مشكلة أطول تسلسل ثانوي بيتوني هي العثور على أطول تسلسل فرعي لتسلسل معين بحيث يكون أولًا متزايدًا ثم يتناقص. يعتبر التسلسل الذي تم فرزه بترتيب متزايد Bitonic مع الجزء المتناقص فارغًا. وبالمثل، يعتبر تسلسل الأمر المتناقص Bitonic مع كون الجزء المتزايد فارغًا. أمثلة:
Input: [1 11 2 10 4 5 2 1] Output: [1 2 10 4 2 1] OR [1 11 10 5 2 1] OR [1 2 4 5 2 1] Input: [12 11 40 5 3 1] Output: [12 11 5 3 1] OR [12 40 5 3 1] Input: [80 60 30 40 20 10] Output: [80 60 30 20 10] OR [80 60 40 20 10]
في سابق لقد ناقشنا بعد ذلك مشكلة أطول تسلسل بيتوني. ومع ذلك، فإن المنشور يغطي فقط الكود المتعلق بإيجاد الحد الأقصى لمجموع التسلسل اللاحق المتزايد ولكن ليس لبناء التسلسل اللاحق. سنناقش في هذا المنشور كيفية إنشاء أطول تسلسل بيتوني بحد ذاته. دع arr[0..n-1] يكون مصفوفة الإدخال. نحن نحدد المتجه LIS بحيث يكون LIS[i] بحد ذاته متجهًا يخزن أطول تسلسل لاحق لـ arr[0..i] الذي ينتهي بـ arr[i]. لذلك بالنسبة للفهرس i LIS[i] يمكن كتابته بشكل متكرر كـ -
LIS[0] = {arr[O]} LIS[i] = {Max(LIS[j])} + arr[i] where j < i and arr[j] < arr[i] = arr[i] if there is no such j
نحدد أيضًا متجه LDS بحيث يكون LDS[i] بحد ذاته متجهًا يخزن أطول تسلسل تناقصي لـ arr[i..n] الذي يبدأ بـ arr[i]. لذلك بالنسبة للفهرس i يمكن كتابة LDS[i] بشكل متكرر كـ -
ج رقم عشوائي
LDS[n] = {arr[n]} LDS[i] = arr[i] + {Max(LDS[j])} where j > i and arr[j] < arr[i] = arr[i] if there is no such j
على سبيل المثال للمصفوفة [1 11 2 10 4 5 2 1]
LIS[0]: 1 LIS[1]: 1 11 LIS[2]: 1 2 LIS[3]: 1 2 10 LIS[4]: 1 2 4 LIS[5]: 1 2 4 5 LIS[6]: 1 2 LIS[7]: 1
LDS[0]: 1 LDS[1]: 11 10 5 2 1 LDS[2]: 2 1 LDS[3]: 10 5 2 1 LDS[4]: 4 2 1 LDS[5]: 5 2 1 LDS[6]: 2 1 LDS[7]: 1
لذلك يمكن أن يكون أطول تسلسل بيتوني
LIS[1] + LDS[1] = [1 11 10 5 2 1] OR LIS[3] + LDS[3] = [1 2 10 5 2 1] OR LIS[5] + LDS[5] = [1 2 4 5 2 1]
وفيما يلي تنفيذ الفكرة المذكورة أعلاه -
C++/* Dynamic Programming solution to print Longest Bitonic Subsequence */ #include using namespace std; // Utility function to print Longest Bitonic // Subsequence void print(vector<int>& arr int size) { for(int i = 0; i < size; i++) cout << arr[i] << ' '; } // Function to construct and print Longest // Bitonic Subsequence void printLBS(int arr[] int n) { // LIS[i] stores the length of the longest // increasing subsequence ending with arr[i] vector<vector<int>> LIS(n); // initialize LIS[0] to arr[0] LIS[0].push_back(arr[0]); // Compute LIS values from left to right for (int i = 1; i < n; i++) { // for every j less than i for (int j = 0; j < i; j++) { if ((arr[j] < arr[i]) && (LIS[j].size() > LIS[i].size())) LIS[i] = LIS[j]; } LIS[i].push_back(arr[i]); } /* LIS[i] now stores Maximum Increasing Subsequence of arr[0..i] that ends with arr[i] */ // LDS[i] stores the length of the longest // decreasing subsequence starting with arr[i] vector<vector<int>> LDS(n); // initialize LDS[n-1] to arr[n-1] LDS[n - 1].push_back(arr[n - 1]); // Compute LDS values from right to left for (int i = n - 2; i >= 0; i--) { // for every j greater than i for (int j = n - 1; j > i; j--) { if ((arr[j] < arr[i]) && (LDS[j].size() > LDS[i].size())) LDS[i] = LDS[j]; } LDS[i].push_back(arr[i]); } // reverse as vector as we're inserting at end for (int i = 0; i < n; i++) reverse(LDS[i].begin() LDS[i].end()); /* LDS[i] now stores Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i] */ int max = 0; int maxIndex = -1; for (int i = 0; i < n; i++) { // Find maximum value of size of LIS[i] + size // of LDS[i] - 1 if (LIS[i].size() + LDS[i].size() - 1 > max) { max = LIS[i].size() + LDS[i].size() - 1; maxIndex = i; } } // print all but last element of LIS[maxIndex] vector print(LIS[maxIndex] LIS[maxIndex].size() - 1); // print all elements of LDS[maxIndex] vector print(LDS[maxIndex] LDS[maxIndex].size()); } // Driver program int main() { int arr[] = { 1 11 2 10 4 5 2 1 }; int n = sizeof(arr) / sizeof(arr[0]); printLBS(arr n); return 0; }
Java /* Dynamic Programming solution to print Longest Bitonic Subsequence */ import java.util.*; class GFG { // Utility function to print Longest Bitonic // Subsequence static void print(Vector<Integer> arr int size) { for (int i = 0; i < size; i++) System.out.print(arr.elementAt(i) + ' '); } // Function to construct and print Longest // Bitonic Subsequence static void printLBS(int[] arr int n) { // LIS[i] stores the length of the longest // increasing subsequence ending with arr[i] @SuppressWarnings('unchecked') Vector<Integer>[] LIS = new Vector[n]; for (int i = 0; i < n; i++) LIS[i] = new Vector<>(); // initialize LIS[0] to arr[0] LIS[0].add(arr[0]); // Compute LIS values from left to right for (int i = 1; i < n; i++) { // for every j less than i for (int j = 0; j < i; j++) { if ((arr[i] > arr[j]) && LIS[j].size() > LIS[i].size()) { for (int k : LIS[j]) if (!LIS[i].contains(k)) LIS[i].add(k); } } LIS[i].add(arr[i]); } /* * LIS[i] now stores Maximum Increasing Subsequence * of arr[0..i] that ends with arr[i] */ // LDS[i] stores the length of the longest // decreasing subsequence starting with arr[i] @SuppressWarnings('unchecked') Vector<Integer>[] LDS = new Vector[n]; for (int i = 0; i < n; i++) LDS[i] = new Vector<>(); // initialize LDS[n-1] to arr[n-1] LDS[n - 1].add(arr[n - 1]); // Compute LDS values from right to left for (int i = n - 2; i >= 0; i--) { // for every j greater than i for (int j = n - 1; j > i; j--) { if (arr[j] < arr[i] && LDS[j].size() > LDS[i].size()) for (int k : LDS[j]) if (!LDS[i].contains(k)) LDS[i].add(k); } LDS[i].add(arr[i]); } // reverse as vector as we're inserting at end for (int i = 0; i < n; i++) Collections.reverse(LDS[i]); /* * LDS[i] now stores Maximum Decreasing Subsequence * of arr[i..n] that starts with arr[i] */ int max = 0; int maxIndex = -1; for (int i = 0; i < n; i++) { // Find maximum value of size of // LIS[i] + size of LDS[i] - 1 if (LIS[i].size() + LDS[i].size() - 1 > max) { max = LIS[i].size() + LDS[i].size() - 1; maxIndex = i; } } // print all but last element of LIS[maxIndex] vector print(LIS[maxIndex] LIS[maxIndex].size() - 1); // print all elements of LDS[maxIndex] vector print(LDS[maxIndex] LDS[maxIndex].size()); } // Driver Code public static void main(String[] args) { int[] arr = { 1 11 2 10 4 5 2 1 }; int n = arr.length; printLBS(arr n); } } // This code is contributed by // sanjeev2552
