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feat: add solutions to lc problem: No.2702 (doocs#1564)
No.2702.Minimum Operations to Make Numbers Non-positive
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‎solution/2700-2799/2702.Minimum Operations to Make Numbers Non-positive/README.md

Lines changed: 144 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -51,34 +51,175 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二分查找**
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我们注意到，如果一个操作次数 $t$ 能够使得所有的数都小于等于 $0$，那么对于任意 $t' > t$，操作次数 $t'$ 也能够使得所有的数都小于等于 $0$。因此我们可以使用二分查找的方法找到最小的操作次数。
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我们定义二分查找的左边界 $l=0$，右边界 $r=\max(nums)$。每一次二分查找，我们找到中间值 $mid=\lfloor\frac{l+r}{2}\rfloor$，然后判断是否存在一种操作方法使得操作次数不超过 $mid$，使得所有的数都小于等于 $0$。如果存在，那么我们就更新右边界 $r = mid$，否则我们就更新左边界 $l = mid + 1$。最终当 $l=r$ 时，我们就找到了最小的操作次数，返回 $l$ 即可。
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问题的关键在于如何判断是否存在一种操作方法使得操作次数不超过 $t$，使得所有的数都小于等于 $0$。我们可以使用贪心的方法来判断是否存在这样的操作方法。
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我们遍历数组中的每一个数 $v$，如果 $v \leq t \times y$，那么我们不需要进行任何操作。否则，我们需要的操作次数为 $\lceil\frac{v - t \times y}{x - y}\rceil$。我们将所有的操作次数相加，如果小于等于 $t$，那么就说明存在一种操作方法使得操作次数不超过 $t$，使得所有的数都小于等于 $0$。
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时间复杂度 $O(n \times \log M)$，其中 $n$ 和 $M$ 分别是数组的长度和数组中的最大值。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minOperations(self, nums: List[int], x: int, y: int) -> int:
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def check(t: int) -> bool:
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cnt = 0
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for v in nums:
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if v > t * y:
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cnt += ceil((v - t * y) / (x - y))
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return cnt <= t
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l, r = 0, max(nums)
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while l < r:
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mid = (l + r) >> 1
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if check(mid):
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r = mid
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else:
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l = mid + 1
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return l
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
69-
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class Solution {
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private int[] nums;
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private int x;
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private int y;
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public int minOperations(int[] nums, int x, int y) {
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this.nums = nums;
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this.x = x;
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this.y = y;
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int l = 0, r = 0;
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for (int v : nums) {
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r = Math.max(r, v);
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}
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while (l < r) {
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int mid = (l + r) >>> 1;
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if (check(mid)) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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private boolean check(int t) {
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long cnt = 0;
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for (int v : nums) {
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if (v > (long) t * y) {
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cnt += (v - (long) t * y + x - y - 1) / (x - y);
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}
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}
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return cnt <= t;
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}
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}
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```
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### **C++**
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```cpp
75-
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class Solution {
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public:
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int minOperations(vector<int>& nums, int x, int y) {
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int l = 0, r = *max_element(nums.begin(), nums.end());
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auto check = [&](int t) {
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long long cnt = 0;
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for (int v : nums) {
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if (v > 1LL * t * y) {
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cnt += (v - 1LL * t * y + x - y - 1) / (x - y);
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}
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}
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return cnt <= t;
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};
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while (l < r) {
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int mid = (l + r) >> 1;
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if (check(mid)) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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};
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```
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### **Go**
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```go
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func minOperations(nums []int, x int, y int) int {
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var l, r int
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for _, v := range nums {
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r = max(r, v)
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}
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check := func(t int) bool {
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cnt := 0
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for _, v := range nums {
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if v > t*y {
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cnt += (v - t*y + x - y - 1) / (x - y)
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}
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}
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return cnt <= t
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}
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for l < r {
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mid := (l + r) >> 1
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if check(mid) {
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r = mid
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} else {
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l = mid + 1
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}
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}
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return l
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function minOperations(nums: number[], x: number, y: number): number {
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let l = 0;
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let r = Math.max(...nums);
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const check = (t: number): boolean => {
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let cnt = 0;
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for (const v of nums) {
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if (v > t * y) {
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cnt += Math.ceil((v - t * y) / (x - y));
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}
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}
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return cnt <= t;
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};
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while (l < r) {
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const mid = (l + r) >> 1;
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if (check(mid)) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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```
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### **...**

‎solution/2700-2799/2702.Minimum Operations to Make Numbers Non-positive/README_EN.md

