Merge pull request #999 from 0xff-dev/1590

Add solution and test-cases for problem 1590

leetcode/1501-1600/1590.Make-Sum-Divisible-by-P/README.md

@@ -1,28 +1,35 @@
 # [1590.Make Sum Divisible by P][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
+Given an array of positive integers `nums`, remove the **smallest** subarray (possibly **empty**) such that the **sum** of the remaining elements is divisible by `p`. It is **not** allowed to remove the whole array.
+
+Return the length of the smallest subarray that you need to remove, or `-1` if it's impossible.
+
+A **subarray** is defined as a contiguous block of elements in the array.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: nums = [3,1,4,2], p = 6
+Output: 1
+Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Make Sum Divisible by P
-```go
 ```
+Input: nums = [6,3,5,2], p = 9
+Output: 2
+Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.
+```
+
+**Example 3:**
 
+```
+Input: nums = [1,2,3], p = 3
+Output: 0
+Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.
+```
 
 ## 结语
 

leetcode/1501-1600/1590.Make-Sum-Divisible-by-P/Solution.go

@@ -1,5 +1,35 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(nums []int, p int) int {
+	n := len(nums)
+	sum := 0
+	for _, num := range nums {
+		sum = (sum + num) % p
+	}
+
+	target := sum % p
+	if target == 0 {
+		return 0
+	}
+
+	cache := make(map[int]int)
+	cache[0] = -1
+	tmpSum := 0
+	ans := n
+
+	for i := 0; i < n; i++ {
+		tmpSum = (tmpSum + nums[i]) % p
+		needed := (tmpSum - target + p) % p
+
+		if idx, found := cache[needed]; found {
+			ans = min(ans, i-idx)
+		}
+
+		cache[tmpSum] = i
+	}
+
+	if ans == n {
+		return -1
+	}
+	return ans
 }

leetcode/1501-1600/1590.Make-Sum-Divisible-by-P/Solution_test.go

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		nums   []int
+		p      int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{3, 1, 4, 2}, 6, 1},
+		{"TestCase2", []int{6, 3, 5, 2}, 9, 2},
+		{"TestCase3", []int{1, 2, 3}, 3, 0},
 	}
 
 	//	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.p)
 			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.p)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }