Posición de un elemento después de una clasificación estable
Pruébalo en GfG Practice 
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Producción
#practiceLinkDiv { mostrar: ninguno !importante; } Dada una matriz de números enteros que puede contener elementos duplicados, se nos proporciona un elemento de esta matriz, debemos indicar la posición final de este elemento en la matriz si se aplica un algoritmo de clasificación estable.
Ejemplos:
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 Clasificación y posición estables ¡Pruébalo!
Una forma sencilla de resolver este problema es utilizar cualquier algoritmo de clasificación estable como Orden de inserción Ordenar va etc. y luego obtener el nuevo índice del elemento dado, pero podemos resolver este problema sin ordenar la matriz.
Como la posición de un elemento en una matriz ordenada la deciden solo aquellos elementos que son más pequeños que el elemento dado. Contamos todos los elementos de la matriz más pequeños que el elemento dado y para aquellos elementos que son iguales a los elementos dados, los elementos que ocurren antes del índice de los elementos dados se incluirán en el recuento de elementos más pequeños, esto asegurará la estabilidad del índice del resultado.
El código simple para implementar el enfoque anterior se implementa a continuación:
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>
Producción
4
Complejidad del tiempo: En) donde n es el tamaño de la matriz.
Espacio auxiliar: O(1)
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