feat: add solutions to lc problem: No.1785

No.1785.Minimum Elements to Add to Form a Given Sum

Author: yanglbme

solution/1700-1799/1784.Check if Binary String Has at Most One Segment of Ones/README.md

@@ -42,24 +42,20 @@
 
 **方法一：0 后面不能有 1**
 
-注意到字符串 $s$ 不含前导零，说明 $s$ 以 "1" 开头，若字符串后面出现 "01"，则不满足题意。
+注意到字符串 $s$ 不含前导零，说明 $s$ 以 '1' 开头。
+
+若字符串 $s$ 存在 "01" 串，那么 $s$ 就变成形如 "1...01..." 的字符串，此时 $s$ 出现了至少两个连续的 '1' 片段，不满足题意，返回 `false`。
+
+若字符串 $s$ 不存在 "01" 串，那么 $s$ 只能是形如 "1..1000..." 的字符串，此时 $s$ 只有一个连续的 '1' 片段，满足题意，返回 `true`。
+
+因此，只需要判断字符串 $s$ 是否存在 "01" 串即可。
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
-```python
-class Solution:
-    def checkOnesSegment(self, s: str) -> bool:
-        for i, c in enumerate(s):
-            if c == '0':
-                if s[:i].count('1') and s[i + 1 :].count('1'):
-                    return False
-        return True
-```
-
 ```python
 class Solution:
     def checkOnesSegment(self, s: str) -> bool:

solution/1700-1799/1784.Check if Binary String Has at Most One Segment of Ones/README_EN.md

@@ -36,16 +36,6 @@
 
 ### **Python3**
 
-```python
-class Solution:
-    def checkOnesSegment(self, s: str) -> bool:
-        for i, c in enumerate(s):
-            if c == '0':
-                if s[:i].count('1') and s[i + 1 :].count('1'):
-                    return False
-        return True
-```
-
 ```python
 class Solution:
     def checkOnesSegment(self, s: str) -> bool:

solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/README.md

@@ -44,22 +44,77 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：贪心**
+
+先计算数组元素总和 $s$，然后计算 $s$ 与 $goal$ 的差值 $d$。
+
+那么需要添加的元素数量为 $d$ 的绝对值除以 $limit$ 向上取整，即 $\lceil \frac{|d|}{limit} \rceil$。
+
+注意，本题中数组元素的数据范围为 $[-10^6, 10^6]$，元素个数最大为 $10^5$，总和 $s$ 以及差值 $d$ 可能会超过 $32$ 位整数的表示范围，因此需要使用 $64$ 位整数。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
+        d = abs(sum(nums) - goal)
+        return (d + limit - 1) // limit
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int minElements(int[] nums, int limit, int goal) {
+        long s = 0;
+        for (int v : nums) {
+            s += v;
+        }
+        long d = Math.abs(s - goal);
+        return (int) ((d + limit - 1) / limit);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minElements(vector<int>& nums, int limit, int goal) {
+        long long s = accumulate(nums.begin(), nums.end(), 0ll);
+        long long d = abs(s - goal);
+        return (d + limit - 1) / limit;
+    }
+};
+```
 
+### **Go**
+
+```go
+func minElements(nums []int, limit int, goal int) int {
+	s := 0
+	for _, v := range nums {
+		s += v
+	}
+	d := abs(s - goal)
+	return (d + limit - 1) / limit
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
 ```
 
 ### **...**

solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/README_EN.md

@@ -43,13 +43,58 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
+        d = abs(sum(nums) - goal)
+        return (d + limit - 1) // limit
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int minElements(int[] nums, int limit, int goal) {
+        long s = 0;
+        for (int v : nums) {
+            s += v;
+        }
+        long d = Math.abs(s - goal);
+        return (int) ((d + limit - 1) / limit);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minElements(vector<int>& nums, int limit, int goal) {
+        long long s = accumulate(nums.begin(), nums.end(), 0ll);
+        long long d = abs(s - goal);
+        return (d + limit - 1) / limit;
+    }
+};
+```
 
+### **Go**
+
+```go
+func minElements(nums []int, limit int, goal int) int {
+	s := 0
+	for _, v := range nums {
+		s += v
+	}
+	d := abs(s - goal)
+	return (d + limit - 1) / limit
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
 ```
 
 ### **...**

solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/Solution.cpp

@@ -0,0 +1,8 @@
+class Solution {
+public:
+    int minElements(vector<int>& nums, int limit, int goal) {
+        long long s = accumulate(nums.begin(), nums.end(), 0ll);
+        long long d = abs(s - goal);
+        return (d + limit - 1) / limit;
+    }
+};

solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/Solution.go

@@ -0,0 +1,15 @@
+func minElements(nums []int, limit int, goal int) int {
+	s := 0
+	for _, v := range nums {
+		s += v
+	}
+	d := abs(s - goal)
+	return (d + limit - 1) / limit
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}

solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/Solution.java

@@ -0,0 +1,10 @@
+class Solution {
+    public int minElements(int[] nums, int limit, int goal) {
+        long s = 0;
+        for (int v : nums) {
+            s += v;
+        }
+        long d = Math.abs(s - goal);
+        return (int) ((d + limit - 1) / limit);
+    }
+}

solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/Solution.py

@@ -0,0 +1,4 @@
+class Solution:
+    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
+        d = abs(sum(nums) - goal)
+        return (d + limit - 1) // limit