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1afb6eb
Update Readme.md
wisdompeak db731b4
Update Readme.md
wisdompeak 58610eb
Create 2366.Minimum-Replacements-to-Sort-the-Array_v2.cpp
wisdompeak 45d30b7
Create 2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar.cpp
wisdompeak 77503de
Update 115.Distinct-Subsequences.cpp
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Update Readme.md
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Update 2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar.cpp
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28 changes: 28 additions & 0 deletions
...Minimum-Replacements-to-Sort-the-Array/2366.Minimum-Replacements-to-Sort-the-Array_v2.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -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; | ||
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| if (y%x==0) | ||
| { | ||
| ret += y/x-1; | ||
| nums[i] = x; | ||
| } | ||
| else | ||
| { | ||
| int k = y/x+1; | ||
| ret += y/x; | ||
| nums[i] = y/k; | ||
| } | ||
| } | ||
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| return ret; | ||
| } | ||
| }; |
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38 changes: 38 additions & 0 deletions
...tions-to-Make-Arrays-Similar/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -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); | ||
| } | ||
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| return helper(even1, even2) + helper(odd1, odd2); | ||
| } | ||
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| LL helper(vector<LL>&nums, vector<LL>&target) | ||
| { | ||
| sort(target.begin(), target.end()); | ||
| sort(nums.begin(), nums.end()); | ||
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| LL count = 0; | ||
| for (int i=0; i<nums.size(); i++) | ||
| if (nums[i] > target[i]) | ||
| count += (nums[i]-target[i])/2; | ||
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| return count; | ||
| } | ||
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| }; |
13 changes: 13 additions & 0 deletions
Greedy/2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,13 @@ | ||
| ### 2449.Minimum-Number-of-Operations-to-Make-Arrays-Similar | ||
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| 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。 | ||
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| 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。 | ||
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| 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。 | ||
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| 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。 |
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