,..max\_sum]$. If it is, return $1$, otherwise return $0$.
+
+The time complexity is $O(10 \times n \times max\_sum)$, and the space complexity is $O(n \times max\_sum)$. Here, $n$ represents the length of $num$.
+
+Similar problems:
+
+-   [2801. Count Stepping Numbers in Range](/solution/2800-2899/2801.Count%20Stepping%20Numbers%20in%20Range/README_EN.md)
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -52,19 +66,19 @@ class Solution:
         @cache
         def dfs(pos: int, s: int, limit: bool) -> int:
             if pos >= len(num):
-                return 1 if min_sum <= s <= max_sum else 0
+                return int(min_sum <= s <= max_sum)
             up = int(num[pos]) if limit else 9
             return (
                 sum(dfs(pos + 1, s + i, limit and i == up) for i in range(up + 1)) % mod
             )
 
         mod = 10**9 + 7
         num = num2
-        ans = dfs(0, 0, True)
+        a = dfs(0, 0, True)
         dfs.cache_clear()
         num = str(int(num1) - 1)
-        ans -= dfs(0, 0, True)
-        return ans % mod
+        b = dfs(0, 0, True)
+        return (a - b) % mod
 ```
 
 ### **Java**
@@ -84,11 +98,11 @@ class Solution {
         max = max_sum;
         num = num2;
         f = new Integer[23][220];
-        int ans = dfs(0, 0, true);
+        int a = dfs(0, 0, true);
         num = new BigInteger(num1).subtract(BigInteger.ONE).toString();
         f = new Integer[23][220];
-        ans = (ans - dfs(0, 0, true) + mod) % mod;
-        return ans;
+        int b = dfs(0, 0, true);
+        return (a - b + mod) % mod;
     }
 
     private int dfs(int pos, int s, boolean limit) {
@@ -141,8 +155,8 @@ public:
             return ans;
         };
 
-        int ans = dfs(0, 0, true);
-        for (int i = num1.size() - 1; i >= 0; --i) {
+        int a = dfs(0, 0, true);
+        for (int i = num1.size() - 1; ~i; --i) {
             if (num1[i] == '0') {
                 num1[i] = '9';
             } else {
@@ -152,8 +166,8 @@ public:
         }
         num = num1;
         memset(f, -1, sizeof(f));
-        ans -= dfs(0, 0, true);
-        return (ans + mod) % mod;
+        int b = dfs(0, 0, true);
+        return (a - b + mod) % mod;
     }
 };
 ```
@@ -194,7 +208,7 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 		}
 		return ans
 	}
-	ans := dfs(0, 0, true)
+	a := dfs(0, 0, true)
 	t := []byte(num1)
 	for i := len(t) - 1; i >= 0; i-- {
 		if t[i] != '0' {
@@ -210,8 +224,8 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 			f[i][j] = -1
 		}
 	}
-	ans -= dfs(0, 0, true)
-	return (ans%mod + mod) % mod
+	b := dfs(0, 0, true)
+	return (a - b + mod) % mod
 }
 ```
 
@@ -220,9 +234,7 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 ```ts
 function count(num1: string, num2: string, min_sum: number, max_sum: number): number {
     const mod = 1e9 + 7;
-    let f: number[][] = Array(23)
-        .fill(0)
-        .map(() => Array(220).fill(-1));
+    const f: number[][] = Array.from({ length: 23 }, () => Array(220).fill(-1));
     let num = num2;
     const dfs = (pos: number, s: number, limit: boolean): number => {
         if (pos >= num.length) {
@@ -241,13 +253,11 @@ function count(num1: string, num2: string, min_sum: number, max_sum: number): nu
         }
         return ans;
     };
-    let ans = dfs(0, 0, true);
+    const a = dfs(0, 0, true);
     num = (BigInt(num1) - 1n).toString();
-    f = Array(23)
-        .fill(0)
-        .map(() => Array(220).fill(-1));
-    ans = (ans - dfs(0, 0, true) + mod) % mod;
-    return ans;
+    f.forEach(v => v.fill(-1));
+    const b = dfs(0, 0, true);
+    return (a - b + mod) % mod;
 }
 ```
 

solution/2700-2799/2719.Count of Integers/Solution.cpp

@@ -25,8 +25,8 @@ class Solution {
             return ans;
         };
 
-        int ans = dfs(0, 0, true);
-        for (int i = num1.size() - 1; i >= 0; --i) {
+        int a = dfs(0, 0, true);
+        for (int i = num1.size() - 1; ~i; --i) {
             if (num1[i] == '0') {
                 num1[i] = '9';
             } else {
@@ -36,7 +36,7 @@ class Solution {
         }
         num = num1;
         memset(f, -1, sizeof(f));
-        ans -= dfs(0, 0, true);
-        return (ans + mod) % mod;
+        int b = dfs(0, 0, true);
+        return (a - b + mod) % mod;
     }
 };

solution/2700-2799/2719.Count of Integers/Solution.go

@@ -31,7 +31,7 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 		}
 		return ans
 	}
-	ans := dfs(0, 0, true)
+	a := dfs(0, 0, true)
 	t := []byte(num1)
 	for i := len(t) - 1; i >= 0; i-- {
 		if t[i] != '0' {
@@ -47,6 +47,6 @@ func count(num1 string, num2 string, min_sum int, max_sum int) int {
 			f[i][j] = -1
 		}
 	}
-	ans -= dfs(0, 0, true)
-	return (ans%mod + mod) % mod
+	b := dfs(0, 0, true)
+	return (a - b + mod) % mod
 }

solution/2700-2799/2719.Count of Integers/Solution.java

@@ -12,11 +12,11 @@ public int count(String num1, String num2, int min_sum, int max_sum) {
         max = max_sum;
         num = num2;
         f = new Integer[23][220];
-        int ans = dfs(0, 0, true);
+        int a = dfs(0, 0, true);
         num = new BigInteger(num1).subtract(BigInteger.ONE).toString();
         f = new Integer[23][220];
-        ans = (ans - dfs(0, 0, true) + mod) % mod;
-        return ans;
+        int b = dfs(0, 0, true);
+        return (a - b + mod) % mod;
     }
 
     private int dfs(int pos, int s, boolean limit) {

solution/2700-2799/2719.Count of Integers/Solution.py

@@ -3,16 +3,16 @@ def count(self, num1: str, num2: str, min_sum: int, max_sum: int) -> int:
         @cache
         def dfs(pos: int, s: int, limit: bool) -> int:
             if pos >= len(num):
-                return 1 if min_sum <= s <= max_sum else 0
+                return int(min_sum <= s <= max_sum)
             up = int(num[pos]) if limit else 9
             return (
                 sum(dfs(pos + 1, s + i, limit and i == up) for i in range(up + 1)) % mod
             )
 
         mod = 10**9 + 7
         num = num2
-        ans = dfs(0, 0, True)
+        a = dfs(0, 0, True)
         dfs.cache_clear()
         num = str(int(num1) - 1)
-        ans -= dfs(0, 0, True)
-        return ans % mod
+        b = dfs(0, 0, True)
+        return (a - b) % mod