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[pull] master from wisdompeak:master #66

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Merged
pull merged 8 commits into AlgorithmAndLeetCode:master from wisdompeak:master
Oct 16, 2022
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View file Open in desktop
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class Solution {
public:
int minimizeArrayValue(vector<int>& nums)
{
int left = nums[0], right = 1e9;
while (left < right)
{
int mid = left+(right-left)/2;

long long buff = 0;
int flag = true;
for (int i=0; i<nums.size(); i++)
{
int x = nums[i];
if (x > mid)
buff -= (x-mid);
else
buff += (mid-x);
if (buff < 0)
{
flag = false;
break;
}
}

if (flag)
right = mid;
else
left = mid+1;
}

return left;

}
};
15 changes: 15 additions & 0 deletions Binary_Search/2439.Minimize-Maximum-of-Array/Readme.md
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### 2439.Minimize-Maximum-of-Array

本题的大体思路就是,如果有一个数字特别大,那么我们希望它能与之前的数字们一起"抹匀",来尽量减少最大值。由于更靠前的数字,只能与更少数量的伙伴一起"抹匀",所以最终呈现出来的最优解形态,应该是piece-wise constant的递减数列。这个数列的第一个元素的大小,就是最终答案。

那么这个数字最小是什么呢?并不太容易直接求出来。但是如果我们猜测一个,是容易判断它是否成立的。

我们首先猜测最终答案是x。明显,这个x必然不会小于nums[0],因为nums[0]没有机会向前分摊数值。如果x>nums[0],那么这意味着后面的元素有机会向nums[0]分摊一些数值,我们记做"缓冲值":`buff = x-nums[0]`.

接下来我们看nums[1]。如果`nums[1]>x`,那么不得不让它往前分摊数值,最多能够分摊多少呢?显然就是buff。于是如果`buff> nums[1]-x`,那么就OK,同时`buff-=nums[1]-x`;否则就直接返回失败。反之,如果`nums[1]<x`,同样意味着后面的元素有机会往前分摊数值,这部分分摊可以均匀承担在nums[0]与nums[1]上,所以`buff+= x-nums[1]`。

由此可见,我们需要做的就是不停的比较x与nums[i],根据孰大孰小和增加或者减少buffer。当buffer降为0以下的时候,说明无法实现前面若干个元素"抹匀"成x(或更小),返回失败。否则返回成功。

通过二分搜值的讨论,我们可以很容易求出满足条件的最小值。

注意本题一定有解(什么都不做就可以返回数组最大值),因此二分搜值的收链解就是最优解。
4 changes: 2 additions & 2 deletions Dynamic_Programming/416.Partition-Equal-Subset-Sum/Readme.md
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Expand Up @@ -7,15 +7,15 @@
这就是01背包问题的基本思想。如果dp的空间大小合理,那么我们就可以来解决之前DFS所无法处理的复杂度。基本的模板如下:
```
for (auto x: nums) // 遍历物品
for (auto s= 0 to sum/2) // 遍历容量
for (auto s= sum/2 to 0) // 遍历容量
if dp'[s-x] = true
dp[s] = true // 如果考察x之前,已经能够凑出s-x,那么加上x这个数字就一定能凑出和为x的subset。
```

此外还有另外一种dp的写法
```
for (auto x: nums) // 遍历物品
for (auto s= 0 to sum/2) // 遍历容量
for (auto s= sum/2 to 0) // 遍历容量
if dp'[s] = true
dp[s+x] = true // 如果考察x之前,已经能够凑出s,那么加上x这个数字就一定能凑出和为s+x的subset。
```
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class Solution {
public:
vector<int> productQueries(int n, vector<vector<int>>& queries)
{
vector<int>powers;
for (int i=0; i<32; i++)
{
if (n%2!=0)
powers.push_back(i);
n/=2;
if (n==0) break;
}

vector<int>presum(powers.size());
for (int i=0; i<powers.size(); i++)
{
if (i==0) presum[i] = powers[i];
else presum[i] = presum[i-1] + powers[i];
}

vector<long>twos(32*32,1);
long M = 1e9+7;
for (int i=1; i<32*32; i++)
twos[i] = twos[i-1] * 2 % M;

vector<int>rets;
for (auto& query : queries)
{
int l = query[0], r = query[1];
int diff = presum[r] - (l==0?0:presum[l-1]);
rets.push_back(twos[diff]);
}
return rets;
}
};
5 changes: 5 additions & 0 deletions Others/2438.Range-Product-Queries-of-Powers/Readme.md
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### 2438.Range-Product-Queries-of-Powers

对于区间乘积,虽然我们可以借鉴区间求和的思路,采用前缀积相除的方法。但是本题涉及到对大数取模,应该注意到`(a/b) mod M != (a mod M) / (b mod M)`,而引入逆元的话,又显得比较繁琐。所以这道题的切入点应该在别处。

因为所有相乘的元素都是2的幂,显然我们知道这个性质,`2^a * 2^b = 2^(a+b)`,所以可以将区间的乘积转化为"指数"区间的求和,再求一次幂即可。
2 changes: 2 additions & 0 deletions Readme.md
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Expand Up @@ -113,6 +113,7 @@
[2137.Pour-Water-Between-Buckets-to-Make-Water-Levels-Equal](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2137.Pour-Water-Between-Buckets-to-Make-Water-Levels-Equal) (M)
[2141.Maximum-Running-Time-of-N-Computers](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2141.Maximum-Running-Time-of-N-Computers) (M+)
[2226.Maximum-Candies-Allocated-to-K-Children](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2226.Maximum-Candies-Allocated-to-K-Children) (M)
[2439.Minimize-Maximum-of-Array](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/2439.Minimize-Maximum-of-Array) (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-)
Expand Down Expand Up @@ -1334,6 +1335,7 @@
[1906.Minimum-Absolute-Difference-Queries](https://github.com/wisdompeak/LeetCode/tree/master/Others/1906.Minimum-Absolute-Difference-Queries) (M+)
[2245.Maximum-Trailing-Zeros-in-a-Cornered-Path](https://github.com/wisdompeak/LeetCode/tree/master/Others/2245.Maximum-Trailing-Zeros-in-a-Cornered-Path) (M)
[2281.Sum-of-Total-Strength-of-Wizards](https://github.com/wisdompeak/LeetCode/tree/master/Others/2281.Sum-of-Total-Strength-of-Wizards) (H)
[2438.Range-Product-Queries-of-Powers](https://github.com/wisdompeak/LeetCode/tree/master/Others/2438.Range-Product-Queries-of-Powers) (M+)
* ``2D Presum``
1314.Matrix-Block-Sum (M)
[1292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold](https://github.com/wisdompeak/LeetCode/tree/master/Binary_Search/1292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold) (H-)
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