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Pronađite je li podniz u obliku planine ili ne

Isprobajte na GfG Practice ' title= #practiceLinkDiv { display: none !important; }

Dobili smo niz cijelih brojeva i raspon koji trebamo pronaći da li podniz koji spada u ovaj raspon ima vrijednosti u obliku planine ili ne. Kaže se da su sve vrijednosti podniza u obliku planine ako sve vrijednosti rastu ili opadaju ili prvo rastu, a zatim opadaju. 
Formalnije podniz [a1 a2 a3…aN] kaže se da je u obliku planine ako postoji cijeli broj K 1<= K <= N such that 
a1<= a2 <= a3 .. <= aK >= a(K+1) >= a(K+2) …. >= aN  

Primjeri:  

  Input : Arr[]   = [2 3 2 4 4 6 3 2] Range = [0 2]   Output :    Yes   Explanation:   The output is yes  subarray is [2 3 2] so subarray first increases and then decreases   Input:    Arr[] = [2 3 2 4 4 6 3 2] Range = [2 7]   Output:   Yes   Explanation:   The output is yes  subarray is [2 4 4 6 3 2] so subarray first increases and then decreases   Input:   Arr[]= [2 3 2 4 4 6 3 2] Range = [1 3]   Output:   no   Explanation:   The output is no subarray is [3 2 4] so subarray is not in the form above stated
Recommended Practice Problem planinskog podniza Probajte!

Otopina:  



