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 the 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
Note: The above algorithm only works if and only if at least one positive number should be present otherwise it does not work i.e if an Array contains all negative numbers it doesn’t work.
Program:
C++
// C++ program to print largest contiguous array sum
#include<iostream>
#include<climits>
usingnamespacestd;
intmaxSubArraySum(inta[], intsize)
{
intmax_so_far = INT_MIN, max_ending_here = 0;
for(inti = 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;
}
returnmax_so_far;
}
/*Driver program to test maxSubArraySum*/
intmain()
{
inta[] = {-2, -3, 4, -1, -2, 1, 5, -3};
intn = sizeof(a)/sizeof(a[0]);
intmax_sum = maxSubArraySum(a, n);
cout << "Maximum contiguous sum is "<< max_sum;
return0;
}
Java
importjava.io.*;
// Java program to print largest contiguous array sum
# Python program to find maximum contiguous subarray
# Function to find the maximum contiguous subarray
fromsys importmaxint
defmaxSubArraySum(a,size):
max_so_far =-maxint -1
max_ending_here =0
fori inrange(0, size):
max_ending_here =max_ending_here +a[i]
if(max_so_far < max_ending_here):
max_so_far =max_ending_here
ifmax_ending_here < 0:
max_ending_here =0
returnmax_so_far
# Driver function to check the above function
a =[-2, -3, 4, -1, -2, 1, 5, -3]
print"Maximum contiguous sum is", maxSubArraySum(a,len(a))
#This code is contributed by _Devesh Agrawal_
C#
// C# program to print largest
// contiguous array sum
usingSystem;
classGFG
{
staticintmaxSubArraySum(int[]a)
{
intsize = a.Length;
intmax_so_far = int.MinValue,
max_ending_here = 0;
for(inti = 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;
}
returnmax_so_far;
}
// Driver code
publicstaticvoidMain ()
{
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
<?php
// PHP program to print largest
// contiguous array sum
functionmaxSubArraySum($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;
}
// Driver code
$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;
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program to find maximum
// contiguous subarray
// Function to find the maximum
// contiguous subarray
functionmaxSubArraySum(a, size)
{
varmaxint = Math.pow(2, 53)
varmax_so_far = -maxint - 1
varmax_ending_here = 0
for(vari = 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
}
returnmax_so_far
}
// Driver code
vara = [ -2, -3, 4, -1, -2, 1, 5, -3 ]
document.write("Maximum contiguous sum is",
maxSubArraySum(a, a.length))
// This code is contributed by AnkThon
</script>
Output
Maximum contiguous sum is 7
Time Complexity: O(n) Auxiliary Space: O(1)
Another approach:
C++
intmaxSubarraySum(intarr[], intsize)
{
intmax_ending_here = 0, max_so_far = INT_MIN;
for(inti = 0; i < size; i++) {
// include current element to previous subarray only
// when it can add to a bigger number than itself.
if(arr[i] <= max_ending_here + arr[i]) {
max_ending_here += arr[i];
}
// Else start the max subarray from current element
else{
max_ending_here = arr[i];
}
if(max_ending_here > max_so_far)
max_so_far = max_ending_here;
}
returnmax_so_far;
} // contributed by Vipul Raj
Java
staticintmaxSubArraySum(inta[],intsize)
{
intmax_so_far = a[0], max_ending_here = 0;
for(inti = 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 */
elseif(max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
returnmax_so_far;
}
// This code is contributed by ANKITRAI1
Python
defmaxSubArraySum(a,size):
max_so_far =a[0]
max_ending_here =0
fori inrange(0, size):
max_ending_here =max_ending_here +a[i]
ifmax_ending_here < 0:
max_ending_here =0
# Do not compare for all elements. Compare only
# when max_ending_here > 0
elif(max_so_far < max_ending_here):
max_so_far =max_ending_here
returnmax_so_far
C#
staticintmaxSubArraySum(int[] a, intsize)
{
intmax_so_far = a[0], max_ending_here = 0;
for(inti = 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 */
elseif(max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
returnmax_so_far;
}
// This code is contributed
// by ChitraNayal
PHP
<?php
functionmaxSubArraySum(&$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)
$max_ending_here= 0;
/* Do not compare for all elements.
Compare only when max_ending_here > 0 */
elseif($max_so_far< $max_ending_here)
$max_so_far= $max_ending_here;
}
return$max_so_far;
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// JavaScript Program to implement
// the above approach
functionmaxSubarraySum(arr, size)
{
let max_ending_here = 0, max_so_far = Number.MIN_VALUE;
for(let i = 0; i < size; i++) {
// include current element to previous subarray only
// when it can add to a bigger number than itself.
