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Grootste som aaneengesloten subarray (algoritme van Kadane)

Gegeven een array arr[] van grootte N . De taak is om de som van de aangrenzende subarray binnen a te vinden arr[] met het grootste bedrag.

Voorbeeld:



Invoer: arr = {-2,-3,4,-1,-2,1,5,-3}
Uitgang: 7
Uitleg: De subarray {4,-1, -2, 1, 5} heeft de grootste som 7.

Invoer: arr = {2}
Uitgang: 2
Uitleg: De subarray {2} heeft de grootste som 1.

Invoer: arr = {5,4,1,7,8}
Uitgang: 25
Uitleg: De subarray {5,4,1,7,8} heeft de grootste som 25.



kadane-algoritme

Aanbevolen probleem Probleem oplossen

Het idee van Kadane's algoritme is het behouden van een variabele max_ending_hier waarin de maximale som van de aaneengesloten subarray wordt opgeslagen die eindigt op de huidige index en een variabele max_so_far slaat de maximale som op van aaneengesloten subarrays die tot nu toe zijn gevonden, elke keer dat er een positieve somwaarde in zit max_ending_hier vergelijk het met max_so_far en bijwerken max_so_far als het groter is dan max_so_far .

Dus de belangrijkste Intuïtie achter Het algoritme van Kadane is,



  • De subarray met een negatieve som wordt weggegooid ( door max_ending_here = 0 in code toe te wijzen ).
  • We dragen de subarray totdat deze een positieve som oplevert.

Pseudocode van Kadane’s algoritme:

Initialiseren:
max_so_far = INT_MIN
max_ending_hier = 0

Lus voor elk element van de array

(a) max_ending_hier = max_ending_hier + a[i]
(b) if(max_so_far
max_so_far = max_ending_hier
(c) if(max_ending_hier <0)
max_ending_hier = 0
retourneer max_so_far

Illustratie van het algoritme van Kadane:

Laten we het voorbeeld nemen: {-2, -3, 4, -1, -2, 1, 5, -3}

Opmerking : in de afbeelding wordt max_so_far weergegeven door Max_Som en max_ending_hier door Curr_Som


Voor i=0 geldt a[0] = -2

wat is een gebruikersnaam
  • max_ending_hier = max_ending_hier + (-2)
  • Stel max_ending_here = 0 in omdat max_ending_here <0
  • en stel max_so_far = -2 in

Voor i=1 geldt a[1] = -3

  • max_ending_hier = max_ending_hier + (-3)
  • Omdat max_ending_here = -3 en max_so_far = -2, blijft max_so_far -2
  • Stel max_ending_here = 0 in omdat max_ending_here <0

Voor i=2, a[2] = 4

  • max_ending_hier = max_ending_hier + (4)
  • max_ending_hier = 4
  • max_so_far is bijgewerkt naar 4 omdat max_ending_her groter is dan max_so_far, wat tot nu toe -2 was

Voor i=3 geldt a[3] = -1

  • max_ending_hier = max_ending_hier + (-1)
  • max_ending_hier = 3

Voor i=4 geldt a[4] = -2

  • max_ending_hier = max_ending_hier + (-2)
  • max_ending_hier = 1

Voor i=5 is a[5] = 1

  • max_ending_hier = max_ending_hier + (1)
  • max_ending_hier = 2

Voor i=6 is a[6] = 5

  • max_ending_hier = max_ending_hier + (5)
  • max_ending_hier =
  • max_so_far is bijgewerkt naar 7 omdat max_ending_here groter is dan max_so_far

Voor i=7 geldt a[7] = -3

  • max_ending_hier = max_ending_hier + (-3)
  • max_ending_hier = 4

Volg de onderstaande stappen om het idee te implementeren:

  • Initialiseer de variabelen max_so_far = INT_MIN en max_ending_hier = 0
  • Voer een for-lus uit vanaf 0 naar N-1 en voor elke index i :
    • Voeg de arr[i] toe naar max_ending_hier.
    • Als max_so_far kleiner is dan max_ending_here, update dan max_so_far naar max_ending_hier .
    • Als max_ending_here <0, update dan max_ending_here = 0
  • Retourneer max_so_far

Hieronder vindt u de implementatie van de bovenstaande aanpak.

C++ 
// C++ program to print largest contiguous array sum #include  using namespace std; int maxSubArraySum(int a[], int size) {  int max_so_far = INT_MIN, max_ending_here = 0;  for (int i = 0; i < size; i++) {  max_ending_here = max_ending_here + a[i];  if (max_so_far < max_ending_here)  max_so_far = max_ending_here;  if (max_ending_here < 0)  max_ending_here = 0;  }  return max_so_far; } // Driver Code int main() {  int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };  int n = sizeof(a) / sizeof(a[0]);  // Function Call  int max_sum = maxSubArraySum(a, n);  cout << 'Maximum contiguous sum is ' << max_sum;  return 0; }>
Java 
// Java program to print largest contiguous array sum import java.io.*; import java.util.*; class Kadane {  // Driver Code  public static void main(String[] args)  {  int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };  System.out.println('Maximum contiguous sum is '  + maxSubArraySum(a));  }  // Function Call  static int maxSubArraySum(int a[])  {  int size = a.length;  int max_so_far = Integer.MIN_VALUE, max_ending_here  = 0;  for (int i = 0; i < size; i++) {  max_ending_here = max_ending_here + a[i];  if (max_so_far < max_ending_here)  max_so_far = max_ending_here;  if (max_ending_here < 0)  max_ending_here = 0;  }  return max_so_far;  } }>
Python 
def GFG(a, size): max_so_far = float('-inf') # Use float('-inf') instead of maxint max_ending_here = 0 for i in range(0, size): max_ending_here = max_ending_here + a[i] if max_so_far < max_ending_here: max_so_far = max_ending_here if max_ending_here < 0: max_ending_here = 0 return max_so_far # Driver function to check the above function a = [-2, -3, 4, -1, -2, 1, 5, -3] print('Maximum contiguous sum is', GFG(a, len(a)))>
C# 
// C# program to print largest // contiguous array sum using System; class GFG {  static int maxSubArraySum(int[] a)  {  int size = a.Length;  int max_so_far = int.MinValue, max_ending_here = 0;  for (int i = 0; i < size; i++) {  max_ending_here = max_ending_here + a[i];  if (max_so_far < max_ending_here)  max_so_far = max_ending_here;  if (max_ending_here < 0)  max_ending_here = 0;  }  return max_so_far;  }  // Driver code  public static void Main()  {  int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };  Console.Write('Maximum contiguous sum is '  + maxSubArraySum(a));  } } // This code is contributed by Sam007_>
Javascript 
>
PHP 
 // PHP program to print largest // contiguous array sum function maxSubArraySum($a, $size) { $max_so_far = PHP_INT_MIN; $max_ending_here = 0; for ($i = 0; $i < $size; $i++) { $max_ending_here = $max_ending_here + $a[$i]; if ($max_so_far < $max_ending_here) $max_so_far = $max_ending_here; if ($max_ending_here < 0) $max_ending_here = 0; } return $max_so_far; } // Driver code $a = array(-2, -3, 4, -1, -2, 1, 5, -3); $n = count($a); $max_sum = maxSubArraySum($a, $n); echo 'Maximum contiguous sum is ' , $max_sum; // This code is contributed by anuj_67. ?>>
Uitvoer 
Maximum contiguous sum is 7>

Tijdcomplexiteit: OP)
Hulpruimte: O(1)

Druk de grootste som aaneengesloten subarray af:

Om de subarray af te drukken met de maximale som die het idee is om te behouden begin index van maximum_sum_ending_hier tegen de huidige index, zodat wanneer dan ook maximale_som_so_far wordt bijgewerkt met maximum_sum_ending_hier vervolgens kunnen de startindex en de eindindex van de subarray worden bijgewerkt begin En huidige index .

Volg de onderstaande stappen om het idee te implementeren:

  • Initialiseer de variabelen S , begin, En einde met 0 En max_so_far = INT_MIN en max_ending_hier = 0
  • Voer een for-lus uit vanaf 0 naar N-1 en voor elke index i :
    • Voeg de arr[i] toe naar max_ending_hier.
    • Als max_so_far kleiner is dan max_ending_here, update dan max_so_far naar max_ending_here en update begin naar S En einde naar i .
    • Als max_ending_here <0, update dan max_ending_here = 0 en S met ik+1 .
  • Waarden uit index afdrukken begin naar einde .

