Recuento de superadores de cada elemento en la matriz
Dada una serie de números enteros distintos arr[] a superar de un elemento arr[i] es cualquier elemento arr[j] tal que j > i y arr[j] > arr[i]. Encuentre el número de superadores para cada elemento de la matriz.
Ejemplos:
Aporte: arreglo[] = [2 7 5 3 8 1]
Producción: [4 1 1 1 0 0]
Explicación: Para 2 hay 4 elementos mayores a su derecha: [7 5 3 8]
Para 7 hay 1 elemento mayor a su derecha: [8]
Para 5 hay 1 elemento mayor a su derecha: [8]
Para 3 hay 1 elemento mayor a su derecha: [8]
Para 8 no hay mayor elemento a su derecha: [0]
Para 1 no hay elemento mayor a su derecha: [0]Aporte: arreglo[] = [4 5 1]
Producción: [1 0 0]
Explicación: Para 4 hay 1 elemento mayor a su derecha: [5]
Para 5 no hay mayor elemento a su derecha: [0]
Para 1 no hay elemento mayor a su derecha: [0]
Tabla de contenido
- [Enfoque ingenuo] - Uso de dos bucles anidados: O(n^2) tiempo y O(1) espacio
- [Enfoque esperado] - Uso del paso de combinación de ordenación de combinación - O(n Log n) Tiempo y O(n) Espacio
[Enfoque ingenuo] - Uso de dos bucles anidados: O(n^2) tiempo y O(1) espacio
El enfoque más simple es iterar a través de la matriz y para cada elemento contar el número de elementos a su bien eso es mayor que que eso.
C++ #include #include using namespace std ; vector < int > findSurpasser ( vector < int > & arr ) { int n = arr . size (); // array to store surpasser counts vector < int > res ( n 0 ); for ( int i = 0 ; i < n ; i ++ ) { // Find surpasser for arr[i] int count = 0 ; for ( int j = i + 1 ; j < n ; j ++ ) { if ( arr [ j ] > arr [ i ]) count ++ ; } res [ i ] = count ; } return res ; } int main () { vector < int > arr = { 2 7 5 3 8 1 }; vector < int > res = findSurpasser ( arr ); for ( int count : res ) cout < < count < < ' ' ; return 0 ; }
C #include #include int * findSurpasser ( int arr [] int n ) { // Array to store surpasser counts int * res = ( int * ) malloc ( n * sizeof ( int )); for ( int i = 0 ; i < n ; i ++ ) { // Find surpasser for arr[i] int count = 0 ; for ( int j = i + 1 ; j < n ; j ++ ) { if ( arr [ j ] > arr [ i ]) count ++ ; } res [ i ] = count ; } return res ; } int main () { int arr [] = { 2 7 5 3 8 1 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); int * res = findSurpasser ( arr n ); for ( int i = 0 ; i < n ; i ++ ) printf ( '%d ' res [ i ]); return 0 ; }
Java import java.util.ArrayList ; class GfG { static ArrayList < Integer > findSurpasser ( int [] arr ) { int n = arr . length ; // List to store surpasser counts ArrayList < Integer > res = new ArrayList <> (); for ( int i = 0 ; i < n ; i ++ ) { // Find surpasser for arr[i] int count = 0 ; for ( int j = i + 1 ; j < n ; j ++ ) { if ( arr [ j ] > arr [ i ] ) count ++ ; } res . add ( count ); } return res ; } public static void main ( String [] args ) { int [] arr = { 2 7 5 3 8 1 }; ArrayList < Integer > res = findSurpasser ( arr ); for ( int count : res ) System . out . print ( count + ' ' ); } }
Python def findSurpasser ( arr ): n = len ( arr ) # List to store surpasser counts res = [ 0 ] * n for i in range ( n ): # Find surpasser for arr[i] count = 0 for j in range ( i + 1 n ): if arr [ j ] > arr [ i ]: count += 1 res [ i ] = count return res if __name__ == '__main__' : arr = [ 2 7 5 3 8 1 ] result = findSurpasser ( arr ) for val in result : print ( val end = ' ' )
