Noskaidrojiet, vai apakšgrupa ir kalna formā vai nē
Izmēģiniet to GfG Practice 
#practiceLinkDiv { display: none !important; }
#practiceLinkDiv { display: none !important; } Mums ir dots veselu skaitļu masīvs un diapazons, kas mums ir nepieciešams, lai noskaidrotu, vai apakšgrupai, kas ietilpst šajā diapazonā, ir vērtības kalna formā vai nav. Tiek uzskatīts, ka visas apakšgrupas vērtības ir kalna formā, ja visas vērtības palielinās vai samazinās, vai arī vispirms palielinās un pēc tam samazinās.
Formālāk apakšgrupa [a1 a2 a3…aN] tiek uzskatīts, ka tas ir kalna formā, ja pastāv vesels skaitlis K 1 <= K <= N such that
a1 <= a2 <= a3 .. <= aK >= a(K+1) >= a(K+2) …. >= aN
Piemēri:
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 Mountain Subarray problēma Izmēģiniet to!
Risinājums:
- Izveidojiet divas papildu garuma atstarpes n pa kreisi un pareizi un papildu mainīgais lastptr
- Inicializēt pa kreisi[0] = 0 un lastptr = 0
- Pārvietojiet sākotnējo masīvu no otrā indeksa līdz beigām
- Katram indeksam pārbaudiet, vai tas ir lielāks par iepriekšējo elementu, ja jā, tad atjauniniet lastptr ar pašreizējo indeksu.
- Katram indeksu veikalam lastptr iekšā pa kreisi [i]
- inicializēt pa labi[N-1] = N-1 un lastptr = N-1
- Pārvietojiet sākotnējo masīvu no otrā pēdējā indeksa līdz sākumam
- Katram indeksam pārbaudiet, vai tas ir lielāks par nākamo elementu, ja jā, tad atjauniniet lastptr ar pašreizējo indeksu.
- Katram indeksu veikalam lastptr iekšā pareizi [i]
- Tagad apstrādājiet vaicājumus
- katram vaicājumam l r ja pa labi[l] >= pa kreisi[r] pēc tam izdrukājiet jā cits nē
// 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
Ir nepieciešami tikai divi šķērsojumi, tāpēc laika sarežģītība ir O(n).
Ir vajadzīgas divas papildu atstarpes ar garumu n, lai telpas sarežģītība būtu O(n).
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