Java Program for Largest Sum Contiguous Subarray - GeeksforGeeks

29 Nov, 2021
Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers that has the largest sum. 
 
Kadane’s Algorithm:
Initialize:
    max_so_far = INT_MIN
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
  (c) if(max_ending_here < 0)
            max_ending_here = 0
return max_so_far
Explanation: The simple idea of Kadane's algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far 
    Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4
Program: 
Java
import java.io.*;// Java program to print largest contiguous array sumimport java.util.*;  class Kadane{    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));    }      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;    }}
Output:
Maximum contiguous sum is 7
Another approach:
Java
static int maxSubArraySum(int a[],int size) {           int max_so_far = a[0], max_ending_here = 0;       for (int i = 0; i < size; i++)     {         max_ending_here = max_ending_here + a[i];        if (max_ending_here < 0)             max_ending_here = 0;                   /* Do not compare for all           elements. Compare only            when max_ending_here > 0 */        else if (max_so_far < max_ending_here)             max_so_far = max_ending_here;               }     return max_so_far; }   // This code is contributed by ANKITRAI1
Time Complexity: O(n) 
Algorithmic Paradigm: Dynamic ProgrammingFollowing is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in the array are negative. 
Java
// Java program to print largest contiguous// array sumimport java.io.*;  class GFG {      static int maxSubArraySum(int a[], int size)    {    int max_so_far = a[0];    int curr_max = a[0];      for (int i = 1; i < size; i++)    {           curr_max = Math.max(a[i], curr_max+a[i]);        max_so_far = Math.max(max_so_far, curr_max);    }    return max_so_far;    }      /* Driver program to test maxSubArraySum */    public static void main(String[] args)    {    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};    int n = a.length;       int max_sum = maxSubArraySum(a, n);    System.out.println("Maximum contiguous sum is "                        + max_sum);    }}  // This code is contributed by Prerna Saini
Output: 
Maximum contiguous sum is 7
To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.  
Java
// Java program to print largest // contiguous array sumclass 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
Output: 
Maximum contiguous sum is 7
Starting index 2
Ending index 6
Kadane's Algorithm can be viewed both as a greedy and DP. As we can see that we are keeping a running sum of integers and when it becomes less than 0, we reset it to 0 (Greedy Part). This is because continuing with a negative sum is way more worse than restarting with a new range. Now it can also be viewed as a DP, at each stage we have 2 choices: Either take the current element and continue with previous sum OR restart a new range. These both choices are being taken care of in the implementation. 
Time Complexity: O(n)
Auxiliary Space: O(1)
Now try the below question Given an array of integers (possibly some elements negative), write a C program to find out the *maximum product* possible by multiplying 'n' consecutive integers in the array where n ≤ ARRAY_SIZE. Also, print the starting point of the maximum product subarray.
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