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Merged
pull merged 13 commits into AlgorithmAndLeetCode:master from wisdompeak:master
Oct 31, 2022
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3fbfafd
Update 2448.Minimum-Cost-to-Make-Array-Equal_v1.cpp
wisdompeak Oct 31, 2022
389bf20
Create 907.Sum-of-Subarray-Minimums_v2.cpp
wisdompeak Oct 31, 2022
2b247e3
Update Readme.md
wisdompeak Oct 31, 2022
d24a8b1
Create 2453.Destroy-Sequential-Targets.cpp
wisdompeak Oct 31, 2022
fd03047
Update Readme.md
wisdompeak Oct 31, 2022
72e4e89
Create Readme.md
wisdompeak Oct 31, 2022
2c2ef32
Create 2454.Next-Greater-Element-IV.cpp
wisdompeak Oct 31, 2022
3115ac5
Update Readme.md
wisdompeak Oct 31, 2022
4286e56
Create Readme.md
wisdompeak Oct 31, 2022
93ef9c0
Create 2457.Minimum-Addition-to-Make-Integer-Beautiful.cpp
wisdompeak Oct 31, 2022
f912ebf
Update Readme.md
wisdompeak Oct 31, 2022
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Create Readme.md
wisdompeak Oct 31, 2022
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Update 2457.Minimum-Addition-to-Make-Integer-Beautiful.cpp
wisdompeak Oct 31, 2022
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Original file line number Diff line number Diff line change
@@ -0,0 +1,42 @@
class Solution {
public:
int DigitSum(long long x)
{
int sum = 0;
while (x>0)
{
sum += x%10;
x/=10;
}
return sum;
}

long long makeIntegerBeautiful(long long n, int target)
{
string ret;
int carry = 0;
while (n > 0)
{
if (DigitSum(n) <= target) break;

int cur = n%10;
int d;
if (cur != 0)
{
ret.push_back('0' + (10-cur));
carry = 1;
}
else
{
ret.push_back('0');
carry = 0;
}

n = n/10 + carry;
}

if(ret.empty()) return 0;
reverse(ret.begin(), ret.end());
return stoll(ret);
}
};
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Original file line number Diff line number Diff line change
@@ -0,0 +1,5 @@
### 2457.Minimum-Addition-to-Make-Integer-Beautiful

很明显,想要以最小的代价来降低digit sum,必然是从低位往高位,逐个加上一个"互补"的数字,使得将该位"清零"。即原数的某位上是2的话,你必然补上8,使得digit sum能够降低2.

这里特别需要注意的是进位。例如原数是232,你补上一个8之后,你下一个考虑的十位数其实是4而不是3.
Original file line number Diff line number Diff line change
Expand Up @@ -17,7 +17,7 @@ class Solution {
for (int i=0; i<arr.size(); i++)
{
curCost += arr[i].second;
if (curCost >= totalCost/2)
if (curCost >= totalCost*1.0/2)
{
k = i;
break;
Expand Down
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@@ -0,0 +1,32 @@
using LL = long long;
class Solution {
public:
int destroyTargets(vector<int>& nums, int space)
{
int len = 0;
int ret = 0;

sort(nums.rbegin(), nums.rend());

unordered_map<int,int>dp;

for (int i=0; i<nums.size(); i++)
{
int r = nums[i]%space;
dp[r] = dp[r] + 1;

if (dp[r] > len)
{
ret = nums[i];
len = dp[r];
}
else if (dp[r] == len)
{
ret = nums[i];
}

}

return ret;
}
};
5 changes: 5 additions & 0 deletions Others/2453.Destroy-Sequential-Targets/Readme.md
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### 2453.Destroy-Sequential-Targets

很明显,能够构成序列的位置必然是间隔为space的等差数列。不同的等差数列之间仅仅区别于offset,这个offset就是关于space的余数。例如,space如果是3,那么就有三种等差数列{0,3,6,9...},{1,4,7,10...},{2,5,8,11...}。

