diff --git a/Dynamic_Programming/115.Distinct-Subsequences/115.Distinct-Subsequences.cpp b/Dynamic_Programming/115.Distinct-Subsequences/115.Distinct-Subsequences.cpp
index e4266191e..dfba2ce32 100644
--- a/Dynamic_Programming/115.Distinct-Subsequences/115.Distinct-Subsequences.cpp
+++ b/Dynamic_Programming/115.Distinct-Subsequences/115.Distinct-Subsequences.cpp
@@ -20,10 +20,9 @@ class Solution {
 for (int j=1; j<=n; j++) { if (s[i]==t[j]) - dp[i][j] = dp[i-1][j] + dp[i-1][j-1]; + dp[i][j] = min(LLONG_MAX/2, dp[i-1][j] + dp[i-1][j-1]); else dp[i][j] = dp[i-1][j]; - //cout<<i<<" "<<j<<" "<<dp[i][j]<<endl; } return dp[m][n]; } diff --git a/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/2366.Minimum-Replacements-to-Sort-the-Array_v2.cpp b/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/2366.Minimum-Replacements-to-Sort-the-Array_v2.cpp new file mode 100644 index 000000000..e68e842b5 --- /dev/null +++ b/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/2366.Minimum-Replacements-to-Sort-the-Array_v2.cpp @@ -0,0 +1,28 @@ +using LL = long long; +class Solution { +public: + long long minimumReplacement(vector<int>& nums) 
+ {
+ LL ret = 0;
+ for (int i = nums.size()-2; i>=0; i--)
+ {
+ LL x = nums[i+1];
+ LL y = nums[i];
+ if (y<=x) continue; + + if (y%x==0) + { + ret += y/x-1; + nums[i] = x; + } + else + { + int k = y/x+1; + ret += y/x; + nums[i] = y/k; + } + } + + return ret; + } +}; diff --git a/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/Readme.md b/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/Readme.md index 6ad7847fb..040c70502 100644 --- a/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/Readme.md +++ b/Greedy/2366.Minimum-Replacements-to-Sort-the-Array/Readme.md @@ -1,5 +1,7 @@ ### 2366.Minimum-Replacements-to-Sort-the-Array +#### 解法1 + 我们从后往前看,对于最后一个数,我们肯定不会拆分。一旦将其拆分的话变小的话,那么前面的数就有更大的概率需要拆得更小。 接着假设最后一个数是x,倒数第二个数是y。如果y小于等于x,那么最后两个元素已经是递增关系,y就不用拆分了,理由同上。如果y大于x,那么就必须拆分y,那么怎么拆分呢? @@ -21,3 +23,6 @@ p <= (x-d) / (k+1) 此时这是不是最优的操作呢?并不是。如果`d2 < x2`,其实我们可以将k个x2里面的一部分(而不是整体)拿出1来再贡献给d2,必然可以使得d2再拉至于x2-1平齐的高度。这是因为之前我们知道,如果k个x2每人都再贡献1出来,会导致`d2`会比`x2-1`还大。所以这意味着,如果贡献出部分的1出来,就能让`d2`与`x2-1`持平。在这种情况下,`x2-1`就是拆分出来的k+1份里的最小值。 于是,这个回合结束我们将nums[i]赋值为`x2`(如果`d2==x2`)或者`x2-1`(如果`d2<x2`)。下一个回合就是针对nums[i],来策划对于nums[i-1]的拆分。完全重复上述的过程。 + +#### 解法2 +还有一种写起来更简单的解法。我们已经知道,我们试图将y做最少的拆分,且每一份不能拆过x。显然,我们拆分的份数就是`k = ⌈y/x⌉`,其中`⌈.⌉`表示一个分数的上整界。如果你还知道一个结论(或者直觉):将一个数y拆成k份,一定可以表示成若干个`⌈y/k⌉`与若干个`⌊y/k⌋`之和,那么显然这就是最为平均的一种分法(因为任何元素之间彼此不超过1)。根据解法1的分析,我们会把nums[i]重置为这种分法得到的最大的最小值值`⌊y/k⌋`。 diff --git a/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar.cpp b/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar.cpp new file mode 100644 index 000000000..a766e4111 --- /dev/null +++ b/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar.cpp @@ -0,0 +1,38 @@ +using LL = long long; +class Solution { +public: + long long makeSimilar(vector<int>& nums, vector<int>& target) 
+ { 
+ vector<ll>odd1, odd2, even1, even2;
+ for (auto x: nums)
+ {
+ if (x%2==0)
+ even1.push_back(x);
+ else
+ odd1.push_back(x);
+ }
+ for (auto x: target)
+ {
+ if (x%2==0)
+ even2.push_back(x);
+ else
+ odd2.push_back(x);
+ }
+ 
+ return helper(even1, even2) + helper(odd1, odd2); 
+ }
+ 
+ LL helper(vector<ll>&nums, vector<ll>&target)
+ {
+ sort(target.begin(), target.end());
+ sort(nums.begin(), nums.end());
+ 
+ LL count = 0;
+ for (int i=0; i<nums.size(); i++) + if (nums[i]> target[i])
+ count += (nums[i]-target[i])/2;
+ 
+ return count;
+ }
+ 
+};
diff --git a/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/Readme.md b/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/Readme.md
new file mode 100644
index 000000000..63a124368
--- /dev/null
+++ b/Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/Readme.md
@@ -0,0 +1,13 @@
+### 2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar
+
+很显然题目的意思是,nums经过一系列操作之后,需要变成targets。于是nums和targets的数组元素之和必然相等,否则无法实现+2/-2的守恒。
+
+另外,我们发现,偶数无论如何也无法操作成奇数,反之亦然。所以知道,需要将奇数偶数分开处理,即nums里的奇数需要多少操作转化为targets里的奇数,同理nums里的偶数需要多少操作否转化为targets里的偶数。
+
+接下来,我们考虑只含有奇数的nums数组和只含有奇数的targets数组。很明显,我们必然会把nums[0]转化为targets[0],将nums[1]转化为targets[1],依次类推。这样能使得每对元素差的绝对值之和最小。简单的证明可以从两对开始研究。假设有`nums[i]<nums[j]`和`targets[x]<targets[y]`,那么无非三种情况
+1. nums[i]<nums[j]<targets[x]<targets[y]
+2. nums[i]<targets[x]<nums[j]<targets[y]
+3. nums[i]<targets[x]<targets[y]<nums[j]
+画个线段图就发现,无论哪一种,都有`abs(nums[i]-targets[x])+ abs(nums[j]-targets[y]) <= abs(nums[i]-targets[y])+ abs(nums[j]-targets[x])`。将这个结论推广至任意两对元素之间的判断,就可以得出前面的结论。
+
+有了这个策略之后,我们就可以计算出每对元素之间,需要多少次+的操作,或者多少次-的操作。但实际上我们只需要记录+的操作即可,因为题目保证最终全局(即包括了奇数数组和偶数数组)而言+和-的操作一定守恒。
diff --git a/Readme.md b/Readme.md
index 795ee99d1..bad089cdf 100644
--- a/Readme.md
+++ b/Readme.md
@@ -1145,6 +1145,7 @@
 [2365.Task-Scheduler-II](https://github.com/wisdompeak/LeetCode/tree/master/Greedy/2365.Task-Scheduler-II) (M) 
 [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+) 
 * ``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) 
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