C# using System ; using System.Collections.Generic ; class GfG { static List < int > findSurpasser ( int [] arr ) { int n = arr . Length ; // List to store surpasser counts List < int > res = new List < int > (); for ( int i = 0 ; i < n ; i ++ ) { // Find surpasser for arr[i] int count = 0 ; for ( int j = i + 1 ; j < n ; j ++ ) { if ( arr [ j ] > arr [ i ]) count ++ ; } res . Add ( count ); } return res ; } static void Main ( string [] args ) { int [] arr = { 2 7 5 3 8 1 }; List < int > result = findSurpasser ( arr ); foreach ( int count in result ) Console . Write ( count + ' ' ); } }
JavaScript function findSurpasser ( arr ) { const n = arr . length ; // array to store surpasser counts let res = new Array ( n ). fill ( 0 ); for ( let i = 0 ; i < n ; i ++ ) { // Find surpasser for arr[i] let count = 0 ; for ( let j = i + 1 ; j < n ; j ++ ) { if ( arr [ j ] > arr [ i ]) count ++ ; } res [ i ] = count ; } return res ; } // Driver Code const arr = [ 2 7 5 3 8 1 ]; const result = findSurpasser ( arr ); console . log ( result . join ( ' ' ));
Producción
4 1 1 1 0 0
[Enfoque esperado] - Uso del paso de combinación de ordenación de combinación - O(n Log n) Tiempo y O(n) Espacio
En este enfoque utilizamos el paso de fusión de Paseos . Durante la fusión si el con elemento en la mitad izquierda es menor que el jth elemento en la mitad derecha significa que todos restante elementos en la mitad derecha (desde j hasta el final) son mayor que que el con elemento en la mitad izquierda (ya que ambas mitades son ordenado ). Por lo tanto sumamos el número de elementos restantes en la mitad derecha a la superar el conteo del con elemento de la mitad izquierda. Este proceso se repite para todos los elementos de la mitad izquierda a medida que se fusionan con la mitad derecha.
Dado que todos los elementos de la matriz son distinto usamos un mapa hash para mantener almacenado el recuento de superadores para cada elemento. Una vez completada la clasificación de combinación, llenamos la matriz de resultados con los recuentos de superadores manteniendo el mismo orden que la entrada original.
C++ #include #include #include using namespace std ; // Merge function to sort the array and update surpasser counts int merge ( vector < int > & arr int lo int mid int hi unordered_map < int int > & m ) { int n1 = mid - lo + 1 ; int n2 = hi - mid ; vector < int > left ( n1 ) right ( n2 ); // Copy data into temporary arrays left[] and right[] for ( int i = 0 ; i < n1 ; i ++ ) left [ i ] = arr [ lo + i ]; for ( int j = 0 ; j < n2 ; j ++ ) right [ j ] = arr [ mid + 1 + j ]; int i = 0 j = 0 k = lo ; // Merging two halves while ( i < n1 && j < n2 ) { // All elements in right[j..n2] are greater than left[i] // So add n2 - j in surpasser count of left[i] if ( left [ i ] < right [ j ]) { m [ left [ i ]] += n2 - j ; arr [ k ++ ] = left [ i ++ ]; } else { arr [ k ++ ] = right [ j ++ ]; } } // Copy remaining elements of left[] if any while ( i < n1 ) arr [ k ++ ] = left [ i ++ ]; // Copy remaining elements of right[] if any while ( j < n2 ) arr [ k ++ ] = right [ j ++ ]; } void mergeSort ( vector < int > & arr int lo int hi unordered_map < int int > & m ) { if ( lo < hi ) { int mid = lo + ( hi - lo ) / 2 ; // Sort left and right half mergeSort ( arr lo mid m ); mergeSort ( arr mid + 1 hi m ); // Merge them merge ( arr lo mid hi m ); } } vector < int > findSurpasser ( vector < int >& arr ) { int n = arr . size (); // Map to store surpasser counts unordered_map < int int > m ; // Duplicate array to perform merge Sort // so that orginial array is not modified vector < int > dup = arr ; mergeSort ( dup 0 n - 1 m ); // Store surpasser counts in result array // in the same order as given array vector < int > res ( n ); for ( int i = 0 ; i < n ; i ++ ) res [ i ] = m [ arr [ i ]]; return res ; } int main () { vector < int > arr = { 2 7 5 3 8 1 }; vector < int > res = findSurpasser ( arr ); for ( int count : res ) cout < < count < < ' ' ; return 0 ; }