Hieronder vindt u de implementatie van bovenstaande aanpak:

C++ 
// C++ program to print largest contiguous array sum #include  #include  using namespace std; void maxSubArraySum(int a[], int size) {  int max_so_far = INT_MIN, max_ending_here = 0,  start = 0, end = 0, s = 0;  for (int i = 0; i < size; i++) {  max_ending_here += a[i];  if (max_so_far < max_ending_here) {  max_so_far = max_ending_here;  start = s;  end = i;  }  if (max_ending_here < 0) {  max_ending_here = 0;  s = i + 1;  }  }  cout << 'Maximum contiguous sum is ' << max_so_far  << endl;  cout << 'Starting index ' << start << endl  << 'Ending index ' << end << endl; } /*Driver program to test maxSubArraySum*/ int main() {  int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };  int n = sizeof(a) / sizeof(a[0]);  maxSubArraySum(a, n);  return 0; }>
Java 
// Java program to print largest // contiguous array sum import java.io.*; import java.util.*; class GFG {  static void maxSubArraySum(int a[], int size)  {  int max_so_far = Integer.MIN_VALUE,  max_ending_here = 0, start = 0, end = 0, s = 0;  for (int i = 0; i < size; i++) {  max_ending_here += a[i];  if (max_so_far < max_ending_here) {  max_so_far = max_ending_here;  start = s;  end = i;  }  if (max_ending_here < 0) {  max_ending_here = 0;  s = i + 1;  }  }  System.out.println('Maximum contiguous sum is '  + max_so_far);  System.out.println('Starting index ' + start);  System.out.println('Ending index ' + end);  }  // Driver code  public static void main(String[] args)  {  int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };  int n = a.length;  maxSubArraySum(a, n);  } } // This code is contributed by prerna saini>
Python 
# Python program to print largest contiguous array sum from sys import maxsize # Function to find the maximum contiguous subarray # and print its starting and end index def maxSubArraySum(a, size): max_so_far = -maxsize - 1 max_ending_here = 0 start = 0 end = 0 s = 0 for i in range(0, size): max_ending_here += a[i] if max_so_far < max_ending_here: max_so_far = max_ending_here start = s end = i if max_ending_here < 0: max_ending_here = 0 s = i+1 print('Maximum contiguous sum is %d' % (max_so_far)) print('Starting Index %d' % (start)) print('Ending Index %d' % (end)) # Driver program to test maxSubArraySum a = [-2, -3, 4, -1, -2, 1, 5, -3] maxSubArraySum(a, len(a))>
C# 
// C# program to print largest // contiguous array sum using System; class GFG {  static void maxSubArraySum(int[] a, int size)  {  int max_so_far = int.MinValue, max_ending_here = 0,  start = 0, end = 0, s = 0;  for (int i = 0; i < size; i++) {  max_ending_here += a[i];  if (max_so_far < max_ending_here) {  max_so_far = max_ending_here;  start = s;  end = i;  }  if (max_ending_here < 0) {  max_ending_here = 0;  s = i + 1;  }  }  Console.WriteLine('Maximum contiguous '  + 'sum is ' + max_so_far);  Console.WriteLine('Starting index ' + start);  Console.WriteLine('Ending index ' + end);  }  // Driver code  public static void Main()  {  int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };  int n = a.Length;  maxSubArraySum(a, n);  } } // This code is contributed // by anuj_67.>
Javascript 
>
PHP 
 // PHP program to print largest  // contiguous array sum function maxSubArraySum($a, $size) { $max_so_far = PHP_INT_MIN; $max_ending_here = 0; $start = 0; $end = 0; $s = 0; for ($i = 0; $i < $size; $i++) { $max_ending_here += $a[$i]; if ($max_so_far < $max_ending_here) { $max_so_far = $max_ending_here; $start = $s; $end = $i; } if ($max_ending_here < 0) { $max_ending_here = 0; $s = $i + 1; } } echo 'Maximum contiguous sum is '. $max_so_far.'
'; echo 'Starting index '. $start . '
'. 'Ending index ' . $end . '
'; } // Driver Code $a = array(-2, -3, 4, -1, -2, 1, 5, -3); $n = sizeof($a); maxSubArraySum($a, $n); // This code is contributed // by ChitraNayal ?>>
Uitvoer 
Maximum contiguous sum is 7 Starting index 2 Ending index 6>

Tijdcomplexiteit: Op)
Hulpruimte: O(1)

Grootste som aaneengesloten subarray gebruikt Dynamisch programmeren :

Voor elke index i slaat DP[i] de maximaal mogelijke Largest Sum Contiguous Subarray op, eindigend op index i, en daarom kunnen we DP[i] berekenen met behulp van de genoemde toestandsovergang:

  • DP[i] = max(DP[i-1] + arr[i] , arr[i] )

Hieronder vindt u de implementatie:

