Esbrineu si un subbarrat té forma de muntanya o no
Prova-ho a GfG Practice 
#practiceLinkDiv { mostrar: cap !important; }
#practiceLinkDiv { mostrar: cap !important; } 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 statedRecommended Practice Problema de Mountain Subarray Prova-ho!
Solució:
- Creeu dos espais addicionals de longitud n esquerra i dret i una variable addicional darrer ptr
- Inicialitzar esquerra[0] = 0 i darrer ptr = 0
- Travessa la matriu original des del segon índex fins al final
- Per a cada índex, comproveu si és més gran que l'element anterior, en cas afirmatiu, actualitzeu el darrer ptr amb l'índex actual.
- Per a cada índex emmagatzema el darrer ptr en esquerra [i]
- inicialitzar dreta[N-1] = N-1 i darrer ptr = N-1
- Travessa la matriu original des del penúltim índex fins al començament
- Per a cada índex, comproveu si és més gran que el següent element, en cas afirmatiu, actualitzeu el darrer ptr amb l'índex actual.
- Per a cada índex emmagatzema el darrer ptr en correcte [i]
- Ara processeu les consultes
- per a cada consulta l r si dreta[l] >= esquerra[r] després imprimir sí altra cosa no
// 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 form n ' ; else cout < < 'Subarray is not in mountain form n ' ; L = 1 ; R = 3 ; if ( isSubarrayMountainForm ( arr left right L R )) cout < < 'Subarray is in mountain form n ' ; else cout < < 'Subarray is not in mountain form n ' ; 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 ( 1 N ): # 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>
Subarray is in mountain form Subarray is not in mountain form
Només es necessiten dos recorreguts, de manera que la complexitat temporal és O(n).
Es requereixen dos espais addicionals de longitud n, de manera que la complexitat espacial és O(n).
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