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pull merged 44 commits into AlgorithmAndLeetCode:master from wisdompeak:master
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Create 2322.Minimum-Score-After-Removals-on-a-Tree.cpp
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Original file line number Diff line number Diff line change
@@ -1,41 +1,31 @@
class Solution {
public:
int maximumMinimumPath(vector<vector<int>>& A)
int maximumMinimumPath(vector<vector<int>>& grid)
{
int left = INT_MAX;
int right = INT_MIN;
int M = A.size();
int N = A[0].size();
for (int i=0; i<M; i++)
for (int j=0; j<N; j++)
{
left = min(left,A[i][j]);
right = max(right,A[i][j]);
}

int left = 0, right = 1e9;
while (left<right)
{
int mid = right-(right-left)/2;

if (check(A,mid))
if (check(grid,mid))
left = mid;
else
right = mid-1;
}
return left;
}

bool check(vector<vector<int>> A, int K)
bool check(vector<vector<int>> grid, int K)
{
if (A[0][0]<K) return false;
if (grid[0][0]<K) return false;
auto dir = vector<pair<int,int>>({{1,0},{-1,0},{0,1},{0,-1}});

int M = A.size();
int N = A[0].size();
int M = grid.size(), N = grid[0].size();

queue<pair<int,int>>q;
q.push({0,0});
A[0][0] = -1;
vector<vector<int>>visited(M, vector<int>(N));
visited[0][0] = 1;

while (q.size()>0)
{
Expand All @@ -47,19 +37,16 @@ class Solution {
{
int i = x+dir[k].first;
int j = y+dir[k].second;
if (i<0||i>=M||j<0||j>=N)
continue;
if (A[i][j]==-1)
continue;
if (A[i][j]<K) continue;
if (i<0||i>=M||j<0||j>=N) continue;
if (visited[i][j]) continue;
if (grid[i][j]<K) continue;
if (i==M-1 && j==N-1) return true;
A[i][j]=-1;
visited[i][j] = 1;

q.push({i,j});
}
}

return false;

}
};
Original file line number Diff line number Diff line change
@@ -0,0 +1,37 @@
using AI3 = array<int,3>;
class Solution {
vector<pair<int,int>>dir = {{1,0},{-1,0},{0,1},{0,-1}};
public:
int maximumMinimumPath(vector<vector<int>>& grid)
{
int M = grid.size();
int N = grid[0].size();
priority_queue<AI3>pq;

pq.push({grid[0][0], 0,0});
vector<vector<int>>visited(M, vector<int>(N,0));
vector<vector<int>>rets(M, vector<int>(N,0));

while (pq.size()>0)
{
auto [d, x, y] = pq.top();
pq.pop();
if (visited[x][y]==1) continue;
rets[x][y] = d;
visited[x][y] = 1;
if (x==M-1 && y==N-1)
return d;

for (int k=0; k<4; k++)
{
int i = x+dir[k].first;
int j = y+dir[k].second;
if (i<0||i>=M||j<0||j>=N)
continue;
pq.push({min(d, grid[i][j]), i, j});
}
}

return -1;
}
};
16 changes: 12 additions & 4 deletions Binary_Search/1102.Path-With-Maximum-Minimum-Value/Readme.md
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -1,11 +1,11 @@
### 1102.Path-With-Maximum-Minimum-Value

此题是二分法的非常精彩的应用.
#### 解法1:二分搜索

我们想,如果这个maximum score是x,那么意味着存在一条路径,里面不能有任何小于x的格子.因此,如果给定x,我们尝试用BFS的方法从左上角走到右下角.如果能抵达,说明至少存在一条成功的路,他们所有的元素都不会小于x,而且x还可能有提升的空间.相反,如果不能走到,说明从左上到右下被一些列小于x的各自给阻断了,因此我们对于maximum score的预期必须下调,至少得小于x.
我们想,如果这个maximum score是x,那么意味着存在一条路径,里面不能有任何小于x的格子。因此,如果给定x,我们尝试用BFS的方法从左上角走到右下角,约定不能经过任何小于x的格子。如果能抵达,说明至少存在一条成功的路,其沿途元素都不会小于x,故x是一个可行解,我们可以调高对maximum score的预期。相反,如果不能走到,说明从左上到右下肯定会被一系列小于x的格子给阻断了,因此我们对于maximum score的预期必须下调,至少得小于x.

