Posició d'un element després de l'ordenació estable
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#practiceLinkDiv { mostrar: cap !important; } Donada una matriu d'enters que pot contenir elements duplicats, se'ns dóna un element d'aquesta matriu, hem d'indicar la posició final d'aquest element a la matriu si s'aplica un algorisme d'ordenació estable.
Exemples:
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 Classificació i posició estables Prova-ho!
Una manera senzilla de resoldre aquest problema és utilitzar qualsevol algorisme d'ordenació estable com Ordenació d'inserció L'ordenació va etc. i després obteniu el nou índex de l'element donat, però podem resoldre aquest problema sense ordenar la matriu.
Com que la posició d'un element en una matriu ordenada només es decideix per aquells elements que són més petits que l'element donat. Comptem tots els elements de la matriu més petits que l'element donat i per a aquells elements que són iguals als elements d'un element donat que es produeixen abans que l'índex d'elements donats s'inclou en el recompte d'elements més petits, això assegurarà l'estabilitat de l'índex del resultat.
El codi senzill per implementar l'enfocament anterior s'implementa a continuació:
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>
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4
Complexitat temporal: O(n) on n és la mida de la matriu.
Espai auxiliar: O(1)
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