Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers that has the largest sum.
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 = INT_MIN
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
and set max_so_far = -2
for i=1, a[1] = -3
max_ending_here = max_ending_here + (-3)
Since max_ending_here = -3 and max_so_far = -2, max_so_far will remain -2
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 -2 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:
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 = [ - 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 = 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 ;
?>
|
Javascript
<script>
function maxSubarraySum(arr, size)
{
let max_ending_here = 0, max_so_far = Number.MIN_VALUE;
for (let 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;
}
</script>
|
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 );
?>
|
Javascript
<script>
function maxSubArraySum(a , size) {
var max_so_far = Number.MIN_VALUE, 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;
}
}
document.write( "Maximum contiguous sum is " + max_so_far);
document.write( "<br/>Starting index " + start);
document.write( "<br/>Ending index " + end);
}
var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ];
var n = a.length;
maxSubArraySum(a, n);
</script>
|
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.