所以二分的策略就非常清楚了:
```
```cpp
while (left<right)
{
int mid = right-(right-left)/2;
Expand All @@ -17,5 +17,13 @@
```
因为我们在收缩范围的时候,永远是将可行解放在闭区间[left,right]内,不可行解排除在闭区间外.所以当left和right收敛的时候,一定是一个可行解,直接返回left即可.

#### 解法2:Dijkstra
此题还可以用Dijkstra来解决,不过这是一个与常规Dijkstra问题完全对偶的场景。

[Leetcode Link](https://leetcode.com/problems/path-with-maximum-minimum-value)
Dijkstra通常用在求最短路径的问题,在非负边权的图里,经过的节点越多,往往意味着路径会更长。所以在小顶堆的PQ中越早弹出来的节点,对应的就是该节点的最短路径。

本题反其道而行之,求的是一个"最长路径"的问题。但这道题定义的"路径长度"不是元素和,而是沿途元素里面的最小值。所以经过的格子越多,"路径长度"反而会越小。所以结合大顶堆(而不是通常的小顶堆),也能够将这道题解出来。

注:此题和[1631](https://github.com/wisdompeak/LeetCode/tree/master/Union_Find/1631.Path-With-Minimum-Effort)非常相似。

[Leetcode Link](https://leetcode.com/problems/path-with-maximum-minimum-value)
Original file line number Diff line number Diff line change
@@ -1,37 +1,30 @@
using LL = long long;
class Solution {
class Solution {
public:
long long maxRunTime(int n, vector<int>& batteries)
{
LL left = 0, right = LLONG_MAX/2;
// sort(batteries.rbegin(), batteries.rend());

LL left = 0, right = LLONG_MAX/n;
while (left < right)
{
LL mid = right-(right-left)/2;
if (checkOK(mid, batteries, n))
left = mid;
if (checkOK(mid, n, batteries))
left = mid;
else
right = mid-1;
right = mid-1;
}
return left;
return left;
}

bool checkOK(LL T, vector<int>&nums, int n)
bool checkOK(LL T, LL n, vector<int>& batteries)
{
int count = 0;
LL cur = 0;
for (int i=0; i<nums.size(); i++)
LL sum = 0;
for (auto x: batteries)
{
cur+=min((LL)nums[i], T);
while (cur >= T)
{
count++;
cur-=T;
}
if (count >= n)
sum += min((LL)x, T);
if (sum >= T*n)
return true;
}
return false;
}
};

Original file line number Diff line number Diff line change
Expand Up @@ -35,37 +35,36 @@ class Solution {
numSmaller[i] += iter-(sorted+a);
}

int i=a, j=mid+1, p = 0;
while (i<=mid && j<=b)
{
if (sorted[i]<=sorted[j])
{
temp[p] = sorted[i];
i++;
}
else
{
temp[p] = sorted[j];
j++;
}
p++;
}
while (i<=mid)
{
temp[p] = sorted[i];
i++;
p++;
}
while (j<=b)
{
temp[p] = sorted[j];
j++;
p++;
}
for (int i=0; i<b-a+1; i++)
sorted[a+i] = temp[i];
inplace_merge(sorted+a, sorted+mid+1, sorted+b+1);

// int i=a, j=mid+1, p = 0;
// while (i<=mid && j<=b)
// {
// if (sorted[i]<=sorted[j])
// {
// temp[p] = sorted[i];
// i++;
// }
// else
// {
// temp[p] = sorted[j];
// j++;
// }
// p++;
// }
// while (i<=mid)
// {
// temp[p] = sorted[i];
// i++;
// p++;
// }
// while (j<=b)
// {
// temp[p] = sorted[j];
// j++;
// p++;
// }
// for (int i=0; i<b-a+1; i++)
// sorted[a+i] = temp[i];
}
};


// X X 3 X 3 Y Y Y
Original file line number Diff line number Diff line change
Expand Up @@ -30,7 +30,6 @@ class Solution {
}

// 将两段已经有序的数组段start~mid,mid+1~end合起来排序。
// 如果写归并排序的code会更快一些。这里就偷懒了,直接用sort函数。
sort(sortedNums.begin()+start,sortedNums.begin()+end+1);
inplace_merge(sortedNums.begin()+start, sortedNums.begin()+mid+1, sortedNums.begin()+end+1);
}
};
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -10,5 +10,7 @@

最后注意,本题需要三个数组,nums, sortedNums, count。原来的数据存在nums, 归并排序后的数组存在sortedNums, count[i]对应的是nums[i]的 number of smaller elements to the right.