Java import java.util.List ; import java.util.Map ; import java.util.HashMap ; import java.util.ArrayList ; class GfG { // Merge function to sort the array // and update surpasser counts static void merge ( int [] arr int lo int mid int hi Map < Integer Integer > m ) { int n1 = mid - lo + 1 ; int n2 = hi - mid ; int [] left = new int [ n1 ] ; int [] right = new int [ n2 ] ; // Copy data into temporary arrays left[] and right[] for ( int i = 0 ; i < n1 ; i ++ ) left [ i ] = arr [ lo + i ] ; for ( int j = 0 ; j < n2 ; j ++ ) right [ j ] = arr [ mid + 1 + j ] ; int i = 0 j = 0 k = lo ; // Merging two halves while ( i < n1 && j < n2 ) { // All elements in right[j..n2] are greater than left[i] // So add n2 - j to surpasser count of left[i] if ( left [ i ] < right [ j ] ) { m . put ( left [ i ] m . getOrDefault ( left [ i ] 0 ) + n2 - j ); arr [ k ++] = left [ i ++] ; } else { arr [ k ++] = right [ j ++] ; } } // Copy remaining elements of left[] if any while ( i < n1 ) arr [ k ++] = left [ i ++] ; // Copy remaining elements of right[] if any while ( j < n2 ) arr [ k ++] = right [ j ++] ; } static void mergeSort ( int [] arr int lo int hi Map < Integer Integer > m ) { if ( lo < hi ) { int mid = lo + ( hi - lo ) / 2 ; // Sort left and right halves mergeSort ( arr lo mid m ); mergeSort ( arr mid + 1 hi m ); // Merge them merge ( arr lo mid hi m ); } } static ArrayList < Integer > findSurpasser ( int [] arr ) { int n = arr . length ; // Map to store surpasser counts Map < Integer Integer > m = new HashMap <> (); // Duplicate array to perform merge sort int [] dup = arr . clone (); mergeSort ( dup 0 n - 1 m ); // Store surpasser counts in result list ArrayList < Integer > res = new ArrayList <> (); for ( int i = 0 ; i < n ; i ++ ) res . add ( m . getOrDefault ( arr [ i ] 0 )); return res ; } public static void main ( String [] args ) { int [] arr = { 2 7 5 3 8 1 }; ArrayList < Integer > res = findSurpasser ( arr ); for ( int count : res ) System . out . print ( count + ' ' ); } }
Python def merge ( arr lo mid hi m ): n1 = mid - lo + 1 n2 = hi - mid left = arr [ lo : lo + n1 ] right = arr [ mid + 1 : mid + 1 + n2 ] i = j = 0 k = lo # Merging two halves while i < n1 and j < n2 : # All elements in right[j..n2] are greater than left[i] # So add n2 - j in surpasser count of left[i] if left [ i ] < right [ j ]: m [ left [ i ]] += n2 - j arr [ k ] = left [ i ] i += 1 else : arr [ k ] = right [ j ] j += 1 k += 1 # Copy remaining elements of left[] if any while i < n1 : arr [ k ] = left [ i ] i += 1 k += 1 # Copy remaining elements of right[] if any while j < n2 : arr [ k ] = right [ j ] j += 1 k += 1 def mergeSort ( arr lo hi m ): if lo < hi : mid = lo + ( hi - lo ) // 2 # Sort left and right half mergeSort ( arr lo mid m ) mergeSort ( arr mid + 1 hi m ) # Merge them merge ( arr lo mid hi m ) def findSurpasser ( arr ): n = len ( arr ) # Map to store surpasser counts m = { key : 0 for key in arr } # Duplicate array to perform merge Sort # so that original array is not modified dup = arr [:] mergeSort ( dup 0 n - 1 m ) # Store surpasser counts in result array # in the same order as given array res = [ m [ arr [ i ]] for i in range ( n )] return res if __name__ == '__main__' : arr = [ 2 7 5 3 8 1 ] result = findSurpasser ( arr ) for val in result : print ( val end = ' ' )
C# using System ; using System.Collections.Generic ; class GfG { // Merge function to sort the array // and update surpasser counts static void merge ( int [] arr int lo int mid int hi Dictionary < int int > m ) { int n1 = mid - lo + 1 ; int n2 = hi - mid ; int [] left = new int [ n1 ]; int [] right = new int [ n2 ]; // Copy data into temporary arrays for ( int i = 0 ; i < n1 ; i ++ ) left [ i ] = arr [ lo + i ]; for ( int j = 0 ; j < n2 ; j ++ ) right [ j ] = arr [ mid + 1 + j ]; int i1 = 0 j1 = 0 k = lo ; // Merge the two halves while ( i1 < n1 && j1 < n2 ) { // Count surpassers if ( left [ i1 ] < right [ j1 ]) { if ( ! m . ContainsKey ( left [ i1 ])) m [ left [ i1 ]] = 0 ; m [ left [ i1 ]] += n2 - j1 ; arr [ k ++ ] = left [ i1 ++ ]; } else { arr [ k ++ ] = right [ j1 ++ ]; } } // Copy remaining elements while ( i1 < n1 ) arr [ k ++ ] = left [ i1 ++ ]; while ( j1 < n2 ) arr [ k ++ ] = right [ j1 ++ ]; } static void mergeSort ( int [] arr int lo int hi Dictionary < int int > m ) { if ( lo < hi ) { int mid = lo + ( hi - lo ) / 2 ; // Recursive sort mergeSort ( arr lo mid m ); mergeSort ( arr mid + 1 hi m ); // Merge sorted halves merge ( arr lo mid hi m ); } } static List < int > findSurpasser ( int [] arr ) { int n = arr . Length ; Dictionary < int int > m = new Dictionary < int int > (); int [] dup = new int [ n ]; Array . Copy ( arr dup n ); mergeSort ( dup 0 n - 1 m ); List < int > res = new List < int > (); for ( int i = 0 ; i < n ; i ++ ) { res . Add ( m . ContainsKey ( arr [ i ]) ? m [ arr [ i ]] : 0 ); } return res ; } static void Main () { int [] arr = { 2 7 5 3 8 1 }; List < int > res = findSurpasser ( arr ); foreach ( int count in res ) Console . Write ( count + ' ' ); } }
JavaScript function merge ( arr lo mid hi m ) { const n1 = mid - lo + 1 ; const n2 = hi - mid ; const left = []; const right = []; // Copy data into temporary arrays left[] and right[] for ( let i = 0 ; i < n1 ; i ++ ) left [ i ] = arr [ lo + i ]; for ( let j = 0 ; j < n2 ; j ++ ) right [ j ] = arr [ mid + 1 + j ]; let i = 0 j = 0 k = lo ; // Merging two halves while ( i < n1 && j < n2 ) { // All elements in right[j..n2] are greater than left[i] // So add n2 - j in surpasser count of left[i] if ( left [ i ] < right [ j ]) { m [ left [ i ]] = ( m [ left [ i ]] || 0 ) + n2 - j ; arr [ k ++ ] = left [ i ++ ]; } else { arr [ k ++ ] = right [ j ++ ]; } } // Copy remaining elements of left[] if any while ( i < n1 ) arr [ k ++ ] = left [ i ++ ]; // Copy remaining elements of right[] if any while ( j < n2 ) arr [ k ++ ] = right [ j ++ ]; } function mergeSort ( arr lo hi m ) { if ( lo < hi ) { const mid = Math . floor ( lo + ( hi - lo ) / 2 ); // Sort left and right half mergeSort ( arr lo mid m ); mergeSort ( arr mid + 1 hi m ); // Merge them merge ( arr lo mid hi m ); } } function findSurpasser ( arr ) { const n = arr . length ; // Map to store surpasser counts const m = {}; // Duplicate array to perform merge Sort // so that original array is not modified const dup = arr . slice (); mergeSort ( dup 0 n - 1 m ); // Store surpasser counts in result array // in the same order as given array const res = []; for ( let i = 0 ; i < n ; i ++ ) res . push ( m [ arr [ i ]] || 0 ); return res ; } // Driver Code const arr = [ 2 7 5 3 8 1 ]; const res = findSurpasser ( arr ); console . log ( res . join ( ' ' ));
Producción
4 1 1 1 0 0
Artículos relacionados:
- Recuento de inversiones
- Paseos
