Annettu joukko arr[] kooltaan N . Tehtävänä on löytää vierekkäisen alitaulukon summa arr[] suurimmalla summalla.

Esimerkki:

Syöte: arr = {-2,-3,4,-1,-2,1,5,-3}
Lähtö: 7
Selitys: Aliryhmässä {4,-1, -2, 1, 5} on suurin summa 7.

Syöte: arr = {2}
Lähtö: 2
Selitys: Aliryhmässä {2} on suurin summa 1.

Syöte: arr = {5,4,1,7,8}
Lähtö: 25
Selitys: Aliryhmässä {5,4,1,7,8} on suurin summa 25.

Ajatus Kadanen algoritmi on ylläpitää muuttujaa max_ending_ here joka tallentaa suurimman summan jatkuvaan alitaulukkoon, joka päättyy nykyiseen indeksiin ja muuttujaan max_tofar_far tallentaa tähän mennessä löydettyjen vierekkäisten aliryhmien enimmäissumman, aina kun kentässä on positiivinen summa max_ending_ here vertaa siihen max_tofar_far ja päivittää max_tofar_far jos se on suurempi kuin max_tofar_far .

Pääasia siis Intuitio takana Kadanen algoritmi On,

• Negatiivisen summan alitaulukko hylätään ( määrittämällä koodissa max_ending_here = 0 ).
• Kannamme alijärjestelmää, kunnes se antaa positiivisen summan.

Kadanen algoritmin pseudokoodi:

Alustaa:
max_so_far = INT_MIN
max_ending_ here = 0

Silmukka jokaiselle taulukon elementille

(a) max_ending_here = max_ending_here + a[i]
(b) if(max_so_far
max_so_far = max_ending_ here
(c) jos(max_loppu_tähän <0)
max_ending_ here = 0
palauttaa max_so_far

Kuva Kadanen algoritmista:

Otetaan esimerkki: {-2, -3, 4, -1, -2, 1, 5, -3}

Huomautus : kuvassa max_so_far edustaa Max_Sum ja max_ending_ here by Curr_Sum

Jos i=0, a[0] = -2

merkkijono taulukkoon java

• max_ending_here = max_ending_here + (-2)
• Aseta max_ending_here = 0, koska max_ending_here <0
• ja aseta max_so_far = -2

Jos i=1, a[1] = -3

• max_ending_here = max_ending_here + (-3)
• Koska max_ending_here = -3 ja max_so_far = -2, max_so_far pysyy -2
• Aseta max_ending_here = 0, koska max_ending_here <0

Jos i=2, a[2] = 4

• max_ending_here = max_ending_here + (4)
• max_ending_here = 4
• max_so_far päivitetään arvoon 4, koska max_ending_here on suurempi kuin max_so_far, joka oli tähän asti -2

Jos i=3, a[3] = -1

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

Jos i=4, a[4] = -2

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

Jos i=5, a[5] = 1

• max_ending_here = max_ending_here + (1)
• max_ending_here = 2

Jos i=6, a[6] = 5

• max_ending_here = max_ending_here + (5)
• max_ending_here =
• max_so_far päivitetään arvoon 7, koska max_ending_here on suurempi kuin max_so_far

Jos i=7, a[7] = -3

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

Toteuta idea noudattamalla alla olevia ohjeita:

• Alusta muuttujat max_tofar_far = INT_MIN ja max_ending_ here = 0
• Suorita for-silmukka alkaen 0 to N-1 ja jokaiselle indeksille i :
  • Lisää arr[i] max_ending_here.
  • Jos max_so_far on pienempi kuin max_ending_here, päivitä max_so_far to max_ending_here .
  • Jos max_ending_here <0, päivitä max_ending_here = 0
• Palautus max_so_far

Alla on yllä olevan lähestymistavan toteutus.

// 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 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;  } }>

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# 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_>

>>PHP>>  
  Lähtö  
Maximum contiguous sum is 7>
  
    Aika monimutkaisuus:    PÄÄLLÄ)  
    Aputila:    O(1)
 
Tulosta suurimman summan vierekkäinen alataulukko:
 
Tulosta alitaulukko maksimisummalla ideana on säilyttää    alkaa    sisällysluettelo    max_sum_ending_ here    nykyisellä indeksillä niin että milloin tahansa    maksimi_summa_toistaiseksi    on päivitetty    max_sum_ending_ here    sitten alitaulukon aloitusindeksi ja loppuindeksi voidaan päivittää    alkaa    ja    nykyinen indeksi    .
 
Toteuta idea noudattamalla seuraavia ohjeita:
 
Alusta muuttujat    s    ,    alkaa,    ja    loppu    kanssa    0    ja    max_tofar_far    = INT_MIN ja    max_ending_ here    = 0
Suorita for-silmukka alkaen    0    to    N-1    ja jokaiselle indeksille    i    :
Lisää arr[i]         max_ending_here.
Jos max_so_far on pienempi kuin max_ending_here, päivitä         max_so_far to max_ending_here ja päivitä    alkaa    to    s    ja    loppu    to    i    .
Jos max_ending_here <0, päivitä max_ending_here = 0 ja    s    kanssa    i+1    .
Tulosta arvot hakemistosta    alkaa    to    loppu    .
Alla on yllä olevan lähestymistavan toteutus:
mamta kulkarni
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>>  
  Lähtö      Aika monimutkaisuus:    Päällä)  
    Aputila:    O(1)
 
Suurin Summa Contiguous Subray käyttää Dynaaminen ohjelmointi :
 
Jokaiselle indeksille i DP[i] tallentaa suurimman mahdollisen suurimman summan jatkuvan alialueen, joka päättyy indeksiin i, ja siksi voimme laskea DP[i]:n käyttämällä mainittua tilasiirtymää:
 
DP[i] = max(DP[i-1] + arr[i] , arr[i] )
Alla toteutus:
C++ 
// C++ program to print largest contiguous array sum #include  using namespace std; void maxSubArraySum(int a[], int size) {  vector dp(koko, 0);  dp[0] = a[0];  int ans = dp[0];  for (int i = 1; i< 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);>
   
  Lähtö    
    Harjoitusongelma:     
  
Kun annetaan joukko kokonaislukuja (mahdollisesti jotkut elementit negatiivisia), kirjoita C-ohjelma selvittääksesi *maksimitulon*, joka on mahdollista kertomalla 'n' peräkkäistä kokonaislukua taulukossa, jossa n ? ARRAY_SIZE. Tulosta myös suurimman tuotteen alitaulukon aloituskohta.