    Pristup:Problem ima više upita pa za svaki upit rješenje treba izračunati uz najmanju moguću vremensku složenost. Dakle, stvorite dva dodatna razmaka duljine izvornog niza. Za svaki element pronađite posljednji indeks na lijevoj strani koji raste, tj. veći je od svog prethodnog elementa, a pronađite element na desnoj strani pohranit će prvi indeks na desnoj strani koji se smanjuje, tj. veći je od svog sljedećeg elementa. Ako se te vrijednosti mogu izračunati za svaki indeks u konstantnom vremenu, tada se za svaki zadani raspon odgovor može dati u konstantnom vremenu.Algoritam: 
    1. Napravite dva dodatna razmaka dužine n lijevo i pravo i dodatnu varijablu lastptr
    2. Inicijalizirati lijevo[0] = 0 i lastptr = 0
    3. Prelazi izvorni niz od drugog indeksa do kraja
    4. Za svaki indeks provjerite je li veći od prethodnog elementa, ako da, ažurirajte lastptr s trenutnim indeksom.
    5. Za svaki indeks pohranite lastptr u lijevo[i]
    6. inicijalizirati desno [N-1] = N-1 i lastptr = N-1
    7. Prelazi izvorni niz od pretposljednjeg indeksa do početka
    8. Za svaki indeks provjerite je li veći od sljedećeg elementa, ako da, ažurirajte lastptr s trenutnim indeksom.
    9. Za svaki indeks pohranite lastptr u točno[i]
    10. Sada obradite upite
    11. za svaki upit l r ako desno[l] >= lijevo[r] zatim ispisati Da drugo Ne
    Implementacija:
C++ 
// C++ program to check whether a subarray is in // mountain form or not #include    using namespace std; // Utility method to construct left and right array int preprocess(int arr[] int N int left[] int right[]) {  // Initialize first left index as that index only  left[0] = 0;  int lastIncr = 0;  for (int i = 1; i < N; i++)  {  // if current value is greater than previous  // update last increasing  if (arr[i] > arr[i - 1])  lastIncr = i;  left[i] = lastIncr;  }  // Initialize last right index as that index only  right[N - 1] = N - 1;  int firstDecr = N - 1;  for (int i = N - 2; i >= 0; i--)  {  // if current value is greater than next  // update first decreasing  if (arr[i] > arr[i + 1])  firstDecr = i;  right[i] = firstDecr;  } } // Method returns true if arr[L..R] is in mountain form bool isSubarrayMountainForm(int arr[] int left[]  int right[] int L int R) {  // return true only if right at starting range is  // greater than left at ending range  return (right[L] >= left[R]); } // Driver code to test above methods int main() {  int arr[] = {2 3 2 4 4 6 3 2};  int N = sizeof(arr) / sizeof(int);  int left[N] right[N];  preprocess(arr N left right);  int L = 0;  int R = 2;  if (isSubarrayMountainForm(arr left right L R))  cout << 'Subarray is in mountain formn';  else  cout << 'Subarray is not in mountain formn';  L = 1;  R = 3;  if (isSubarrayMountainForm(arr left right L R))  cout << 'Subarray is in mountain formn';  else  cout << 'Subarray is not in mountain formn';  return 0; }
Java 
// Java program to check whether a subarray is in // mountain form or not class SubArray {  // Utility method to construct left and right array  static void preprocess(int arr[] int N int left[] int right[])  {  // initialize first left index as that index only  left[0] = 0;  int lastIncr = 0;    for (int i = 1; i < N; i++)  {  // if current value is greater than previous  // update last increasing  if (arr[i] > arr[i - 1])  lastIncr = i;  left[i] = lastIncr;  }    // initialize last right index as that index only  right[N - 1] = N - 1;  int firstDecr = N - 1;    for (int i = N - 2; i >= 0; i--)  {  // if current value is greater than next  // update first decreasing  if (arr[i] > arr[i + 1])  firstDecr = i;  right[i] = firstDecr;  }  }    // method returns true if arr[L..R] is in mountain form  static boolean isSubarrayMountainForm(int arr[] int left[]  int right[] int L int R)  {  // return true only if right at starting range is  // greater than left at ending range  return (right[L] >= left[R]);  }    public static void main(String[] args)  {  int arr[] = {2 3 2 4 4 6 3 2};  int N = arr.length;  int left[] = new int[N];  int right[] = new int[N];  preprocess(arr N left right);  int L = 0;  int R = 2;    if (isSubarrayMountainForm(arr left right L R))  System.out.println('Subarray is in mountain form');  else  System.out.println('Subarray is not in mountain form');    L = 1;  R = 3;    if (isSubarrayMountainForm(arr left right L R))  System.out.println('Subarray is in mountain form');  else  System.out.println('Subarray is not in mountain form');  } } // This Code is Contributed by Saket Kumar
Python3 
# Python 3 program to check whether a subarray is in # mountain form or not # Utility method to construct left and right array def preprocess(arr N left right): # initialize first left index as that index only left[0] = 0 lastIncr = 0 for i in range(1N): # if current value is greater than previous # update last increasing if (arr[i] > arr[i - 1]): lastIncr = i left[i] = lastIncr # initialize last right index as that index only right[N - 1] = N - 1 firstDecr = N - 1 i = N - 2 while(i >= 0): # if current value is greater than next # update first decreasing if (arr[i] > arr[i + 1]): firstDecr = i right[i] = firstDecr i -= 1 # method returns true if arr[L..R] is in mountain form def isSubarrayMountainForm(arr left right L R): # return true only if right at starting range is # greater than left at ending range return (right[L] >= left[R]) # Driver code  if __name__ == '__main__': arr = [2 3 2 4 4 6 3 2] N = len(arr) left = [0 for i in range(N)] right = [0 for i in range(N)] preprocess(arr N left right) L = 0 R = 2 if (isSubarrayMountainForm(arr left right L R)): print('Subarray is in mountain form') else: print('Subarray is not in mountain form') L = 1 R = 3 if (isSubarrayMountainForm(arr left right L R)): print('Subarray is in mountain form') else: print('Subarray is not in mountain form') # This code is contributed by # Surendra_Gangwar
C# 
// C# program to check whether  // a subarray is in mountain  // form or not using System; class GFG {    // Utility method to construct   // left and right array  static void preprocess(int []arr int N   int []left int []right)  {  // initialize first left   // index as that index only  left[0] = 0;  int lastIncr = 0;    for (int i = 1; i < N; i++)  {  // if current value is   // greater than previous  // update last increasing  if (arr[i] > arr[i - 1])  lastIncr = i;  left[i] = lastIncr;  }    // initialize last right   // index as that index only  right[N - 1] = N - 1;  int firstDecr = N - 1;    for (int i = N - 2; i >= 0; i--)  {  // if current value is   // greater than next  // update first decreasing  if (arr[i] > arr[i + 1])  firstDecr = i;  right[i] = firstDecr;  }  }    // method returns true if  // arr[L..R] is in mountain form  static bool isSubarrayMountainForm(int []arr int []left  int []right int L int R)  {  // return true only if right at   // starting range is greater   // than left at ending range  return (right[L] >= left[R]);  }      // Driver Code  static public void Main ()  {  int []arr = {2 3 2 4  4 6 3 2};  int N = arr.Length;  int []left = new int[N];  int []right = new int[N];  preprocess(arr N left right);    int L = 0;  int R = 2;    if (isSubarrayMountainForm(arr left   right L R))  Console.WriteLine('Subarray is in ' +   'mountain form');  else  Console.WriteLine('Subarray is not ' +   'in mountain form');    L = 1;  R = 3;    if (isSubarrayMountainForm(arr left   right L R))  Console.WriteLine('Subarray is in ' +   'mountain form');  else  Console.WriteLine('Subarray is not ' +   'in mountain form');  } } // This code is contributed by aj_36
JavaScript 
<script>  // Javascript program to check whether   // a subarray is in mountain   // form or not    // Utility method to construct   // left and right array  function preprocess(arr N left right)  {  // initialize first left   // index as that index only  left[0] = 0;  let lastIncr = 0;    for (let i = 1; i < N; i++)  {  // if current value is   // greater than previous  // update last increasing  if (arr[i] > arr[i - 1])  lastIncr = i;  left[i] = lastIncr;  }    // initialize last right   // index as that index only  right[N - 1] = N - 1;  let firstDecr = N - 1;    for (let i = N - 2; i >= 0; i--)  {  // if current value is   // greater than next  // update first decreasing  if (arr[i] > arr[i + 1])  firstDecr = i;  right[i] = firstDecr;  }  }    // method returns true if  // arr[L..R] is in mountain form  function isSubarrayMountainForm(arr left right L R)  {  // return true only if right at   // starting range is greater   // than left at ending range  return (right[L] >= left[R]);  }    let arr = [2 3 2 4 4 6 3 2];  let N = arr.length;  let left = new Array(N);  let right = new Array(N);  preprocess(arr N left right);  let L = 0;  let R = 2;  if (isSubarrayMountainForm(arr left right L R))  document.write('Subarray is in ' + 'mountain form' + '
');  else  document.write('Subarray is not ' + 'in mountain form' + '
');  L = 1;  R = 3;  if (isSubarrayMountainForm(arr left right L R))  document.write('Subarray is in ' + 'mountain form');  else  document.write('Subarray is not ' + 'in mountain form');   </script>
    Izlaz: 
Subarray is in mountain form Subarray is not in mountain form
    Analiza složenosti: 
      Vremenska složenost:Na). 
      Potrebna su samo dva obilaska tako da je vremenska složenost O(n).Složenost prostora:Na). 
      Potrebna su dva dodatna prostora duljine n tako da je složenost prostora O(n).


 

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