C++ 
// C++ program to print largest contiguous array sum #include  using namespace std; void maxSubArraySum(int a[], int size) {  vector dp(grootte, 0);  dp[0] = a[0];  int ans = dp[0];  voor (int i = 1; ik< size; i++) {  dp[i] = max(a[i], a[i] + dp[i - 1]);  ans = max(ans, dp[i]);  }  cout << ans; } /*Driver program to test maxSubArraySum*/ int main() {  int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };  int n = sizeof(a) / sizeof(a[0]);  maxSubArraySum(a, n);  return 0; }>
Java 
import java.util.Arrays; public class Main {  // Function to find the largest contiguous array sum  public static void maxSubArraySum(int[] a) {  int size = a.length;  int[] dp = new int[size]; // Create an array to store intermediate results  dp[0] = a[0]; // Initialize the first element of the intermediate array with the first element of the input array  int ans = dp[0]; // Initialize the answer with the first element of the intermediate array  for (int i = 1; i < size; i++) {  // Calculate the maximum of the current element and the sum of the current element and the previous result  dp[i] = Math.max(a[i], a[i] + dp[i - 1]);  // Update the answer with the maximum value encountered so far  ans = Math.max(ans, dp[i]);  }  // Print the maximum contiguous array sum  System.out.println(ans);  }  public static void main(String[] args) {  int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };  maxSubArraySum(a); // Call the function to find and print the maximum contiguous array sum  } } // This code is contributed by shivamgupta310570>
Python 
# Python program for the above approach def max_sub_array_sum(a, size): # Create a list to store intermediate results dp = [0] * size # Initialize the first element of the list with the first element of the array dp[0] = a[0] # Initialize the answer with the first element of the array ans = dp[0] # Loop through the array starting from the second element for i in range(1, size): # Choose the maximum value between the current element and the sum of the current element # and the previous maximum sum (stored in dp[i - 1]) dp[i] = max(a[i], a[i] + dp[i - 1]) # Update the overall maximum sum ans = max(ans, dp[i]) # Print the maximum contiguous subarray sum print(ans) # Driver program to test max_sub_array_sum if __name__ == '__main__': # Sample array a = [-2, -3, 4, -1, -2, 1, 5, -3] # Get the length of the array n = len(a) # Call the function to find the maximum contiguous subarray sum max_sub_array_sum(a, n) # This code is contributed by Susobhan Akhuli>
C# 
using System; class MaxSubArraySum {  // Function to find and print the maximum sum of a  // subarray  static void FindMaxSubArraySum(int[] arr, int size)  {  // Create an array to store the maximum sum of  // subarrays  int[] dp = new int[size];  // Initialize the first element of dp with the first  // element of arr  dp[0] = arr[0];  // Initialize a variable to store the final result  int ans = dp[0];  // Iterate through the array to find the maximum sum  for (int i = 1; i < size; i++) {  // Calculate the maximum sum ending at the  // current position  dp[i] = Math.Max(arr[i], arr[i] + dp[i - 1]);  // Update the final result with the maximum sum  // found so far  ans = Math.Max(ans, dp[i]);  }  // Print the maximum sum of the subarray  Console.WriteLine(ans);  }  // Driver program to test FindMaxSubArraySum  static void Main()  {  // Example array  int[] arr = { -2, -3, 4, -1, -2, 1, 5, -3 };  // Calculate and print the maximum subarray sum  FindMaxSubArraySum(arr, arr.Length);  } }>
Javascript 
// Javascript program to print largest contiguous array sum // Function to find the largest contiguous array sum function maxSubArraySum(a) {  let size = a.length;  // Create an array to store intermediate results  let dp = new Array(size);  // Initialize the first element of the intermediate array with the first element of the input array  dp[0] = a[0];  // Initialize the answer with the first element of the intermediate array  let ans = dp[0];    for (let i = 1; i < size; i++) {  // Calculate the maximum of the current element and the sum of the current element and the previous result  dp[i] = Math.max(a[i], a[i] + dp[i - 1]);  // Update the answer with the maximum value encountered so far  ans = Math.max(ans, dp[i]);  }  // Print the maximum contiguous array sum  console.log(ans); } let a = [-2, -3, 4, -1, -2, 1, 5, -3]; // Call the function to find and print the maximum contiguous array sum maxSubArraySum(a);>
Uitvoer 
7>


Oefenprobleem:

selectie sorteren in Java

Gegeven een array van gehele getallen (mogelijk zijn sommige elementen negatief), schrijf dan een C-programma om het *maximale product* te vinden dat mogelijk is door ‘n’ opeenvolgende gehele getallen in de array te vermenigvuldigen waarbij n ? ARRAY_SIZE. Druk ook het startpunt van de maximale productsubarray af.