Largest Sum Contiguous Subarray (Kadane’s Algorithm)
Given an array arr[] of size N. The task is to find the sum of the contiguous subarray within a arr[] with the largest sum.Â
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The idea of Kadane’s algorithm is to maintain a variable max_ending_here that stores the maximum sum contiguous subarray ending at current index and a variable max_so_far stores the maximum sum of contiguous subarray found so far, Everytime there is a positive-sum value in max_ending_here compare it with max_so_far and update max_so_far if it is greater than max_so_far.
So the main Intuition behind Kadane’s Algorithm is,Â
- The subarray with negative sum is discarded (by assigning max_ending_here = 0 in code).
- We carry subarray till it gives positive sum.
Pseudocode of 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
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Illustration of Kadane’s Algorithm:
Lets take the example: {-2, -3, 4, -1, -2, 1, 5, -3}
Note: in the image max_so_far is represented by Max_Sum and max_ending_here by Curr_Sum
For i=0, Â a[0] = Â -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, Â a[1] = Â -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, Â a[2] = Â 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, Â a[3] = Â -1
- max_ending_here = max_ending_here + (-1)
- max_ending_here = 3
For i=4, Â a[4] = Â -2
- max_ending_here = max_ending_here + (-2)
- max_ending_here = 1
For i=5, Â a[5] = Â 1
- max_ending_here = max_ending_here + (1)
- max_ending_here = 2
For i=6, Â a[6] = Â 5
- max_ending_here = max_ending_here + (5)
- max_ending_here =
- max_so_far is updated to 7 because max_ending_here is greater than max_so_far
For i=7, Â a[7] = Â -3
- max_ending_here = max_ending_here + (-3)
- max_ending_here = 4
Follow the below steps to Implement the idea:
- Initialize the variables max_so_far = INT_MIN and max_ending_here = 0
- Run a for loop from 0 to N-1 and for each index i:Â
- Add the arr[i] to max_ending_here.
- If  max_so_far is less than max_ending_here then update max_so_far to max_ending_here.
- If max_ending_here < 0 then update max_ending_here = 0
- Return max_so_far
Below is the Implementation of the above approach.
C++
#include <bits/stdc++.h>
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));
}
}
|
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>
|
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 ;
?>
|
Output
Maximum contiguous sum is 7
Time Complexity: O(N)
Auxiliary Space: O(1)
Print the Largest Sum Contiguous Subarray:
To print the subarray with the maximum sum the idea is to maintain start index of maximum_sum_ending_here at current index so that whenever maximum_sum_so_far is updated with maximum_sum_ending_here then start index and end index of subarray can be updated with start and current index.
Follow the below steps to implement the idea:
- Initialize the variables s, start, and end with 0 and max_so_far = INT_MIN and max_ending_here = 0
- Run a for loop from 0 to N-1 and for each index i:Â
- Add the arr[i] to max_ending_here.
- If max_so_far is less than max_ending_here then update max_so_far to max_ending_here and update start to s and end to i .
- If max_ending_here < 0 then update max_ending_here = 0 and s with i+1.
- Print values from index start to end.
Below is the Implementation of above approach:
C++
#include <climits>
#include <iostream>
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;
}
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.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);
}
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);
}
}
|
Javascript
<script>
function maxSubArraySum(a , size) {
var max_so_far = Number.MIN_SAFE_INTEGER, 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>
|
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 );
maxSubArraySum( $a , $n );
?>
|
Output
Maximum contiguous sum is 7
Starting index 2
Ending index 6
Time Complexity: O(n)
Auxiliary Space: O(1)
Largest Sum Contiguous Subarray using Dynamic Programming:
For each index i, DP[i] stores the maximum possible Largest Sum Contiguous Subarray ending at index i, and therefore we can calculate DP[i] using the mentioned state transition:
- DP[i] = max(DP[i-1] + arr[i] , arr[i] )
Below is the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void maxSubArraySum( int a[], int size)
{
vector< int > dp(size, 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;
}
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 {
public static void maxSubArraySum( int [] a) {
int size = a.length;
int [] dp = new int [size];
dp[ 0 ] = a[ 0 ];
int ans = dp[ 0 ];
for ( int i = 1 ; i < size; i++) {
dp[i] = Math.max(a[i], a[i] + dp[i - 1 ]);
ans = Math.max(ans, dp[i]);
}
System.out.println(ans);
}
public static void main(String[] args) {
int [] a = { - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 };
maxSubArraySum(a);
}
}
|
Python3
def max_sub_array_sum(a, size):
dp = [ 0 ] * size
dp[ 0 ] = a[ 0 ]
ans = dp[ 0 ]
for i in range ( 1 , size):
dp[i] = max (a[i], a[i] + dp[i - 1 ])
ans = max (ans, dp[i])
print (ans)
if __name__ = = "__main__" :
a = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ]
n = len (a)
max_sub_array_sum(a, n)
|
C#
using System;
class MaxSubArraySum {
static void FindMaxSubArraySum( int [] arr, int size)
{
int [] dp = new int [size];
dp[0] = arr[0];
int ans = dp[0];
for ( int i = 1; i < size; i++) {
dp[i] = Math.Max(arr[i], arr[i] + dp[i - 1]);
ans = Math.Max(ans, dp[i]);
}
Console.WriteLine(ans);
}
static void Main()
{
int [] arr = { -2, -3, 4, -1, -2, 1, 5, -3 };
FindMaxSubArraySum(arr, arr.Length);
}
}
|
Javascript
function maxSubArraySum(a) {
let size = a.length;
let dp = new Array(size);
dp[0] = a[0];
let ans = dp[0];
for (let i = 1; i < size; i++) {
dp[i] = Math.max(a[i], a[i] + dp[i - 1]);
ans = Math.max(ans, dp[i]);
}
console.log(ans);
}
let a = [-2, -3, 4, -1, -2, 1, 5, -3];
maxSubArraySum(a);
|
Practice Problem:Â
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.
Last Updated :
08 Dec, 2023
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