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[pull] master from wisdompeak:master #316

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Merged
pull merged 7 commits into AlgorithmAndLeetCode:master from wisdompeak:master
Mar 17, 2025
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Original file line number Diff line number Diff line change
@@ -0,0 +1,34 @@
class Solution {
int dp[10][1001];
public:
bool isOK(vector<int>& nums)
{
for (int i=0; i<nums.size(); i++)
{
if (dp[i][nums[i]] == false)
return false;
}
return true;
}

int minZeroArray(vector<int>& nums, vector<vector<int>>& queries)
{
for (int i=0; i<nums.size(); i++)
dp[i][0] = true;

if (isOK(nums)) return 0;

for (int k=0; k<queries.size(); k++)
{
int a = queries[k][0], b = queries[k][1], d = queries[k][2];
for (int i=a; i<=b; i++)
{
for (int v=1000; v>=0; v--)
if (v>=d && dp[i][v-d] == true)
dp[i][v] = true;
}
if (isOK(nums)) return k+1;
}
return -1;
}
};
17 changes: 17 additions & 0 deletions Dynamic_Programming/3489.Zero-Array-Transformation-IV/Readme.md
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### 3489.Zero-Array-Transformation-IV

这本质是一个背包问题。每处理一个query,在对应区间内的nums[i]就多得了一次删减的操作。

我们需要查看这些nums[i]在获得这个额外的删减机会之后,是否能连同之前的删减操作,实现置零?很显然,如果该query能够让nums[i]再削减d,那么就取决于nums[i]之前能否削减至d。

我们令dp[i][v]表示如果nums[i]的数值是v,能否最终削减成为零。就有
```
for q: queries
a = q[0], b = q[1], d = q[2];
for (i=a; i=b; i++) {
for (int v=0; v<=1000; v++) {
dp[i][v] = dp[i][v] || d[i][v-d];
}
}
```
最终查看所有的dp[i][nums[i]]是否为true。
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This file was deleted.

2 changes: 1 addition & 1 deletion Readme.md
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Original file line number Diff line number Diff line change
Expand Up @@ -857,6 +857,7 @@
[2518.Number-of-Great-Partitions](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2518.Number-of-Great-Partitions) (H-)
[2585.Number-of-Ways-to-Earn-Points](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2585.Number-of-Ways-to-Earn-Points) (M)
[2902.Count-of-Sub-Multisets-With-Bounded-Sum](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2902.Count-of-Sub-Multisets-With-Bounded-Sum) (H)
[3489.Zero-Array-Transformation-IV](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/3489.Zero-Array-Transformation-IV) (H-)
* ``键盘型``
[650.2-Keys-Keyboard](https://github.com/wisdompeak/LeetCode/blob/master/Dynamic_Programming/650.2-Keys-Keyboard) (M+)
[651.4-Keys-Keyboard](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/651.4-Keys-Keyboard) (M+)
Expand Down Expand Up @@ -1620,7 +1621,6 @@
[2963.Count-the-Number-of-Good-Partitions](https://github.com/wisdompeak/LeetCode/tree/master/Others/2963.Count-the-Number-of-Good-Partitions) (H-)
[3009.Maximum-Number-of-Intersections-on-the-Chart](https://github.com/wisdompeak/LeetCode/tree/master/Others/3009.Maximum-Number-of-Intersections-on-the-Chart) (H)
[3169.Count-Days-Without-Meetings](https://github.com/wisdompeak/LeetCode/tree/master/Others/3169.Count-Days-Without-Meetings) (M)
[3489.Zero-Array-Transformation-IV](https://github.com/wisdompeak/LeetCode/tree/master/Others/3489.Zero-Array-Transformation-IV) (H)
* ``二维差分``
[850.Rectangle-Area-II](https://github.com/wisdompeak/LeetCode/tree/master/Others/850.Rectangle-Area-II) (H)
[2132.Stamping-the-Grid](https://github.com/wisdompeak/LeetCode/tree/master/Others/2132.Stamping-the-Grid) (H)
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2 changes: 1 addition & 1 deletion Recursion/3490.Count-Beautiful-Numbers/Readme.md
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Expand Up @@ -6,7 +6,7 @@
1. 该位置是否贴近上限T。如果pos之前的选择都是贴着上限T,那么在第pos位上,我们的选择上限也只能是T[pos],否则上限可以是9.
2. 该位置是否是先导零。如果pos之前的选择全部都是0,那么在pos位置之前记录的乘积应该强制认作是1。这么做是为了处理这样一种情况:pos之前都是0,并且pos位也想取零。如果没有这条规则,那么递归到后面的乘积就永远是零了。

由此,我们一旦做出了pos位置上的鞠策,在往后递归的时候,也需要相应更新isTight和isLeadingZero这两个状态。
由此,我们一旦做出了pos位置上的决策,在往后递归的时候,也需要相应更新isTight和isLeadingZero这两个状态。

递归需要记忆化的支持。本题记忆化的状态就是递归函数的参数:pos, sum, product, isTight, isLeadingZero。我们可以用tuple作为key,加上有序map来存储访问过的状态。

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