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Esbrineu si un subbarrat té forma de muntanya o no

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Ens donen una matriu de nombres enters i un interval que necessitem per trobar si el subbarrat que cau en aquest rang té valors en forma de muntanya o no. Es diu que tots els valors del subarray tenen forma de muntanya si tots els valors augmenten o decreixen o primer augmenten i després disminueixen. 
Més formalment un subbarray [a1 a2 a3…aN] es diu que té forma de muntanya si existeix un nombre enter K 1<= K <= N such that 
a1<= a2 <= a3 .. <= aK >= a(K+1) >= a(K+2) …. >= aN  

Exemples:  

  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 Problema de Mountain Subarray Prova-ho!

Solució:  



    Enfocament:El problema té múltiples consultes, de manera que per a cada consulta la solució s'ha de calcular amb la menor complexitat de temps possible. Per tant, creeu dos espais addicionals de la longitud de la matriu original. Per a cada element, trobeu l'últim índex del costat esquerre que està creixent, és a dir, més gran que el seu element anterior i trobeu l'element del costat dret, emmagatzemarà el primer índex del costat dret que està decreixent, és a dir, més gran que el seu element següent. Si aquests valors es poden calcular per a cada índex en temps constant, llavors per a cada rang donat la resposta es pot donar en temps constant.Algorisme: 
    1. Creeu dos espais addicionals de longitud n esquerra i dret i una variable addicional darrer ptr
    2. Inicialitzar esquerra[0] = 0 i darrer ptr = 0
    3. Travessa la matriu original des del segon índex fins al final
    4. Per a cada índex, comproveu si és més gran que l'element anterior, en cas afirmatiu, actualitzeu el darrer ptr amb l'índex actual.
    5. Per a cada índex emmagatzema el darrer ptr en esquerra [i]
    6. inicialitzar dreta[N-1] = N-1 i darrer ptr = N-1
    7. Travessa la matriu original des del penúltim índex fins al començament
    8. Per a cada índex, comproveu si és més gran que el següent element, en cas afirmatiu, actualitzeu el darrer ptr amb l'índex actual.
    9. Per a cada índex emmagatzema el darrer ptr en correcte [i]
    10. Ara processeu les consultes
    11. per a cada consulta l r si dreta[l] >= esquerra[r] després imprimir sí altra cosa no
    Implementació:
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>
    Sortida: 
Subarray is in mountain form Subarray is not in mountain form
    Anàlisi de complexitat: 
      Complexitat temporal:O(n). 
      Només es necessiten dos recorreguts, de manera que la complexitat temporal és O(n).Complexitat espacial:O(n). 
      Es requereixen dos espais addicionals de longitud n, de manera que la complexitat espacial és O(n).


 

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