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[pull] master from wisdompeak:master #293

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pull merged 3 commits into AlgorithmAndLeetCode:master from wisdompeak:master
Dec 22, 2024
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Original file line number Diff line number Diff line change
@@ -0,0 +1,61 @@
class Solution {
public:
int minLength(string s, int numOps)
{
vector<int>arr;
vector<int>nums;
for (auto ch: s) nums.push_back(ch-'0');

int n = s.size();
for (int i=0; i<n;)
{
int j = i;
while (j<n && s[j]==s[i])
j++;
arr.push_back(j-i);
i = j;
}

int left = 1, right = n;
while (left < right)
{
int mid = left + (right-left)/2;
if (isOK(arr, nums, mid, numOps))
right = mid;
else
left = mid+1;
}
return left;
}

bool isOK(vector<int>arr, vector<int>&nums, int len, int numOps)
{
if (len==1)
{
int count = 0;
for (int i=0; i<nums.size(); i++)
count += (nums[i]==(i%2));
if (count <= numOps)
return true;

count = 0;
for (int i=0; i<nums.size(); i++)
count += (nums[i]!=(i%2));
if (count <= numOps)
return true;

return false;
}

int count = 0;

for (int i=0; i<arr.size(); i++)
{
int x = arr[i];
count += ceil((x-len)*1.0/(len+1));
if (count>numOps)
return false;
}
return true;
}
};
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Original file line number Diff line number Diff line change
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### 3399.Smallest-Substring-With-Identical-Characters-II

容易发现,只要操作次数越多,就越容易将最长的identical-chracter sbustring长度降下来,所以很明显适合二分搜值的框架。

于是问题转变成给定一个len,问是否能在numOps次flip操作内,使得s里不存在超过长度len的identical substring。

我们将原字符串里进行预处理,分割为一系列由相同字符组成的子串。对于任意一段长度为x的子串,我们至少需要做多少次flip呢?很明显,贪心思想就可以得到最优解,即每隔len个字符,我们就做一次flip。假设最少做t次flip,我们需要满足
```
x <= (len+1) * t + len
```
才能保证x里面不会有超过长度为len的identical substring。于是求得`t>=(x-k)/(k+1)`。不等式右边是小数时,取上界整数。

但是此题有一个坑。如果len是1的话,那么当x是偶数时,会给下一段x带来困扰。比如说s=00001111,我们在处理第一段0000时,会得到贪心的做法做两次flip使得其变成0101。我们发现这样的话下一段的1111其实需要处理的长度是5,给算法带来了极大的不便。比较简单的处理方法就是对len=1的情况特别处理,跳出之前的思维模式,只要考虑将s强制转换为01相间的字符串,计算需要做的flip即可。

那为什么len是2的时候我们就不用担心呢?距离s=000111,如果按照贪心的做法,对于第一段000我们需要做一次flip使得其变成001,似乎依然会影响到下一段的111. 但事实上,对于一段我们不需要变换成001,可以将最右端的1随意往左调动一下,变换成010。结果就是flip的次数不变的前提下,依然可以保证不会出现长度超过2的identical substring。注意,回顾一下二分搜值的框架,我们不要求一定要构造出长度为2的identical substring。

同理当len=3时,也不会出现类似影响下一段的困扰。之前的计算t的表达式依然可以适用。

1 change: 1 addition & 0 deletions Readme.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -147,6 +147,7 @@
[3048.Earliest-Second-to-Mark-Indices-I](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/3048.Earliest-Second-to-Mark-Indices-I) (M+)
[3049.Earliest-Second-to-Mark-Indices-II](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/3049.Earliest-Second-to-Mark-Indices-II) (H)
[3097.Shortest-Subarray-With-OR-at-Least-K-II](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/3097.Shortest-Subarray-With-OR-at-Least-K-II) (M)
[3399.Smallest-Substring-With-Identical-Characters-II](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/3399.Smallest-Substring-With-Identical-Characters-II) (H-)
* ``Find K-th Element``
[215.Kth-Largest-Element-in-an-Array](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/215.Kth-Largest-Element-in-an-Array) (M)
[287.Find-the-Duplicate-Number](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/287.Find-the-Duplicate-Number) (H-)
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