我们将所有的位置逆序排列,对于任意nums[i],令`r = nums[i] % space`,那么说明此位置属于offset为r的序列上,就有`dp[r] += 1`. 最终我们返回最长的dp[r]所对应的最后一个元素。
3 changes: 3 additions & 0 deletions Readme.md
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Expand Up @@ -359,6 +359,7 @@
[1950.Maximum-of-Minimum-Values-in-All-Subarrays](https://github.com/wisdompeak/LeetCode/tree/master/Stack/1950.Maximum-of-Minimum-Values-in-All-Subarrays) (H-)
[1966.Binary-Searchable-Numbers-in-an-Unsorted-Array](https://github.com/wisdompeak/LeetCode/tree/master/Stack/1966.Binary-Searchable-Numbers-in-an-Unsorted-Array) (M+)
[2434.Using-a-Robot-to-Print-the-Lexicographically-Smallest-String](https://github.com/wisdompeak/LeetCode/tree/master/Stack/2434.Using-a-Robot-to-Print-the-Lexicographically-Smallest-String) (H-)
[2454.Next-Greater-Element-IV](https://github.com/wisdompeak/LeetCode/tree/master/Stack/2454.Next-Greater-Element-IV) (H-)
* ``monotonic stack: other usages``
[962.Maximum-Width-Ramp](https://github.com/wisdompeak/LeetCode/tree/master/Stack/962.Maximum-Width-Ramp) (H)
[1130.Minimum-Cost-Tree-From-Leaf-Values](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/1130.Minimum-Cost-Tree-From-Leaf-Values) (H)
Expand Down Expand Up @@ -1148,6 +1149,7 @@
[2366.Minimum-Replacements-to-Sort-the-Array](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/2366.Minimum-Replacements-to-Sort-the-Array) (H-)
[2371.Minimize-Maximum-Value-in-a-Grid](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/2371.Minimize-Maximum-Value-in-a-Grid) (M+)
[2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar) (M+)
[2457.Minimum-Addition-to-Make-Integer-Beautiful](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/2457.Minimum-Addition-to-Make-Integer-Beautiful) (M)
* ``DI Sequence``
[942.DI-String-Match](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/942.DI-String-Match) (M)
[484.Find-Permutation](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/484.Find-Permutation) (M)
Expand Down Expand Up @@ -1287,6 +1289,7 @@
[2337.Move-Pieces-to-Obtain-a-String](https://github.com/wisdompeak/LeetCode/tree/master/Others/2337.Move-Pieces-to-Obtain-a-String) (aka. 777.Swap-Adjacent-in-LR-String) (M+)
[2359.Find-Closest-Node-to-Given-Two-Nodes](https://github.com/wisdompeak/LeetCode/tree/master/Others/2359.Find-Closest-Node-to-Given-Two-Nodes) (M)
[2380.Time-Needed-to-Rearrange-a-Binary-String](https://github.com/wisdompeak/LeetCode/tree/master/Others/2380.Time-Needed-to-Rearrange-a-Binary-String) (H)
[2453.Destroy-Sequential-Targets](https://github.com/wisdompeak/LeetCode/tree/master/Others/2453.Destroy-Sequential-Targets) (M)
* ``结论转移``
[1685.Sum-of-Absolute-Differences-in-a-Sorted-Array](https://github.com/wisdompeak/LeetCode/tree/master/Others/1685.Sum-of-Absolute-Differences-in-a-Sorted-Array) (M)
[2121.Intervals-Between-Identical-Elements](https://github.com/wisdompeak/LeetCode/tree/master/Others/2121.Intervals-Between-Identical-Elements) (M)
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class Solution {
public:
vector<int> secondGreaterElement(vector<int>& nums)
{
stack<int>st1;
stack<int>st2;

vector<int>rets(nums.size(), -1);

for (int i=0; i<nums.size(); i++)
{
while (!st2.empty() && nums[st2.top()] < nums[i])
{
rets[st2.top()] = nums[i];
st2.pop();
}

vector<int>temp;
while (!st1.empty() && nums[st1.top()] < nums[i])
{
temp.push_back(st1.top());
st1.pop();
}

reverse(temp.begin(), temp.end());
for (auto x: temp)
st2.push(x);

st1.push(i);
}

return rets;
}
};
9 changes: 9 additions & 0 deletions Stack/2454.Next-Greater-Element-IV/Readme.md
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### 2454.Next-Greater-Element-IV

我们已经知道,常规的Next Greater Element可以用单调栈实现o(n)的解法。我们维护一个单调递减的栈,如果遇到新元素大于栈顶元素,就意味着栈顶元素遇到了next greater element,于是就可以退栈。

在此题里,栈顶元素遇到了next greater selement,并不意味着它就可以一劳永逸地舍弃。我们需要的是the second greater element,于是我们应该对这些元素进行标记,表示他们已经看到了一次next greater。当它们再次遇到greater element的时候,才能记录答案。

那么如何标记呢?如果把常规的单调栈记做stk1,那么我们可以把遇到过next greater的元素拿出来,放入另外一个单调栈里,记做stk2。每次新来一个元素nums[i],先看stk2的栈顶元素是否小于num[i],是的话就意味着这些栈顶元素遇到了the second greater element,就可以记录答案并退栈了。接下来看stk1的栈顶元素是否小于nums[i],同理,是的话就意味着这些栈顶元素遇到过了next greater element,并将其移至stk2中。

这里要注意一定,将stk1的元素移至stk2的过程中,是否会干扰stk2的单调顺序?是不会的。stk2经过退栈之后,栈顶元素一定是大于nums[i]的;而从stk1转移至stk2的这些元素都是小于nums[i]的,所以我们可以放心将转移的元素都堆在stk1的栈顶,依然能保持stk2的递减性质。
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class Solution {
public:
int sumSubarrayMins(vector<int>& arr)
{
int n = arr.size();
vector<int>nextSmaller(n, n);
vector<int>prevSmaller(n, -1);

stack<int>Stack;
for (int i=0; i<n; i++)
{
while (!Stack.empty() && arr[Stack.top()] > arr[i])
{
nextSmaller[Stack.top()] = i;
Stack.pop();
}

if (!Stack.empty())
prevSmaller[i] = Stack.top();
Stack.push(i);
}


long ret = 0;
long M = 1e9+7;
for (int i=0; i<n; i++)
{
int a = prevSmaller[i];
int b = nextSmaller[i];
long num = arr[i]*(i-a) % M *(b-i) % M;
ret = (ret + num) % M;
}
return ret;
}
};
18 changes: 17 additions & 1 deletion Stack/907.Sum-of-Subarray-Minimums/Readme.md
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Expand Up @@ -8,5 +8,21 @@

所以本题确切的说,是求每个元素的next smaller element,以及previous smaller or equal element. 另外,特别注意:如果一个数没有next smaller element,那么意味着它的左边界是可以到n;如果一个数没有prev smaller/equal element,那么意味着它的左边界是可以到-1.

Update: 事实上,`next smaller element`和`previous smaller or equal element`可以one-pass同时实现。
```cpp
for (int i=0; i<n; i++)
{
while (!Stack.empty() && arr[Stack.top()] > arr[i])
{
nextSmaller[Stack.top()] = i;
Stack.pop();
}

[Leetcode Link](https://leetcode.com/problems/sum-of-subarray-minimums)
if (!Stack.empty())
prevSmaller[i] = Stack.top();
Stack.push(i);
}
```


[Leetcode Link](https://leetcode.com/problems/sum-of-subarray-minimums)

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