补充:inplace_merge(iter1, iter2, iter3)可以实现[iter1,iter2)和[iter2,iter3)两段区间的归并排序(前提是两段各自有序)。

[Leetcode Link](https://leetcode.com/problems/count-of-smaller-numbers-after-self)

[Leetcode Link](https://leetcode.com/problems/count-of-smaller-numbers-after-self)
Original file line number Diff line number Diff line change
@@ -1,6 +1,6 @@
class Solution {
int result;
long temp[10001];
long temp[100005];
public:
int countRangeSum(vector<int>& nums, int lower, int upper)
{
Expand All @@ -26,34 +26,36 @@ class Solution {
result+=p2-p1;
}

int i=a, j=mid+1, p = 0;
while (i<=mid && j<=b)
{
if (nums[i]<=nums[j])
{
temp[p] = nums[i];
i++;
}
else
{
temp[p] = nums[j];
j++;
}
p++;
}
while (i<=mid)
{
temp[p] = nums[i];
i++;
p++;
}
while (j<=b)
{
temp[p] = nums[j];
j++;
p++;
}
for (int i=0; i<b-a+1; i++)
nums[a+i] = temp[i];
inplace_merge(nums.begin()+a, nums.begin()+mid+1, nums.begin()+b+1);

// int i=a, j=mid+1, p = 0;
// while (i<=mid && j<=b)
// {
// if (nums[i]<=nums[j])
// {
// temp[p] = nums[i];
// i++;
// }
// else
// {
// temp[p] = nums[j];
// j++;
// }
// p++;
// }
// while (i<=mid)
// {
// temp[p] = nums[i];
// i++;
// p++;
// }
// while (j<=b)
// {
// temp[p] = nums[j];
// j++;
// p++;
// }
// for (int i=0; i<b-a+1; i++)
// nums[a+i] = temp[i];
}
};
3 changes: 2 additions & 1 deletion Divide_Conquer/327.Count-of-Range-Sum/Readme.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -10,5 +10,6 @@

本题的另一个训练点就是对C++的STL里lower_bound的考察。如何写自定义比较函数是关键。我们需要在右序列中找到下限的位置,希望找到的位置在原序列中是大于等于sums[i]+lower的,所以自定义函数里要写a<b。另一方面,我们需要在右序列中找到上限的位置,希望找到的位置在原序列中是大于sums[i]+upper的,所以自定义函数里要写a<=b。最终pos2-pos1表示原序列里处于[sums[i]+lower,sums[i]+upper]闭区间的个数。

补充:```inplace_merge(iter1, iter2, iter3)```可以实现```[iter1,iter2)```和```[iter2,iter3)```两段区间的归并排序(前提是两段各自有序)。

[Leetcode Link](https://leetcode.com/problems/count-of-range-sum)
[Leetcode Link](https://leetcode.com/problems/count-of-range-sum)
3 changes: 2 additions & 1 deletion Divide_Conquer/493.Reverse-Pairs/Readme.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -6,5 +6,6 @@

另外,两个有序数组的归并排序操作,代码要熟练掌握。

补充:inplace_merge(iter1, iter2, iter3)可以实现[iter1,iter2)和[iter2,iter3)两段区间的归并排序(前提是两段各自有序)。

[Leetcode Link](https://leetcode.com/problems/reverse-pairs)
[Leetcode Link](https://leetcode.com/problems/reverse-pairs)
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -7,17 +7,17 @@ class Solution {
}

int helper(int st, int en, int k, vector<int>& slices)
{
vector<int>f(k+1,0); // f[i]: the maximum gain by the current round if we take i slices, and we do take the current slice.
vector<int>g(k+1,0); // g[i]: the maximum gain by the current round if we take i slices, and we do NOT take the current slice.
{
vector<int>dp0(k+1);
vector<int>dp1(k+1);

for (int i=st; i<=en; i++)
for (int j=min(k,i-st+1); j>=1; j--)
{
g[j] = max(g[j], f[j]);
f[j] = g[j-1] + slices[i];
dp0[j] = max(dp0[j], dp1[j]);
dp1[j] = dp0[j-1] + slices[i];
}

return max(f[k], g[k]);
return max(dp0[k], dp1[k]);
}
};
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