Givet en heltalsmatris är uppgiften att dela upp en heltalsmatris i två delmatriser för att göra deras medelvärden lika om möjligt.
Exempel:
python lista initialisera
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 Naivt förhållningssätt är att köra två slingor och hitta subarrayer vars medelvärden är lika.
Genomförande:
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>
Produktion
From (0 1) to (2 4)
Tidskomplexitet : O(n2)
Hjälputrymme: O(1)
En Effektiv lösning är att hitta summan av matriselement. Initiera Leftsum som noll. Kör en slinga och hitta Leftsum genom att lägga till element av array. För rättsumman subtraherar vi leavesumma från totalsumman, då hittar vi rättsumma och hittar medelvärdet av vänster och högersumma enligt deras index.
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.
Nedan är implementeringen för ovanstående tillvägagångssätt:
C++// 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>
Produktion
From (0 1) to (2 4)
Tidskomplexitet: O(n)
Hjälputrymme: O(1)
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