Elementa pozīcija pēc stabilas kārtošanas
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Izvade
#practiceLinkDiv { display: none !important; } Ņemot vērā veselu skaitļu masīvu, kurā var būt dublēti elementi, mums ir dots šī masīva elements, mums ir jāpaziņo šī elementa galīgā pozīcija masīvā, ja tiek izmantots stabils kārtošanas algoritms.
Piemēri:
Input : arr[] = [3 4 3 5 2 3 4 3 1 5] index = 5 Output : 4 Element initial index – 5 (third 3) After sorting array by stable sorting algorithm we get array as shown below [1(8) 2(4) 3(0) 3(2) 3(5) 3(7) 4(1) 4(6) 5(3) 5(9)] with their initial indices shown in parentheses next to them Element's index after sorting = 4Recommended Practice Stabila šķirošana un pozīcija Izmēģiniet to!
Viens vienkāršs veids, kā atrisināt šo problēmu, ir izmantot jebkuru stabilu šķirošanas algoritmu, piemēram Ievietošanas kārtošana Kārtot iet utt. un pēc tam iegūstiet dotā elementa jauno indeksu, taču mēs varam atrisināt šo problēmu, nekārtojot masīvu.
Kā elementa pozīciju sakārtotā masīvā nosaka tikai tie elementi, kas ir mazāki par doto elementu. Mēs uzskaitām visus masīva elementus, kas ir mazāki par doto elementu, un tiem elementiem, kas ir vienādi ar dotā elementa elementiem, kas atrodas pirms dotā elementa indeksa, tiks iekļauti mazāko elementu skaitā, tas nodrošinās rezultāta indeksa stabilitāti.
Vienkāršs kods, lai ieviestu iepriekš minēto pieeju, ir ieviests tālāk:
C++ // C++ program to get index of array element in // sorted array #include using namespace std ; // Method returns the position of arr[idx] after // performing stable-sort on array int getIndexInSortedArray ( int arr [] int n int idx ) { /* Count of elements smaller than current element plus the equal element occurring before given index*/ int result = 0 ; for ( int i = 0 ; i < n ; i ++ ) { // If element is smaller then increase // the smaller count if ( arr [ i ] < arr [ idx ]) result ++ ; // If element is equal then increase count // only if it occurs before if ( arr [ i ] == arr [ idx ] && i < idx ) result ++ ; } return result ; } // Driver code to test above methods int main () { int arr [] = { 3 4 3 5 2 3 4 3 1 5 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); int idxOfEle = 5 ; cout < < getIndexInSortedArray ( arr n idxOfEle ); return 0 ; }
Java // Java program to get index of array // element in sorted array class ArrayIndex { // Method returns the position of // arr[idx] after performing stable-sort // on array static int getIndexInSortedArray ( int arr [] int n int idx ) { /* Count of elements smaller than current element plus the equal element occurring before given index*/ int result = 0 ; for ( int i = 0 ; i < n ; i ++ ) { // If element is smaller then // increase the smaller count if ( arr [ i ] < arr [ idx ] ) result ++ ; // If element is equal then increase // count only if it occurs before if ( arr [ i ] == arr [ idx ] && i < idx ) result ++ ; } return result ; } // Driver code to test above methods public static void main ( String [] args ) { int arr [] = { 3 4 3 5 2 3 4 3 1 5 }; int n = arr . length ; int idxOfEle = 5 ; System . out . println ( getIndexInSortedArray ( arr n idxOfEle )); } } // This code is contributed by Raghav sharma
Python3 # Python program to get index of array element in # sorted array # Method returns the position of arr[idx] after # performing stable-sort on array def getIndexInSortedArray ( arr n idx ): # Count of elements smaller than current # element plus the equal element occurring # before given index result = 0 for i in range ( n ): # If element is smaller then increase # the smaller count if ( arr [ i ] < arr [ idx ]): result += 1 # If element is equal then increase count # only if it occurs before if ( arr [ i ] == arr [ idx ] and i < idx ): result += 1 return result ; # Driver code to test above methods arr = [ 3 4 3 5 2 3 4 3 1 5 ] n = len ( arr ) idxOfEle = 5 print ( getIndexInSortedArray ( arr n idxOfEle )) # Contributed by: Afzal Ansari
C# // C# program to get index of array // element in sorted array using System ; class ArrayIndex { // Method returns the position of // arr[idx] after performing stable-sort // on array static int getIndexInSortedArray ( int [] arr int n int idx ) { /* Count of elements smaller than current element plus the equal element occurring before given index*/ int result = 0 ; for ( int i = 0 ; i < n ; i ++ ) { // If element is smaller then // increase the smaller count if ( arr [ i ] < arr [ idx ]) result ++ ; // If element is equal then increase // count only if it occurs before if ( arr [ i ] == arr [ idx ] && i < idx ) result ++ ; } return result ; } // Driver code to test above methods public static void Main () { int [] arr = { 3 4 3 5 2 3 4 3 1 5 }; int n = arr . Length ; int idxOfEle = 5 ; Console . WriteLine ( getIndexInSortedArray ( arr n idxOfEle )); } } // This code is contributed by vt_m
PHP // PHP program to get index of // array element in sorted array // Method returns the position of // arr[idx] after performing // stable-sort on array function getIndexInSortedArray ( $arr $n $idx ) { /* Count of elements smaller than current element plus the equal element occurring before given index */ $result = 0 ; for ( $i = 0 ; $i < $n ; $i ++ ) { // If element is smaller then // increase the smaller count if ( $arr [ $i ] < $arr [ $idx ]) $result ++ ; // If element is equal then // increase count only if // it occurs before if ( $arr [ $i ] == $arr [ $idx ] and $i < $idx ) $result ++ ; } return $result ; } // Driver Code $arr = array ( 3 4 3 5 2 3 4 3 1 5 ); $n = count ( $arr ); $idxOfEle = 5 ; echo getIndexInSortedArray ( $arr $n $idxOfEle ); // This code is contributed by anuj_67. ?>
JavaScript < script > // JavaScript program to get index of array // element in sorted array // Method returns the position of // arr[idx] after performing stable-sort // on array function getIndexInSortedArray ( arr n idx ) { /* Count of elements smaller than current element plus the equal element occurring before given index*/ let result = 0 ; for ( let i = 0 ; i < n ; i ++ ) { // If element is smaller then // increase the smaller count if ( arr [ i ] < arr [ idx ]) result ++ ; // If element is equal then increase // count only if it occurs before if ( arr [ i ] == arr [ idx ] && i < idx ) result ++ ; } return result ; } // Driver Code let arr = [ 3 4 3 5 2 3 4 3 1 5 ]; let n = arr . length ; let idxOfEle = 5 ; document . write ( getIndexInSortedArray ( arr n idxOfEle )); // This code is contributed by code_hunt. < /script>
Izvade
4
Laika sarežģītība: O(n) kur n ir masīva lielums.
Palīgtelpa: O(1)
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