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grep komanda Linux
Input : arr[] = {1 5 7 2 0}; Output : (0 1) and (2 4) Subarrays arr[0..1] and arr[2..4] have same average. Input : arr[] = {4 3 5 9 11}; Output : Not possible A Naiva pieeja ir palaist divas cilpas un atrast apakšgrupas, kuru vidējie rādītāji ir vienādi.
Īstenošana:
C++// Simple C++ program to find subarrays // whose averages are equal #include
// Simple Java program to find subarrays // whose averages are equal public class GFG { // Finding two subarrays // with equal average. static void findSubarrays(int[] arr int n) { boolean found = false; int lsum = 0; for (int i = 0; i < n - 1; i++) { lsum += arr[i]; int rsum = 0; for (int j = i + 1; j < n; j++) rsum += arr[j]; // If averages of arr[0...i] and // arr[i+1..n-1] are same. To avoid // floating point problems we compare // 'lsum*(n-i+1)' and 'rsum*(i+1)' // instead of 'lsum/(i+1)' and // 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)) { System.out.println('From (0 ' + i + ') to (' +(i + 1) + ' ' + (n - 1)+ ')'); found = true; } } // If no subarrays found if (found == false) System.out.println( 'Subarrays not ' + 'found'); } // Driver code static public void main (String[] args) { int[] arr = {1 5 7 2 0}; int n = arr.length; findSubarrays(arr n); } } // This code is contributed by Mukul Singh.
# Simple Python 3 program to find subarrays # whose averages are equal # Finding two subarrays with equal average. def findSubarrays(arr n): found = False lsum = 0 for i in range(n - 1): lsum += arr[i] rsum = 0 for j in range(i + 1 n): rsum += arr[j] # If averages of arr[0...i] and # arr[i+1..n-1] are same. To avoid # floating point problems we compare # 'lsum*(n-i+1)' and 'rsum*(i+1)' # instead of 'lsum/(i+1)' and # 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)): print('From' '(' 0 i ')' 'to' '(' i + 1 n - 1 ')') found = True # If no subarrays found if (found == False): print('Subarrays not found') # Driver code if __name__ == '__main__': arr = [1 5 7 2 0] n = len(arr) findSubarrays(arr n) # This code is contributed by ita_c
// Simple C# program to find subarrays // whose averages are equal using System; public class GFG { // Finding two subarrays // with equal average. static void findSubarrays(int []arr int n) { bool found = false; int lsum = 0; for (int i = 0; i < n - 1; i++) { lsum += arr[i]; int rsum = 0; for (int j = i + 1; j < n; j++) rsum += arr[j]; // If averages of arr[0...i] and // arr[i+1..n-1] are same. To avoid // floating point problems we compare // 'lsum*(n-i+1)' and 'rsum*(i+1)' // instead of 'lsum/(i+1)' and // 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)) { Console.WriteLine('From ( 0 ' + i + ') to(' + (i + 1) + ' ' + (n - 1) + ')'); found = true; } } // If no subarrays found if (found == false) Console.WriteLine( 'Subarrays not ' + 'found'); } // Driver code static public void Main () { int []arr = {1 5 7 2 0}; int n = arr.Length; findSubarrays(arr n); } } // This code is contributed by anuj_67.
// Simple PHP program to find subarrays // whose averages are equal // Finding two subarrays // with equal average. function findSubarrays( $arr $n) { $found = false; $lsum = 0; for ( $i = 0; $i < $n - 1; $i++) { $lsum += $arr[$i]; $rsum = 0; for ( $j = $i + 1; $j < $n; $j++) $rsum += $arr[$j]; // If averages of arr[0...i] and // arr[i+1..n-1] are same. To avoid // floating point problems we compare // 'lsum*(n-i+1)' and 'rsum*(i+1)' // instead of 'lsum/(i+1)' and 'rsum/(n-i+1)' if ($lsum * ($n - $i - 1) == $rsum * ($i + 1)) { echo 'From ( 0 ' $i' )'. ' to (' $i + 1' ' $n - 1')n'; $found = true; } } // If no subarrays found if ($found == false) echo 'Subarrays not found' ; } // Driver code $arr = array(1 5 7 2 0); $n = count($arr); findSubarrays($arr $n); // This code is contributed by vt_m ?> <script> // Simple Javascript program to find subarrays // whose averages are equal // Finding two subarrays // with equal average. function findSubarrays(arrn) { let found = false; let lsum = 0; for (let i = 0; i < n - 1; i++) { lsum += arr[i]; let rsum = 0; for (let j = i + 1; j < n; j++) rsum += arr[j]; // If averages of arr[0...i] and // arr[i+1..n-1] are same. To avoid // floating point problems we compare // 'lsum*(n-i+1)' and 'rsum*(i+1)' // instead of 'lsum/(i+1)' and // 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)) { document.write('From (0 ' + i + ') to (' +(i + 1) + ' ' + (n - 1)+ ')'); found = true; } } // If no subarrays found if (found == false) document.write( 'Subarrays not ' + 'found'); } // Driver code let arr=[1 5 7 2 0]; let n = arr.length; findSubarrays(arr n); // This code is contributed by avanitrachhadiya2155 </script>
Izvade
From (0 1) to (2 4)
Laika sarežģītība: O(n2)
Palīgtelpa: O(1)
An Efektīvs risinājums ir atrast masīva elementu summu. Inicializējiet Leftsum kā nulli. Palaidiet cilpu un atrodiet kreiso summu, pievienojot masīva elementus. Tiesību summai mēs no kopējās summas atņemam leftum, tad atrodam rightum un atrodam vidējo kreiso un rightum kā Saskaņā ar to indeksu.
