Elemento padėtis po stabilaus rūšiavimo
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Išvestis
#practiceLinkDiv { display: none !important; } Atsižvelgiant į sveikųjų skaičių masyvą, kuriame gali būti pasikartojančių elementų, mums pateikiamas šio masyvo elementas, turime pasakyti galutinę šio elemento padėtį masyve, jei taikomas stabilus rūšiavimo algoritmas.
Pavyzdžiai:
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 Stabilus rūšiavimas ir padėtis Išbandykite!
Vienas paprastas būdas išspręsti šią problemą yra naudoti bet kokį stabilų rūšiavimo algoritmą, pvz Įterpimo rūšiavimas Rūšiuoti eina ir tt, tada gaukite naują nurodyto elemento indeksą, tačiau šią problemą galime išspręsti nerūšiuodami masyvo.
Elemento vietą surūšiuotame masyve sprendžia tik tie elementai, kurie yra mažesni už nurodytą elementą. Skaičiuojame visus masyvo elementus, mažesnius už nurodytą elementą, o tiems elementams, kurie yra lygūs duotiesiems elementams, atsiradusiems prieš duotų elementų indeksą, bus įtraukti į mažesnių elementų skaičių, tai užtikrins rezultato indekso stabilumą.
Paprastas kodas, skirtas įgyvendinti aukščiau pateiktą metodą, yra įdiegtas toliau:
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>
Išvestis
4
Laiko sudėtingumas: O(n) kur n yra masyvo dydis.
Pagalbinė erdvė: O(1)
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