if(arr[i] <= max_ending_here + arr[i]) {
max_ending_here += arr[i];
}
// Else start the max subarray from current element
else{
max_ending_here = arr[i];
}
if(max_ending_here > max_so_far) {
max_so_far = max_ending_here;
}
}
returnmax_so_far;
}
// This code is contributed by Potta Lokesh
</script>
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>
usingnamespacestd;
intmaxSubArraySum(inta[], intsize)
{
intmax_so_far = a[0];
intcurr_max = a[0];
for(inti = 1; i < size; i++)
{
curr_max = max(a[i], curr_max+a[i]);
max_so_far = max(max_so_far, curr_max);
}
returnmax_so_far;
}
/* Driver program to test maxSubArraySum */
intmain()
{
inta[] = {-2, -3, 4, -1, -2, 1, 5, -3};
intn = sizeof(a)/sizeof(a[0]);
intmax_sum = maxSubArraySum(a, n);
cout << "Maximum contiguous sum is "<< max_sum;
return0;
}
Java
// Java program to print largest contiguous
// array sum
importjava.io.*;
classGFG {
staticintmaxSubArraySum(inta[], intsize)
{
intmax_so_far = a[0];
intcurr_max = a[0];
for(inti = 1; i < size; i++)
{
curr_max = Math.max(a[i], curr_max+a[i]);
max_so_far = Math.max(max_so_far, curr_max);
}
returnmax_so_far;
}
/* Driver program to test maxSubArraySum */
publicstaticvoidmain(String[] args)
{
inta[] = {-2, -3, 4, -1, -2, 1, 5, -3};
intn = a.length;
intmax_sum = maxSubArraySum(a, n);
System.out.println("Maximum contiguous sum is "
+ max_sum);
}
}
// This code is contributed by Prerna Saini
Python
# Python program to find maximum contiguous subarray
defmaxSubArraySum(a,size):
max_so_far =a[0]
curr_max =a[0]
fori inrange(1,size):
curr_max =max(a[i], curr_max +a[i])
max_so_far =max(max_so_far,curr_max)
returnmax_so_far
# Driver function to check the above function
a =[-2, -3, 4, -1, -2, 1, 5, -3]
print"Maximum contiguous sum is", maxSubArraySum(a,len(a))
#This code is contributed by _Devesh Agrawal_
C#
// C# program to print largest
// contiguous array sum
usingSystem;
classGFG
{
staticintmaxSubArraySum(int[]a, intsize)
{
intmax_so_far = a[0];
intcurr_max = a[0];
for(inti = 1; i < size; i++)
{
curr_max = Math.Max(a[i], curr_max+a[i]);
max_so_far = Math.Max(max_so_far, curr_max);
}
returnmax_so_far;
}
// Driver code
publicstaticvoidMain ()
{
int[]a = {-2, -3, 4, -1, -2, 1, 5, -3};
intn = a.Length;
Console.Write("Maximum contiguous sum is "
+ maxSubArraySum(a, n));
}
}
// This code is contributed by Sam007_
PHP
<?php
functionmaxSubArraySum($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;
}
// Driver Code
$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;
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>
Javascript
<script>
// C# program to print largest
// contiguous array sum
functionmaxSubArraySum(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);
}
returnmax_so_far;
}
// Driver code
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
Time complexity: O(n), where n is the size of the array. Auxiliary Space: O(1)
To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.
C++
// C++ program to print largest contiguous array sum
#include<iostream>
#include<climits>
usingnamespacestd;
intmaxSubArraySum(inta[], intsize)
{
intmax_so_far = INT_MIN, max_ending_here = 0,
start =0, end = 0, s=0;
for(inti=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*/
intmain()
{
inta[] = {-2, -3, 4, -1, -2, 1, 5, -3};
intn = sizeof(a)/sizeof(a[0]);
intmax_sum = maxSubArraySum(a, n);
return0;
}
Java
// Java program to print largest
// contiguous array sum
classGFG {
staticvoidmaxSubArraySum(inta[], intsize)
{
intmax_so_far = Integer.MIN_VALUE,
max_ending_here = 0,start = 0,
end = 0, s = 0;
for(inti = 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
publicstaticvoidmain(String[] args)
{
inta[] = { -2, -3, 4, -1, -2, 1, 5, -3};
intn = a.length;
maxSubArraySum(a, n);
}
}
// This code is contributed by prerna saini
Python3
# Python program to print largest contiguous array sum
fromsys importmaxsize
# Function to find the maximum contiguous subarray
# and print its starting and end index
defmaxSubArraySum(a,size):
max_so_far =-maxsize -1
max_ending_here =0
start =0
end =0
s =0
fori inrange(0,size):
max_ending_here +=a[i]
ifmax_so_far < max_ending_here:
max_so_far =max_ending_here
start =s
end =i
ifmax_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
usingSystem;
classGFG
{
staticvoidmaxSubArraySum(int[]a,
intsize)
{
intmax_so_far = int.MinValue,
max_ending_here = 0, start = 0,
end = 0, s = 0;
for(inti = 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
publicstaticvoidMain()
{
int[]a = {-2, -3, 4, -1,
-2, 1, 5, -3};
intn = a.Length;
maxSubArraySum(a, n);
}
}
// This code is contributed
// by anuj_67.
PHP
<?php
// PHP program to print largest
// contiguous array sum
functionmaxSubArraySum($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";
}
// Driver Code
$a= array(-2, -3, 4, -1, -2, 1, 5, -3);
$n= sizeof($a);
$max_sum= maxSubArraySum($a, $n);
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// javascript program to print largest
// contiguous array sum
functionmaxSubArraySum(a , size) {
varmax_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);
}
// Driver code
vara = [ -2, -3, 4, -1, -2, 1, 5, -3 ];
varn = a.length;
maxSubArraySum(a, n);
// This code is contributed by Rajput-Ji
</script>
Output
Maximum contiguous sum is 7
Starting index 2
Ending index 6
Kadane’s Algorithm can be viewed both as 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 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 the previous sum OR restart a new range. 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 if you want to share more information about the topic discussed above.
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