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_farExplanation:
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 = 4Program:
C++
#include<iostream>
#include<climits>
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;
}
int main()
{
int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
int n = sizeof(a)/sizeof(a[0]);
int max_sum = maxSubArraySum(a, n);
cout << "Maximum contiguous sum is " << max_sum;
return 0;
}
|
Java
import java.io.*;
import 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;
}
}
|
Python
from sys import maxint
def maxSubArraySum(a,size):
max_so_far = -maxint - 1
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
a = [-13, -3, -25, -20, -3, -16, -23, -12, -5, -22, -15, -4, -7]
print "Maximum contiguous sum is", maxSubArraySum(a,len(a))
|
C#
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;
}
public static void Main ()
{
int [] a = {-2, -3, 4, -1, -2, 1, 5, -3};
Console.Write("Maximum contiguous sum is " +
maxSubArraySum(a));
}
}
|
PHP
<?php
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;
}
$a = array(-2, -3, 4, -1,
-2, 1, 5, -3);
$n = count($a);
$max_sum = maxSubArraySum($a, $n);
echo "Maximum contiguous sum is " ,
$max_sum;
?>
|
Javascript
<script>
function maxSubArraySum(a, size)
{
var maxint = Math.pow(2, 53)
var max_so_far = -maxint - 1
var max_ending_here = 0
for (var i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here)
max_so_far = max_ending_here
if (max_ending_here < 0)
max_ending_here = 0
}
return max_so_far
}
var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ]
document.write("Maximum contiguous sum is",
maxSubArraySum(a, a.length))
</script>
|
Output:
Maximum contiguous sum is 7
Another approach:
C++
int maxSubarraySum(int arr[], int size)
{
int max_ending_here = 0, max_so_far = INT_MIN;
for (int i = 0; i < size; i++) {
if (arr[i] <= max_ending_here + arr[i]) {
max_ending_here += arr[i];
}
else {
max_ending_here = arr[i];
}
if (max_ending_here > max_so_far)
max_so_far = max_ending_here;
}
return max_so_far;
}
|
Java
static int maxSubArraySum(int a[],int size)
{
int max_so_far = a[0], max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
|
Python
def maxSubArraySum(a,size):
max_so_far = a[0]
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if max_ending_here < 0:
max_ending_here = 0
elif (max_so_far < max_ending_here):
max_so_far = max_ending_here
return max_so_far
|
C#
static int maxSubArraySum(int[] a, int size)
{
int max_so_far = a[0], max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
|
PHP
<?php
function maxSubArraySum(&$a, $size)
{
$max_so_far = $a[0];
$max_ending_here = 0;
for ($i = 0; $i < $size; $i++)
{
$max_ending_here = $max_ending_here + $a[$i];
if ($max_ending_here < 0)
$max_ending_here = 0;
else if ($max_so_far < $max_ending_here)
$max_so_far = $max_ending_here;
}
return $max_so_far;
?>
|
Time Complexity: O(n)
Algorithmic Paradigm: Dynamic Programming
Following is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in the array are negative.
C++
#include<iostream>
using namespace std;
int maxSubArraySum(int a[], int size)
{
int max_so_far = a[0];
int curr_max = a[0];
for (int i = 1; i < size; i++)
{
curr_max = max(a[i], curr_max+a[i]);
max_so_far = max(max_so_far, curr_max);
}
return max_so_far;
}
int main()
{
int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
int n = sizeof(a)/sizeof(a[0]);
int max_sum = maxSubArraySum(a, n);
cout << "Maximum contiguous sum is " << max_sum;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int maxSubArraySum(int a[], int size)
{
int max_so_far = a[0];
int curr_max = a[0];
for (int i = 1; i < size; i++)
{
curr_max = Math.max(a[i], curr_max+a[i]);
max_so_far = Math.max(max_so_far, curr_max);
}
return max_so_far;
}
public static void main(String[] args)
{
int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
int n = a.length;
int max_sum = maxSubArraySum(a, n);
System.out.println("Maximum contiguous sum is "
+ max_sum);
}
}
|
Python
def maxSubArraySum(a,size):
max_so_far =a[0]
curr_max = a[0]
for i in range(1,size):
curr_max = max(a[i], curr_max + a[i])
max_so_far = max(max_so_far,curr_max)
return max_so_far
a = [-2, -3, 4, -1, -2, 1, 5, -3]
print"Maximum contiguous sum is" , maxSubArraySum(a,len(a))
|
C#
using System;
class GFG
{
static int maxSubArraySum(int []a, int size)
{
int max_so_far = a[0];
int curr_max = a[0];
for (int i = 1; i < size; i++)
{
curr_max = Math.Max(a[i], curr_max+a[i]);
max_so_far = Math.Max(max_so_far, curr_max);
}
return max_so_far;
}
public static void Main ()
{
int []a = {-2, -3, 4, -1, -2, 1, 5, -3};
int n = a.Length;
Console.Write("Maximum contiguous sum is "
+ maxSubArraySum(a, n));
}
}
|
PHP
<?php
function maxSubArraySum($a, $size)
{
$max_so_far = $a[0];
$curr_max = $a[0];
for ($i = 1; $i < $size; $i++)
{
$curr_max = max($a[$i],
$curr_max + $a[$i]);
$max_so_far = max($max_so_far,
$curr_max);
}
return $max_so_far;
}
$a = array(-2, -3, 4, -1,
-2, 1, 5, -3);
$n = sizeof($a);
$max_sum = maxSubArraySum($a, $n);
echo "Maximum contiguous sum is " .
$max_sum;
?>
|
Javascript
<script>
function maxSubArraySum(a,size)
{
let max_so_far = a[0];
let curr_max = a[0];
for (let 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;
}
let a = [-2, -3, 4, -1, -2, 1, 5, -3];
let n = a.length;
document.write("Maximum contiguous sum is ",maxSubArraySum(a, n));
</script>
|
Output:
Maximum contiguous sum is 7
To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.
C++
#include<iostream>
#include<climits>
using namespace std;
int 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;
}
int main()
{
int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
int n = sizeof(a)/sizeof(a[0]);
int max_sum = maxSubArraySum(a, n);
return 0;
}
|
Java
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);
}
public static void main(String[] args)
{
int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
int n = a.length;
maxSubArraySum(a, n);
}
}
|
Python3
from sys import maxsize
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))
a = [-2, -3, 4, -1, -2, 1, 5, -3]
maxSubArraySum(a,len(a))
|
C#
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);
}
public static void Main()
{
int []a = {-2, -3, 4, -1,
-2, 1, 5, -3};
int n = a.Length;
maxSubArraySum(a, n);
}
}
|
PHP
<?php
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."\n";
echo "Starting index ". $start . "\n".
"Ending index " . $end . "\n";
}
$a = array(-2, -3, 4, -1, -2, 1, 5, -3);
$n = sizeof($a);
$max_sum = maxSubArraySum($a, $n);
?>
|
Output:
Maximum contiguous sum is 7
Starting index 2
Ending index 6
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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