feat: update solutions to lc problem: No.3375 (#4341)

Author: rain84

solution/3300-3399/3375.Minimum Operations to Make Array Values Equal to K/README.md

@@ -176,16 +176,25 @@ func minOperations(nums []int, k int) int {
 
 ```ts
 function minOperations(nums: number[], k: number): number {
-    const s = new Set<number>();
-    let mi = Infinity;
+    const s = new Set<number>([k]);
     for (const x of nums) {
-        if (x < k) {
-            return -1;
-        }
+        if (x < k) return -1;
+        s.add(x);
+    }
+    return s.size - 1;
+}
+```
+
+#### JavaScript
+
+```js
+function minOperations(nums, k) {
+    const s = new Set([k]);
+    for (const x of nums) {
+        if (x < k) return -1;
         s.add(x);
-        mi = Math.min(mi, x);
     }
-    return s.size - (mi === k ? 1 : 0);
+    return s.size - 1;
 }
 ```
 

solution/3300-3399/3375.Minimum Operations to Make Array Values Equal to K/README_EN.md

@@ -174,16 +174,25 @@ func minOperations(nums []int, k int) int {
 
 ```ts
 function minOperations(nums: number[], k: number): number {
-    const s = new Set<number>();
-    let mi = Infinity;
+    const s = new Set<number>([k]);
     for (const x of nums) {
-        if (x < k) {
-            return -1;
-        }
+        if (x < k) return -1;
+        s.add(x);
+    }
+    return s.size - 1;
+}
+```
+
+#### JavaScript
+
+```js
+function minOperations(nums, k) {
+    const s = new Set([k]);
+    for (const x of nums) {
+        if (x < k) return -1;
         s.add(x);
-        mi = Math.min(mi, x);
     }
-    return s.size - (mi === k ? 1 : 0);
+    return s.size - 1;
 }
 ```
 

solution/3300-3399/3375.Minimum Operations to Make Array Values Equal to K/Solution.js

@@ -0,0 +1,8 @@
+function minOperations(nums, k) {
+    const s = new Set([k]);
+    for (const x of nums) {
+        if (x < k) return -1;
+        s.add(x);
+    }
+    return s.size - 1;
+}

solution/3300-3399/3375.Minimum Operations to Make Array Values Equal to K/Solution.ts

@@ -1,12 +1,8 @@
 function minOperations(nums: number[], k: number): number {
-    const s = new Set<number>();
-    let mi = Infinity;
+    const s = new Set<number>([k]);
     for (const x of nums) {
-        if (x < k) {
-            return -1;
-        }
+        if (x < k) return -1;
         s.add(x);
-        mi = Math.min(mi, x);
     }
-    return s.size - (mi === k ? 1 : 0);
+    return s.size - 1;
 }