Add solution and test-cases for problem 3346

Author: 0xff-dev

leetcode/3301-3400/3346.Maximum-Frequency-of-an-Element-After-Performing-Operations-I/README.md

@@ -1,28 +1,43 @@
 # [3346.Maximum Frequency of an Element After Performing Operations I][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+You are given an integer array `nums` and two integers `k` and `numOperations`.
+
+You must perform an **operation** `numOperations` times on `nums`, where in each operation you:
+
+- Select an index `i` that was **not** selected in any previous operations.
+- Add an integer in the range `[-k, k]` to `nums[i]`.
+
+Return the **maximum** possible frequency of any element in `nums` after performing the **operations**.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
-```
+Input: nums = [1,4,5], k = 1, numOperations = 2
+
+Output: 2
 
-## 题意
-> ...
+Explanation:
 
-## 题解
+We can achieve a maximum frequency of two by:
 
-### 思路1
-> ...
-Maximum Frequency of an Element After Performing Operations I
-```go
+Adding 0 to nums[1]. nums becomes [1, 4, 5].
+Adding -1 to nums[2]. nums becomes [1, 4, 4].
 ```
 
+**Example 2:**
+
+```
+Input: nums = [5,11,20,20], k = 5, numOperations = 1
+
+Output: 2
+
+Explanation:
+
+We can achieve a maximum frequency of two by:
+
+Adding 0 to nums[1].
+```
 
 ## 结语
 

leetcode/3301-3400/3346.Maximum-Frequency-of-an-Element-After-Performing-Operations-I/Solution.go

@@ -1,5 +1,40 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(nums []int, k int, numOperations int) int {
+	const N = 100000 + 2
+
+	freq := make([]int, N)
+	sweep := make([]int, N)
+
+	mm := N
+	MM := 0
+
+	for _, x := range nums {
+		if x >= N {
+			panic("Value in nums exceeds the array limit N")
+		}
+		freq[x]++
+
+		x0 := max(1, x-k)
+
+		xN := min(x+k+1, N-1)
+
+		sweep[x0]++
+		sweep[xN]--
+
+		mm = min(mm, x0)
+		MM = max(MM, xN)
+	}
+
+	ans := 0
+	cnt := 0
+
+	for x := mm; x <= MM; x++ {
+		cnt += sweep[x]
+
+		max_addable := min(cnt-freq[x], numOperations)
+		ans = max(ans, freq[x]+max_addable)
+	}
+
+	return ans
 }

leetcode/3301-3400/3346.Maximum-Frequency-of-an-Element-After-Performing-Operations-I/Solution_test.go

@@ -9,31 +9,31 @@ import (
 func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
-		name   string
-		inputs bool
-		expect bool
+		name             string
+		nums             []int
+		k, numOperations int
+		expect           int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{1, 4, 5}, 1, 2, 2},
+		{"TestCase2", []int{5, 11, 20, 20}, 5, 1, 2},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.nums, c.k, c.numOperations)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v %v",
+					c.expect, got, c.nums, c.k, c.numOperations)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }