Overgangstælling af hvert element i array
Givet en række forskellige heltals arr [] a overgå af et element arr [i] er ethvert element arr [j] således at j> i og arr [j]> arr [i]. Find antallet af overgange for hvert element i matrixen.
Eksempler:
Input: ARR [] = [2 7 5 3 8 1]
Produktion: [4 1 1 1 0 0]
Forklaring: For 2 er der 4 større elementer til højre: [7 5 3 8]
For 7 er der 1 større element til højre: [8]
For 5 er der 1 større element til højre: [8]
For 3 er der 1 større element til højre: [8]
For 8 er der ikke noget større element til højre: [0]
For 1 er der ikke noget større element til højre: [0]Input: arr [] = [4 5 1]
Produktion: [1 0 0]
Forklaring: For 4 er der 1 større element til højre: [5]
For 5 er der ikke noget større element til højre: [0]
For 1 er der ikke noget større element til højre: [0]
Indholdstabel
- [Naiv tilgang] - Brug af to indlejrede løkker - o (n^2) tid og o (1) plads
- [Forventet tilgang] - Brug af Merge Trin of Merge Sort - O (n log n) Tid og O (n) plads
[Naiv tilgang] - Brug af to indlejrede løkker - o (n^2) tid og o (1) plads
Den enkleste tilgang er at iterat gennem arrayet og for hvert element tælle antallet af elementer til dets højre det er større end det.
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 ( ' ' ));
Produktion
4 1 1 1 0 0
[Forventet tilgang] - Brug af Merge Trin of Merge Sort - O (n log n) Tid og O (n) plads
I denne tilgang bruger vi Flet trin af Gåture . Under fusionen, hvis med element i den venstre halvdel er mindre end jth element i højre halvdel betyder det, at alle Resterende elementer i højre halvdel (fra J til slut) er større end med element i venstre halvdel (da begge halvdele er sorteret ). Derfor tilføjer vi antallet af Resterende elementer i højre halvdel til overgå tælling af med element i venstre halvdel. Denne proces gentages for alle elementer i venstre halvdel, da de er fusioneret med den højre halvdel.
Da alle elementer i matrixen er særskilt Vi bruger en hash kort For at holde butikken for overgodtællingen for hvert element. Efter at fletningssorteringen er komplet, fylder vi resultatarrayet med overgange tællinger, der opretholder den samme rækkefølge som det originale input.
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 ( ' ' ));
Produktion
4 1 1 1 0 0
Relaterede artikler:
- Inversionstælling
- Gåture
