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Find ud af, om en undergruppe er i form af et bjerg eller ej

Prøv det på GfG Practice ' title= #practiceLinkDiv { display: ingen !important; }

Vi får et array af heltal og et interval, vi skal finde ud af, om underarrayet, der falder i dette interval, har værdier i form af et bjerg eller ej. Alle værdier af subarrayet siges at være i form af et bjerg, hvis enten alle værdier er stigende eller faldende eller først stigende og derefter faldende. 
Mere formelt en subarray [a1 a2 a3...aN] siges at være i form af et bjerg, hvis der eksisterer et heltal K 1<= K <= N such that 
a1<= a2 <= a3 .. <= aK >= a(K+1) >= a(K+2) …. >= aN  

Eksempler:  

forårsstøvle annotationer 
  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 Mountain Subarray problem Prøv det!

Løsning:  



python sorteret tuple
    Nærme sig:Problemet har flere forespørgsler, så for hver forespørgsel bør løsningen beregnes med mindst mulig tidskompleksitet. Så opret to ekstra mellemrum af længden af ​​det originale array. For hvert element skal du finde det sidste indeks på venstre side, som er stigende, dvs. større end dets forrige element, og find elementet på højre side vil gemme det første indeks på højre side, som er faldende, dvs. større end dets næste element. Hvis disse værdier kan beregnes for hvert indeks i konstant tid, kan svaret gives i konstant tid for hvert givet område.Algoritme: 
    1. Opret to ekstra længder n venstre og højre og en ekstra variabel lastptr
    2. Initialiser venstre[0] = 0 og lastptr = 0
    3. Gå gennem det originale array fra andet indeks til slutningen
    4. For hvert indeks skal du kontrollere, om det er større end det foregående element, hvis ja, så opdater lastptr med det aktuelle indeks.
    5. For hver indeksbutik lastptr i venstre[i]
    6. initialisere højre[N-1] = N-1 og lastptr = N-1
    7. Gå gennem det originale array fra næstsidste indeks til starten
    8. For hvert indeks skal du kontrollere, om det er større end det næste element, hvis ja, så opdater lastptr med det aktuelle indeks.
    9. For hver indeksbutik lastptr i ret[i]
    10. Behandl nu forespørgslerne
    11. for hver forespørgsel l r hvis højre[l] >= venstre[r] udskriv derefter ja andet ingen
    Implementering:
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>
    Produktion: 
Subarray is in mountain form Subarray is not in mountain form
    Kompleksitetsanalyse: 
      Tidskompleksitet:På). 
      Kun to traverseringer er nødvendige, så tidskompleksiteten er O(n).Rumkompleksitet:På). 
      To ekstra rum med længden n er påkrævet, så rummets kompleksitet er O(n).


 

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