Add solution and test-cases for problem 2818

leetcode/2801-2900/2818.Apply-Operations-to-Maximize-Score/README.md

@@ -0,0 +1,49 @@
+# [2818.Apply Operations to Maximize Score][title]
+
+## Description
+
+You are given an array `nums` of `n` positive integers and an integer `k`.
+
+Initially, you start with a score of `1`. You have to maximize your score by applying the following operation at most `k` times:
+
+- Choose any **non-empty** subarray `nums[l, ..., r]` that you haven't chosen previously.
+- Choose an element `x` of `nums[l, ..., r]` with the highest **prime score**. If multiple such elements exist, choose the one with the smallest index.
+- Multiply your score by `x`.
+
+Here, `nums[l, ..., r]` denotes the subarray of `nums` starting at index `l` and ending at the index `r`, both ends being inclusive.
+
+The **prime score** of an integer `x` is equal to the number of distinct prime factors of x. For example, the prime score of `300` is `3` since `300 = 2 * 2 * 3 * 5 * 5`.
+
+Return the **maximum possible score** after applying at most `k` operations.
+
+Since the answer may be large, return it modulo `10^9 + 7`.
+
+**Example 1:**
+
+```
+Input: nums = [8,3,9,3,8], k = 2
+Output: 81
+Explanation: To get a score of 81, we can apply the following operations:
+- Choose subarray nums[2, ..., 2]. nums[2] is the only element in this subarray. Hence, we multiply the score by nums[2]. The score becomes 1 * 9 = 9.
+- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 1, but nums[2] has the smaller index. Hence, we multiply the score by nums[2]. The score becomes 9 * 9 = 81.
+It can be proven that 81 is the highest score one can obtain.
+```
+
+**Example 2:**
+
+```
+Input: nums = [19,12,14,6,10,18], k = 3
+Output: 4788
+Explanation: To get a score of 4788, we can apply the following operations: 
+- Choose subarray nums[0, ..., 0]. nums[0] is the only element in this subarray. Hence, we multiply the score by nums[0]. The score becomes 1 * 19 = 19.
+- Choose subarray nums[5, ..., 5]. nums[5] is the only element in this subarray. Hence, we multiply the score by nums[5]. The score becomes 19 * 18 = 342.
+- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 2, but nums[2] has the smaller index. Hence, we multipy the score by nums[2]. The score becomes 342 * 14 = 4788.
+It can be proven that 4788 is the highest score one can obtain.
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-algorithm][me]
+
+[title]: https://leetcode.com/problems/apply-operations-to-maximize-score/
+[me]: https://github.com/kylesliu/awesome-golang-algorithm

leetcode/2801-2900/2818.Apply-Operations-to-Maximize-Score/Solution.go

@@ -1,5 +1,141 @@
 package Solution
 
-func Solution(x bool) bool {
+import (
+	"container/heap"
+)
+
+const mod2818 = 1_000_000_007
+
+type Pair2818 struct {
+	value int
+	index int
+}
+
+// PriorityQueue implements a priority queue for pairs based on value
+type PriorityQueue []Pair2818
+
+func (pq PriorityQueue) Len() int {
+	return len(pq)
+}
+
+func (pq PriorityQueue) Less(i, j int) bool {
+	return pq[i].value > pq[j].value // Max-heap based on value
+}
+
+func (pq PriorityQueue) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+}
+
+func (pq *PriorityQueue) Push(x interface{}) {
+	*pq = append(*pq, x.(Pair2818))
+}
+
+func (pq *PriorityQueue) Pop() interface{} {
+	old := *pq
+	n := len(old)
+	x := old[n-1]
+	*pq = old[0 : n-1]
 	return x
 }
+
+// Helper function to compute the number of distinct prime factors
+func distinctPrimeFactors(x int) int {
+	count := 0
+	for i := 2; i*i <= x; i++ {
+		if x%i == 0 {
+			count++
+			for x%i == 0 {
+				x /= i
+			}
+		}
+	}
+	if x > 1 {
+		count++ // If x is a prime number larger than sqrt(original x)
+	}
+	return count
+}
+
+func Solution(nums []int, k int) int {
+	n := len(nums)
+	primeScores := make([]int, n)
+
+	for i := 0; i < n; i++ {
+		primeScores[i] = distinctPrimeFactors(nums[i])
+	}
+
+	nextDominant := make([]int, n)
+	prevDominant := make([]int, n)
+	for i := 0; i < n; i++ {
+		nextDominant[i] = n
+		prevDominant[i] = -1
+	}
+
+	stack := []int{}
+
+	for i := 0; i < n; i++ {
+		for len(stack) > 0 && primeScores[stack[len(stack)-1]] < primeScores[i] {
+			topIndex := stack[len(stack)-1]
+			stack = stack[:len(stack)-1]
+			nextDominant[topIndex] = i
+		}
+
+		if len(stack) > 0 {
+			prevDominant[i] = stack[len(stack)-1]
+		}
+
+		stack = append(stack, i)
+	}
+
+	numOfSubarrays := make([]int64, n)
+	for i := 0; i < n; i++ {
+		numOfSubarrays[i] = int64(nextDominant[i]-i) * int64(i-prevDominant[i])
+	}
+
+	pq := &PriorityQueue{}
+	heap.Init(pq)
+
+	// Push each number and its index onto the priority queue
+	for i := 0; i < n; i++ {
+		heap.Push(pq, Pair2818{value: nums[i], index: i})
+	}
+
+	score := int64(1)
+
+	// Process elements while there are operations left
+	for k > 0 {
+		// Get the element with the maximum value from the queue
+		top := heap.Pop(pq).(Pair2818)
+		num := top.value
+		index := top.index
+
+		// Calculate operations to apply on the current element
+		operations := int64(k)
+		if operations > numOfSubarrays[index] {
+			operations = numOfSubarrays[index]
+		}
+
+		// Update the score
+		score = (score * power(num, operations)) % mod2818
+
+		// Reduce the remaining operations count
+		k -= int(operations)
+	}
+
+	return int(score)
+}
+
+// Helper function to compute the power of a number modulo mod2818
+func power(base int, exponent int64) int64 {
+	res := int64(1)
+
+	// Calculate the exponentiation using binary exponentiation
+	for exponent > 0 {
+		if exponent%2 == 1 {
+			res = (res * int64(base)) % mod2818
+		}
+		base = (base * base) % mod2818
+		exponent /= 2
+	}
+
+	return res
+}

leetcode/2801-2900/2818.Apply-Operations-to-Maximize-Score/Solution_test.go

@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		nums   []int
+		k      int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{8, 3, 9, 3, 8}, 2, 81},
+		{"TestCase2", []int{19, 12, 14, 6, 10, 18}, 3, 4788},
 	}
 
 	//	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)
 			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",
+					c.expect, got, c.nums, c.k)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }