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feat: add solutions to lc problem: No.1578
No.1578.Minimum Time to Make Rope Colorful
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‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/README.md

Lines changed: 88 additions & 3 deletions
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@@ -55,27 +55,112 @@ Bob 可以移除下标 2 的蓝色气球。这将花费 3 秒。
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：双指针 + 贪心**
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我们可以用双指针指向当前连续相同颜色的气球的首尾，然后计算出当前连续相同颜色的气球的总时间 $s$，以及最大的时间 $mx$。如果当前连续相同颜色的气球个数大于 $1$，那么我们可以贪心地选择保留时间最大的气球，然后移除其它相同颜色的气球，耗时 $s - mx$，累加到答案中。接下来继续遍历，直到遍历完所有气球。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为气球的个数。
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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 minCost(self, colors: str, neededTime: List[int]) -> int:
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ans = i = 0
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n = len(colors)
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while i < n:
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j = i
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s = mx = 0
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while j < n and colors[j] == colors[i]:
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s += neededTime[j]
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if mx < neededTime[j]:
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mx = neededTime[j]
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j += 1
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if j - i > 1:
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ans += s - mx
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i = j
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return ans
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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 int minCost(String colors, int[] neededTime) {
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int ans = 0;
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int n = neededTime.length;
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for (int i = 0, j = 0; i < n; i = j) {
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j = i;
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int s = 0, mx = 0;
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while (j < n && colors.charAt(j) == colors.charAt(i)) {
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s += neededTime[j];
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mx = Math.max(mx, neededTime[j]);
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++j;
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}
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if (j - i > 1) {
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ans += s - mx;
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}
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}
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return ans;
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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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int minCost(string colors, vector<int>& neededTime) {
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int ans = 0;
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int n = colors.size();
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for (int i = 0, j = 0; i < n; i = j) {
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j = i;
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int s = 0, mx = 0;
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while (j < n && colors[j] == colors[i]) {
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s += neededTime[j];
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mx = max(mx, neededTime[j]);
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++j;
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}
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if (j - i > 1) {
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ans += s - mx;
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}
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}
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return ans;
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}
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};
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```
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### **TypeScript**
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### **Go**
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```go
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func minCost(colors string, neededTime []int) (ans int) {
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n := len(colors)
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for i, j := 0, 0; i < n; i = j {
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j = i
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s, mx := 0, 0
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for j < n && colors[j] == colors[i] {
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s += neededTime[j]
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if mx < neededTime[j] {
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mx = neededTime[j]
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}
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j++
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}
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if j-i > 1 {
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ans += s - mx
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}
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}
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return
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -54,13 +54,94 @@ There are no longer two consecutive balloons of the same color. Total time = 1 +
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### **Python3**
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```python
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class Solution:
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def minCost(self, colors: str, neededTime: List[int]) -> int:
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ans = i = 0
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n = len(colors)
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while i < n:
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j = i
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s = mx = 0
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while j < n and colors[j] == colors[i]:
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s += neededTime[j]
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if mx < neededTime[j]:
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mx = neededTime[j]
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j += 1
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if j - i > 1:
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ans += s - mx
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i = j
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return ans
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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 int minCost(String colors, int[] neededTime) {
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int ans = 0;
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int n = neededTime.length;
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for (int i = 0, j = 0; i < n; i = j) {
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j = i;
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int s = 0, mx = 0;
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while (j < n && colors.charAt(j) == colors.charAt(i)) {
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s += neededTime[j];
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mx = Math.max(mx, neededTime[j]);
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++j;
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}
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if (j - i > 1) {
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ans += s - mx;
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}
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}
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return ans;
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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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int minCost(string colors, vector<int>& neededTime) {
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int ans = 0;
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int n = colors.size();
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for (int i = 0, j = 0; i < n; i = j) {
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j = i;
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int s = 0, mx = 0;
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while (j < n && colors[j] == colors[i]) {
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s += neededTime[j];
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mx = max(mx, neededTime[j]);
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++j;
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}
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if (j - i > 1) {
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ans += s - mx;
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}
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}
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return ans;
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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 minCost(colors string, neededTime []int) (ans int) {
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n := len(colors)
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for i, j := 0, 0; i < n; i = j {
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j = i
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s, mx := 0, 0
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for j < n && colors[j] == colors[i] {
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s += neededTime[j]
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if mx < neededTime[j] {
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mx = neededTime[j]
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}
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j++
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}
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if j-i > 1 {
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ans += s - mx
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}
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}
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return
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}
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```
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### **TypeScript**

‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/Solution.cpp

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class Solution {
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public:
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int minCost(string colors, vector<int>& neededTime) {
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int ans = 0;
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int n = colors.size();
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for (int i = 0, j = 0; i < n; i = j) {
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j = i;
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int s = 0, mx = 0;
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while (j < n && colors[j] == colors[i]) {
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s += neededTime[j];
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mx = max(mx, neededTime[j]);
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++j;
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}
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if (j - i > 1) {
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ans += s - mx;
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}
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}
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return ans;
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}
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};

‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/Solution.go

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@@ -0,0 +1,18 @@
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func minCost(colors string, neededTime []int) (ans int) {
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n := len(colors)
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for i, j := 0, 0; i < n; i = j {
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j = i
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s, mx := 0, 0
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for j < n && colors[j] == colors[i] {
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s += neededTime[j]
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if mx < neededTime[j] {
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mx = neededTime[j]
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}
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j++
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}
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if j-i > 1 {
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ans += s - mx
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}
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}
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return
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}

‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/Solution.java

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@@ -0,0 +1,19 @@
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class Solution {
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public int minCost(String colors, int[] neededTime) {
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int ans = 0;
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int n = neededTime.length;
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for (int i = 0, j = 0; i < n; i = j) {
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j = i;
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int s = 0, mx = 0;
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while (j < n && colors.charAt(j) == colors.charAt(i)) {
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s += neededTime[j];
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mx = Math.max(mx, neededTime[j]);
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++j;
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}
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if (j - i > 1) {
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ans += s - mx;
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}
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}
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return ans;
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}
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}

‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/Solution.py

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@@ -0,0 +1,16 @@
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class Solution:
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def minCost(self, colors: str, neededTime: List[int]) -> int:
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ans = i = 0
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n = len(colors)
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while i < n:
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j = i
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s = mx = 0
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while j < n and colors[j] == colors[i]:
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s += neededTime[j]
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if mx < neededTime[j]:
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mx = neededTime[j]
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j += 1
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if j - i > 1:
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ans += s - mx
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i = j
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return ans

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