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feat: add solutions to lc problem: No.0774
No.0774.Minimize Max Distance to Gas Station
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‎solution/0700-0799/0774.Minimize Max Distance to Gas Station/README.md

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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（浮点数二分）**
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我们二分枚举相邻两个加油站间的距离，找到最小的距离，使得加油站的数量不超过 $k$。
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时间复杂度 $O(n\log M)$。其中 $n$ 为加油站的数量；而 $M$ 为答案的范围，即 $10^8$ 除以答案的精度 $10^{-6}$。
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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 minmaxGasDist(self, stations: List[int], k: int) -> float:
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def check(x):
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return sum(int((b - a) / x) for a, b in pairwise(stations)) <= k
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left, right = 0, 1e8
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while right - left > 1e-6:
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mid = (left + right) / 2
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if check(mid):
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right = mid
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else:
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left = mid
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return left
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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
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class Solution {
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public double minmaxGasDist(int[] stations, int k) {
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double left = 0, right = 1e8;
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while (right - left > 1e-6) {
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double mid = (left + right) / 2.0;
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if (check(mid, stations, k)) {
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right = mid;
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} else {
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left = mid;
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}
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}
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return left;
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}
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private boolean check(double x, int[] stations, int k) {
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int s = 0;
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for (int i = 0; i < stations.length - 1; ++i) {
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s += (int) ((stations[i + 1] - stations[i]) / x);
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}
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return s <= k;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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double minmaxGasDist(vector<int>& stations, int k) {
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double left = 0, right = 1e8;
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auto check = [&](double x) {
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int s = 0;
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for (int i = 0; i < stations.size() - 1; ++i) {
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s += (int) ((stations[i + 1] - stations[i]) / x);
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}
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return s <= k;
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};
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while (right - left > 1e-6) {
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double mid = (left + right) / 2.0;
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if (check(mid)) {
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right = mid;
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} else {
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left = mid;
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}
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}
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return left;
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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 minmaxGasDist(stations []int, k int) float64 {
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check := func(x float64) bool {
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s := 0
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for i, v := range stations[:len(stations)-1] {
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s += int(float64(stations[i+1]-v) / x)
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}
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return s <= k
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}
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var left, right float64 = 0, 1e8
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for right-left > 1e-6 {
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mid := (left + right) / 2.0
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if check(mid) {
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right = mid
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} else {
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left = mid
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}
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}
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return left
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}
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```
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### **...**

‎solution/0700-0799/0774.Minimize Max Distance to Gas Station/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minmaxGasDist(self, stations: List[int], k: int) -> float:
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def check(x):
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return sum(int((b - a) / x) for a, b in pairwise(stations)) <= k
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left, right = 0, 1e8
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while right - left > 1e-6:
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mid = (left + right) / 2
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if check(mid):
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right = mid
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else:
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left = mid
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return left
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```
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### **Java**
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```java
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class Solution {
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public double minmaxGasDist(int[] stations, int k) {
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double left = 0, right = 1e8;
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while (right - left > 1e-6) {
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double mid = (left + right) / 2.0;
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if (check(mid, stations, k)) {
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right = mid;
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} else {
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left = mid;
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}
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}
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return left;
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}
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private boolean check(double x, int[] stations, int k) {
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int s = 0;
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for (int i = 0; i < stations.length - 1; ++i) {
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s += (int) ((stations[i + 1] - stations[i]) / x);
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}
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return s <= k;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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double minmaxGasDist(vector<int>& stations, int k) {
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double left = 0, right = 1e8;
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auto check = [&](double x) {
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int s = 0;
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for (int i = 0; i < stations.size() - 1; ++i) {
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s += (int) ((stations[i + 1] - stations[i]) / x);
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}
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return s <= k;
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};
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while (right - left > 1e-6) {
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double mid = (left + right) / 2.0;
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if (check(mid)) {
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right = mid;
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} else {
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left = mid;
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}
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}
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return left;
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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 minmaxGasDist(stations []int, k int) float64 {
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check := func(x float64) bool {
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s := 0
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for i, v := range stations[:len(stations)-1] {
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s += int(float64(stations[i+1]-v) / x)
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}
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return s <= k
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}
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var left, right float64 = 0, 1e8
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for right-left > 1e-6 {
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mid := (left + right) / 2.0
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if check(mid) {
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right = mid
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} else {
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left = mid
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}
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}
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return left
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}
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```
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### **...**

‎solution/0700-0799/0774.Minimize Max Distance to Gas Station/Solution.cpp

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class Solution {
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public:
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double minmaxGasDist(vector<int>& stations, int k) {
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double left = 0, right = 1e8;
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auto check = [&](double x) {
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int s = 0;
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for (int i = 0; i < stations.size() - 1; ++i) {
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s += (int) ((stations[i + 1] - stations[i]) / x);
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}
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return s <= k;
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};
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while (right - left > 1e-6) {
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double mid = (left + right) / 2.0;
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if (check(mid)) {
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right = mid;
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} else {
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left = mid;
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}
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}
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return left;
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}
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};

‎solution/0700-0799/0774.Minimize Max Distance to Gas Station/Solution.go

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func minmaxGasDist(stations []int, k int) float64 {
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check := func(x float64) bool {
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s := 0
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for i, v := range stations[:len(stations)-1] {
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s += int(float64(stations[i+1]-v) / x)
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}
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return s <= k
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}
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var left, right float64 = 0, 1e8
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for right-left > 1e-6 {
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mid := (left + right) / 2.0
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if check(mid) {
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right = mid
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} else {
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left = mid
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}
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}
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return left
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}

‎solution/0700-0799/0774.Minimize Max Distance to Gas Station/Solution.java

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class Solution {
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public double minmaxGasDist(int[] stations, int k) {
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double left = 0, right = 1e8;
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while (right - left > 1e-6) {
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double mid = (left + right) / 2.0;
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if (check(mid, stations, k)) {
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right = mid;
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} else {
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left = mid;
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}
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}
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return left;
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}
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private boolean check(double x, int[] stations, int k) {
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int s = 0;
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for (int i = 0; i < stations.length - 1; ++i) {
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s += (int) ((stations[i + 1] - stations[i]) / x);
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}
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return s <= k;
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}
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}

‎solution/0700-0799/0774.Minimize Max Distance to Gas Station/Solution.py

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class Solution:
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def minmaxGasDist(self, stations: List[int], k: int) -> float:
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def check(x):
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return sum(int((b - a) / x) for a, b in pairwise(stations)) <= k
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left, right = 0, 1e8
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while right - left > 1e-6:
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mid = (left + right) / 2
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if check(mid):
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right = mid
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else:
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left = mid
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return left

‎solution/0700-0799/0775.Global and Local Inversions/README.md

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根据题意，我们可以发现，一个数组中的局部倒置一定是全局倒置，但是全局倒置不一定是局部倒置。也就是说，全局倒置的数量一定大于等于局部倒置的数量。
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因此，我们枚举每个数 `nums[i]`，其中 $ 2 \leq i \leq n - 1$，维护前缀数组 $nums[0,..i-2]$ 中的最大值，记为 $mx$。如果存在 $mx$ 大于 $nums[i]$，则说明全局倒置的数量大于局部倒置的数量，返回 `false` 即可。
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因此，我们枚举每个数 `nums[i]`，其中 $2 \leq i \leq n - 1$，维护前缀数组 $nums[0,..i-2]$ 中的最大值，记为 $mx$。如果存在 $mx$ 大于 $nums[i]$，则说明全局倒置的数量大于局部倒置的数量，返回 `false` 即可。
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遍历结束后，返回 `true`。
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