1) Compute sum of all array elements. Let this sum be 'sum' 2) Initialize leftsum = 0. 3) Run a loop for i=0 to n-1. a) leftsum = leftsum + arr[i] b) rightsum = sum - leftsum c) If average of left and right are same print current index as output.
Tālāk ir norādīta iepriekš minētās pieejas īstenošana:
ceļš iestatīts JavaC++
// Efficient C++ program for // dividing array to make // average equal #include
// Efficient Java program for // dividing array to make // average equal import java.util.*; class GFG { static void findSubarrays(int arr[] int n) { // Find array sum int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; boolean found = false; int lsum = 0; for (int i = 0; i < n - 1; i++) { lsum += arr[i]; int rsum = sum - lsum; // If averages of arr[0...i] // and arr[i+1..n-1] are same. // To avoid floating point problems // we compare 'lsum*(n-i+1)' // and 'rsum*(i+1)' instead of // 'lsum/(i+1)' and 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)) { System.out.printf('From (%d %d) to (%d %d)n' 0 i i + 1 n - 1); found = true; } } // If no subarrays found if (found == false) System.out.println('Subarrays not found'); } // Driver code static public void main ( String []arg) { int arr[] = {1 5 7 2 0}; int n = arr.length; findSubarrays(arr n); } } // This code is contributed by Princi Singh
# Efficient Python program for # dividing array to make # average equal def findSubarrays(arr n): # Find array sum sum = 0; for i in range(n): sum += arr[i]; found = False; lsum = 0; for i in range(n - 1): lsum += arr[i]; rsum = sum - lsum; # If averages of arr[0...i] # and arr[i + 1..n - 1] are same. # To avoid floating point problems # we compare 'lsum*(n - i + 1)' # and 'rsum*(i + 1)' instead of # 'lsum / (i + 1)' and 'rsum/(n - i + 1)' if (lsum * (n - i - 1) == rsum * (i + 1)): print('From (%d %d) to (%d %d)n'% (0 i i + 1 n - 1)); found = True; # If no subarrays found if (found == False): print('Subarrays not found'); # Driver code if __name__ == '__main__': arr = [ 1 5 7 2 0 ]; n = len(arr); findSubarrays(arr n); # This code is contributed by Rajput-Ji
// Efficient C# program for // dividing array to make // average equal using System; class GFG { static void findSubarrays(int []arr int n) { // Find array sum int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; bool found = false; int lsum = 0; for (int i = 0; i < n - 1; i++) { lsum += arr[i]; int rsum = sum - lsum; // If averages of arr[0...i] // and arr[i+1..n-1] are same. // To avoid floating point problems // we compare 'lsum*(n-i+1)' // and 'rsum*(i+1)' instead of // 'lsum/(i+1)' and 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)) { Console.Write('From ({0} {1}) to ({2} {3})n' 0 i i + 1 n - 1); found = true; } } // If no subarrays found if (found == false) Console.WriteLine('Subarrays not found'); } // Driver code static public void Main ( String []arg) { int []arr = {1 5 7 2 0}; int n = arr.Length; findSubarrays(arr n); } } // This code is contributed by Rajput-Ji
<script> // Efficient Javascript program for // dividing array to make // average equal function findSubarrays(arrn) { // Find array sum let sum = 0; for (let i = 0; i < n; i++) sum += arr[i]; let found = false; let lsum = 0; for (let i = 0; i < n - 1; i++) { lsum += arr[i]; let rsum = sum - lsum; // If averages of arr[0...i] // and arr[i+1..n-1] are same. // To avoid floating point problems // we compare 'lsum*(n-i+1)' // and 'rsum*(i+1)' instead of // 'lsum/(i+1)' and 'rsum/(n-i+1)' if (lsum * (n - i - 1) == rsum * (i + 1)) { document.write( 'From (0 '+i+') to ('+(i+1)+' '+(n-1)+')n' ); found = true; } } // If no subarrays found if (found == false) document.write('Subarrays not found'); } // Driver code let arr=[1 5 7 2 0]; let n = arr.length; findSubarrays(arr n); // This code is contributed by rag2127 </script>
Izvade
From (0 1) to (2 4)
Laika sarežģītība: O(n)
Palīgtelpa: O(1)
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