Lines changed: 132 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -50,25 +50,154 @@ Now, all the numbers in nums are non-positive. Therefore, we return 3.
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### **Python3**
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```python
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class Solution:
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def minOperations(self, nums: List[int], x: int, y: int) -> int:
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def check(t: int) -> bool:
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cnt = 0
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for v in nums:
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if v > t * y:
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cnt += ceil((v - t * y) / (x - y))
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return cnt <= t
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l, r = 0, max(nums)
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while l < r:
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mid = (l + r) >> 1
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if check(mid):
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r = mid
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else:
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l = mid + 1
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return l
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```
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### **Java**
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```java
59-
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class Solution {
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private int[] nums;
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private int x;
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private int y;
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public int minOperations(int[] nums, int x, int y) {
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this.nums = nums;
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this.x = x;
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this.y = y;
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int l = 0, r = 0;
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for (int v : nums) {
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r = Math.max(r, v);
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}
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while (l < r) {
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int mid = (l + r) >>> 1;
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if (check(mid)) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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private boolean check(int t) {
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long cnt = 0;
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for (int v : nums) {
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if (v > (long) t * y) {
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cnt += (v - (long) t * y + x - y - 1) / (x - y);
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}
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}
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return cnt <= t;
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}
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}
60109
```
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### **C++**
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```cpp
65-
114+
class Solution {
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public:
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int minOperations(vector<int>& nums, int x, int y) {
117+
int l = 0, r = *max_element(nums.begin(), nums.end());
118+
auto check = [&](int t) {
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long long cnt = 0;
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for (int v : nums) {
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if (v > 1LL * t * y) {
122+
cnt += (v - 1LL * t * y + x - y - 1) / (x - y);
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}
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}
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return cnt <= t;
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};
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while (l < r) {
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int mid = (l + r) >> 1;
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if (check(mid)) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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};
66138
```
67139
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### **Go**
69141
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```go
143+
func minOperations(nums []int, x int, y int) int {
144+
var l, r int
145+
for _, v := range nums {
146+
r = max(r, v)
147+
}
148+
check := func(t int) bool {
149+
cnt := 0
150+
for _, v := range nums {
151+
if v > t*y {
152+
cnt += (v - t*y + x - y - 1) / (x - y)
153+
}
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}
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return cnt <= t
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}
157+
for l < r {
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mid := (l + r) >> 1
159+
if check(mid) {
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r = mid
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} else {
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l = mid + 1
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}
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}
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return l
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}
167+
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func max(a, b int) int {
169+
if a > b {
170+
return a
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}
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return b
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}
174+
```
71175

176+
### **TypeScript**
177+
178+
```ts
179+
function minOperations(nums: number[], x: number, y: number): number {
180+
let l = 0;
181+
let r = Math.max(...nums);
182+
const check = (t: number): boolean => {
183+
let cnt = 0;
184+
for (const v of nums) {
185+
if (v > t * y) {
186+
cnt += Math.ceil((v - t * y) / (x - y));
187+
}
188+
}
189+
return cnt <= t;
190+
};
191+
while (l < r) {
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const mid = (l + r) >> 1;
193+
if (check(mid)) {
194+
r = mid;
195+
} else {
196+
l = mid + 1;
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}
198+
}
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return l;
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}
72201
```
73202

74203
### **...**

‎solution/2700-2799/2702.Minimum Operations to Make Numbers Non-positive/Solution.cpp

Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
class Solution {
2+
public:
3+
int minOperations(vector<int>& nums, int x, int y) {
4+
int l = 0, r = *max_element(nums.begin(), nums.end());
5+
auto check = [&](int t) {
6+
long long cnt = 0;
7+
for (int v : nums) {
8+
if (v > 1LL * t * y) {
9+
cnt += (v - 1LL * t * y + x - y - 1) / (x - y);
10+
}
11+
}
12+
return cnt <= t;
13+
};
14+
while (l < r) {
15+
int mid = (l + r) >> 1;
16+
if (check(mid)) {
17+
r = mid;
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} else {
19+
l = mid + 1;
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}
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}
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return l;
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}
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};

‎solution/2700-2799/2702.Minimum Operations to Make Numbers Non-positive/Solution.go

Lines changed: 31 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,31 @@
1+
func minOperations(nums []int, x int, y int) int {
2+
var l, r int
3+
for _, v := range nums {
4+
r = max(r, v)
5+
}
6+
check := func(t int) bool {
7+
cnt := 0
8+
for _, v := range nums {
9+
if v > t*y {
10+
cnt += (v - t*y + x - y - 1) / (x - y)
11+
}
12+
}
13+
return cnt <= t
14+
}
15+
for l < r {
16+
mid := (l + r) >> 1
17+
if check(mid) {
18+
r = mid
19+
} else {
20+
l = mid + 1
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}
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}
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return l
24+
}
25+
26+
func max(a, b int) int {
27+
if a > b {
28+
return a
29+
}
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return b
31+
}

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