Python3 # Dynamic Programming solution to print Longest # Bitonic Subsequence def _print(arr: list size: int): for i in range(size): print(arr[i] end=' ') # Function to construct and print Longest # Bitonic Subsequence def printLBS(arr: list n: int): # LIS[i] stores the length of the longest # increasing subsequence ending with arr[i] LIS = [0] * n for i in range(n): LIS[i] = [] # initialize LIS[0] to arr[0] LIS[0].append(arr[0]) # Compute LIS values from left to right for i in range(1 n): # for every j less than i for j in range(i): if ((arr[j] < arr[i]) and (len(LIS[j]) > len(LIS[i]))): LIS[i] = LIS[j].copy() LIS[i].append(arr[i]) # LIS[i] now stores Maximum Increasing # Subsequence of arr[0..i] that ends with # arr[i] # LDS[i] stores the length of the longest # decreasing subsequence starting with arr[i] LDS = [0] * n for i in range(n): LDS[i] = [] # initialize LDS[n-1] to arr[n-1] LDS[n - 1].append(arr[n - 1]) # Compute LDS values from right to left for i in range(n - 2 -1 -1): # for every j greater than i for j in range(n - 1 i -1): if ((arr[j] < arr[i]) and (len(LDS[j]) > len(LDS[i]))): LDS[i] = LDS[j].copy() LDS[i].append(arr[i]) # reverse as vector as we're inserting at end for i in range(n): LDS[i] = list(reversed(LDS[i])) # LDS[i] now stores Maximum Decreasing Subsequence # of arr[i..n] that starts with arr[i] max = 0 maxIndex = -1 for i in range(n): # Find maximum value of size of LIS[i] + size # of LDS[i] - 1 if (len(LIS[i]) + len(LDS[i]) - 1 > max): max = len(LIS[i]) + len(LDS[i]) - 1 maxIndex = i # print all but last element of LIS[maxIndex] vector _print(LIS[maxIndex] len(LIS[maxIndex]) - 1) # print all elements of LDS[maxIndex] vector _print(LDS[maxIndex] len(LDS[maxIndex])) # Driver Code if __name__ == '__main__': arr = [1 11 2 10 4 5 2 1] n = len(arr) printLBS(arr n) # This code is contributed by # sanjeev2552
C# /* Dynamic Programming solution to print longest Bitonic Subsequence */ using System; using System.Linq; using System.Collections.Generic; class GFG { // Utility function to print longest Bitonic // Subsequence static void print(List<int> arr int size) { for (int i = 0; i < size; i++) Console.Write(arr[i] + ' '); } // Function to construct and print longest // Bitonic Subsequence static void printLBS(int[] arr int n) { // LIS[i] stores the length of the longest // increasing subsequence ending with arr[i] List<int>[] LIS = new List<int>[n]; for (int i = 0; i < n; i++) LIS[i] = new List<int>(); // initialize LIS[0] to arr[0] LIS[0].Add(arr[0]); // Compute LIS values from left to right for (int i = 1; i < n; i++) { // for every j less than i for (int j = 0; j < i; j++) { if ((arr[i] > arr[j]) && LIS[j].Count > LIS[i].Count) { foreach (int k in LIS[j]) if (!LIS[i].Contains(k)) LIS[i].Add(k); } } LIS[i].Add(arr[i]); } /* * LIS[i] now stores Maximum Increasing Subsequence * of arr[0..i] that ends with arr[i] */ // LDS[i] stores the length of the longest // decreasing subsequence starting with arr[i] List<int>[] LDS = new List<int>[n]; for (int i = 0; i < n; i++) LDS[i] = new List<int>(); // initialize LDS[n-1] to arr[n-1] LDS[n - 1].Add(arr[n - 1]); // Compute LDS values from right to left for (int i = n - 2; i >= 0; i--) { // for every j greater than i for (int j = n - 1; j > i; j--) { if (arr[j] < arr[i] && LDS[j].Count > LDS[i].Count) foreach (int k in LDS[j]) if (!LDS[i].Contains(k)) LDS[i].Add(k); } LDS[i].Add(arr[i]); } // reverse as vector as we're inserting at end for (int i = 0; i < n; i++) LDS[i].Reverse(); /* * LDS[i] now stores Maximum Decreasing Subsequence * of arr[i..n] that starts with arr[i] */ int max = 0; int maxIndex = -1; for (int i = 0; i < n; i++) { // Find maximum value of size of // LIS[i] + size of LDS[i] - 1 if (LIS[i].Count + LDS[i].Count - 1 > max) { max = LIS[i].Count + LDS[i].Count - 1; maxIndex = i; } } // print all but last element of LIS[maxIndex] vector print(LIS[maxIndex] LIS[maxIndex].Count - 1); // print all elements of LDS[maxIndex] vector print(LDS[maxIndex] LDS[maxIndex].Count); } // Driver Code public static void Main(String[] args) { int[] arr = { 1 11 2 10 4 5 2 1 }; int n = arr.Length; printLBS(arr n); } } // This code is contributed by PrinciRaj1992
JavaScript // Function to print the longest bitonic subsequence function _print(arr size) { for (let i = 0; i<size; i++) { process.stdout.write(arr[i]+' '); } } // Function to construct and print the longest bitonic subsequence function printLBS(arr n) { // LIS[i] stores the length of the longest increasing subsequence ending with arr[i] let LIS = new Array(n); for (let i = 0; i < n; i++) { LIS[i] = []; } // initialize LIS[0] to arr[0] LIS[0].push(arr[0]); // Compute LIS values from left to right for (let i = 1; i < n; i++) { // for every j less than i for (let j = 0; j < i; j++) { if (arr[j] < arr[i] && LIS[j].length > LIS[i].length) { LIS[i] = LIS[j].slice(); } } LIS[i].push(arr[i]); } // LIS[i] now stores the Maximum Increasing Subsequence of arr[0..i] that ends with arr[i] // LDS[i] stores the length of the longest decreasing subsequence starting with arr[i] let LDS = new Array(n); for (let i = 0; i < n; i++) { LDS[i] = []; } // initialize LDS[n-1] to arr[n-1] LDS[n - 1].push(arr[n - 1]); // Compute LDS values from right to left for (let i = n - 2; i >= 0; i--) { // for every j greater than i for (let j = n - 1; j > i; j--) { if (arr[j] < arr[i] && LDS[j].length > LDS[i].length) { LDS[i] = LDS[j].slice(); } } LDS[i].push(arr[i]); } // reverse the LDS vector as we're inserting at the end for (let i = 0; i < n; i++) { LDS[i].reverse(); } // LDS[i] now stores the Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i] let max = 0; let maxIndex = -1; for (let i = 0; i < n; i++) { // Find maximum value of size of LIS[i] + size of LDS[i] - 1 if (LIS[i].length + LDS[i].length - 1 > max) { max = LIS[i].length + LDS[i].length - 1; maxIndex = i; } } // print all but // print all but last element of LIS[maxIndex] array _print(LIS[maxIndex].slice(0 -1) LIS[maxIndex].length - 1); // print all elements of LDS[maxIndex] array _print(LDS[maxIndex] LDS[maxIndex].length); } // Driver program const arr = [1 11 2 10 4 5 2 1]; const n = arr.length; printLBS(arr n);
الإخراج:
1 11 10 5 2 1
تعقيد الوقت من حل البرمجة الديناميكية أعلاه هو O(n2). مساحة مساعدة المستخدم في البرنامج هو O(n2).