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e992d0c
Create 3640.Trionic-Array-II.cpp
wisdompeak cc2dcb7
Create Readme.md
wisdompeak bb94cc3
Update Readme.md
wisdompeak 5ce863c
Create 3639.Minimum-Time-to-Activate-String.cpp
wisdompeak 70f07b9
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43 changes: 43 additions & 0 deletions
Binary_Search/3639.Minimum-Time-to-Activate-String/3639.Minimum-Time-to-Activate-String.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,43 @@ | ||
| using ll = long long; | ||
| class Solution { | ||
| public: | ||
| int minTime(string s, vector<int>& order, int k) { | ||
| int n = s.size(); | ||
| vector<bool>isStar(n, false); | ||
| ll T = ll(n)*(n+1)/2; | ||
| if (k>T) return -1; | ||
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| auto check = [&](int mid) { | ||
| fill(isStar.begin(), isStar.end(), false); | ||
| for (int i=0; i<=mid; i++) | ||
| isStar[order[i]] = true; | ||
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| ll sumNon = 0; | ||
| for (int i=0; i<n; ){ | ||
| if (isStar[i]) { | ||
| i++; | ||
| continue; | ||
| } | ||
| int j = i; | ||
| while (j<n && !isStar[j]) | ||
| j++; | ||
| ll len = j-i; | ||
| sumNon += len*(len+1)/2; | ||
| i = j; | ||
| } | ||
| return T - sumNon >= k; | ||
| }; | ||
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| int lo = 0, hi = n-1, ans = -1; | ||
| while (lo < hi) { | ||
| int mid = lo + (hi-lo)/2; | ||
| if (check(mid)) { | ||
| hi = mid; | ||
| } else { | ||
| lo = mid+1; | ||
| } | ||
| } | ||
| if (check(hi)) return hi; | ||
| else return -1; | ||
| } | ||
| }; |
5 changes: 5 additions & 0 deletions
Binary_Search/3639.Minimum-Time-to-Activate-String/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,5 @@ | ||
| ### 3639.Minimum-Time-to-Activate-String | ||
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| 非常明显的二分搜值。假设运行到某个时刻t,那么我们就得到一个包含若干星号的字符串。我们需要考察该字符串里至少包含一个星号的substring的个数是否超过k。超过的话,就可以尝试减少k,否则需要增加k。 | ||
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| 计算"至少包含一个星号的substring的个数",等效于反向计算"没有任何星号的substring的个数",并且后者更容易计算。对于任何一段连续的、不包含任何星号的子串长度p,那么就有p*(p+1)/2个子串符合条件。我们分割原始字符串为若干段"没有任何星号的区间",分别计算再相加即可。 |
54 changes: 54 additions & 0 deletions
Others/3640.Trionic-Array-II/3640.Trionic-Array-II.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,54 @@ | ||
| using ll = long long; | ||
| class Solution { | ||
| public: | ||
| vector<pair<int, int>>split(vector<int>&arr) { | ||
| int n = arr.size(); | ||
| vector<pair<int, int>> result; | ||
| int i = 0; | ||
| while (i < n) { | ||
| int j = i; | ||
| while (j + 1 < n && arr[j + 1] < arr[j]) { | ||
| j++; | ||
| } | ||
| if (j > i) { | ||
| result.push_back({i, j}); | ||
| } | ||
| i = j+1; | ||
| } | ||
| return result; | ||
| } | ||
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| long long maxSumTrionic(vector<int>& nums) { | ||
| int n = nums.size(); | ||
| vector<pair<int,int>>arr = split(nums); | ||
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| ll ret = LLONG_MIN/2; | ||
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| for (int i=0; i<arr.size(); i++) { | ||
| int x = arr[i].first, y = arr[i].second; | ||
| if (x-1<0) continue; | ||
| if (y+1>=n) continue; | ||
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| ll sum = nums[x-1]; | ||
| ll maxSum1 = nums[x-1]; | ||
| for (int j=x-2; j>=0; j--) { | ||
| if (nums[j]>=nums[j+1]) break; | ||
| sum += nums[j]; | ||
| maxSum1 = max(maxSum1, sum); | ||
| } | ||
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| sum = nums[y+1]; | ||
| ll maxSum2 = nums[y+1]; | ||
| for (int j=y+2; j<n; j++) { | ||
| if (nums[j]<=nums[j-1]) break; | ||
| sum += nums[j]; | ||
| maxSum2 = max(maxSum2, sum); | ||
| } | ||
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| sum = accumulate(nums.begin()+x, nums.begin()+y+1, 0ll); | ||
| ret = max(ret, maxSum1 + maxSum2 + sum); | ||
| } | ||
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| return ret; | ||
| } | ||
| }; |
5 changes: 5 additions & 0 deletions
Others/3640.Trionic-Array-II/Readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,5 @@ | ||
| ### 3640.Trionic-Array-II | ||
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| 因为符合条件的区间必然包含一段完整的递减区间,所以入手点就是拆解nums,遍历其中所有的递减区间。假设其中一个递减区间是[x,y],然后从x-1分别往前、从y+1往后找隔壁递增区间的最大前缀和即可。 | ||
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| 最终返回全局最大的三段和。 |
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