Merge pull request #861 from 0xff-dev/2997

Add solution and test-cases for problem 2997

Author: 6boris

leetcode/2901-3000/2997.Minimum-Number-of-Operations-to-Make-Array-XOR-Equal-to-K/README.md

@@ -1,28 +1,35 @@
 # [2997.Minimum Number of Operations to Make Array XOR Equal to K][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 a **0-indexed** integer array `nums` and a positive integer `k`.
+
+You can apply the following operation on the array **any** number of times:
+
+- Choose **any** element of the array and **flip** a bit in its **binary** representation. Flipping a bit means changing a `0` to `1` or vice versa.
+
+Return the **minimum** number of operations required to make the bitwise `XOR` of all elements of the final array equal to `k`.
+
+**Note** that you can flip leading zero bits in the binary representation of elements. For example, for the number `(101)2` you can flip the fourth bit and obtain `(1101)2`.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: nums = [2,1,3,4], k = 1
+Output: 2
+Explanation: We can do the following operations:
+- Choose element 2 which is 3 == (011)2, we flip the first bit and we obtain (010)2 == 2. nums becomes [2,1,2,4].
+- Choose element 0 which is 2 == (010)2, we flip the third bit and we obtain (110)2 = 6. nums becomes [6,1,2,4].
+The XOR of elements of the final array is (6 XOR 1 XOR 2 XOR 4) == 1 == k.
+It can be shown that we cannot make the XOR equal to k in less than 2 operations.
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Minimum Number of Operations to Make Array XOR Equal to K
-```go
 ```
-
+Input: nums = [2,0,2,0], k = 0
+Output: 0
+Explanation: The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.
+```
 
 ## 结语
 

leetcode/2901-3000/2997.Minimum-Number-of-Operations-to-Make-Array-XOR-Equal-to-K/Solution.go

@@ -1,5 +1,18 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(nums []int, k int) int {
+	mask := 1
+	ans := 0
+	for shift := 0; shift < 32; shift++ {
+		cur := mask << shift
+		kbit := k & cur
+		xor := 0
+		for _, n := range nums {
+			xor ^= (n & cur)
+		}
+		if xor != kbit {
+			ans++
+		}
+	}
+	return ans
 }

leetcode/2901-3000/2997.Minimum-Number-of-Operations-to-Make-Array-XOR-Equal-to-K/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{2, 1, 3, 4}, 1, 2},
+		{"TestCase2", []int{2, 0, 2, 0}, 0, 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.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() {
 }