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.
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,
– the subarray with negative sum is discarded (by assigning max_ending_here = 0 in code).
– we carry subarray till it gives positive sum.
Initialize:
max_so_far = INT_MIN
max_ending_here = 0Loop 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
Illustration:
Lets take the example: {-2, -3, 4, -1, -2, 1, 5, -3}
max_so_far = INT_MIN
max_ending_here = 0for 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 = -2for 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 < 0for 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 nowfor i=3, a[3] = -1
max_ending_here = max_ending_here + (-1)
max_ending_here = 3for i=4, a[4] = -2
max_ending_here = max_ending_here + (-2)
max_ending_here = 1for i=5, a[5] = 1
max_ending_here = max_ending_here + (1)
max_ending_here = 2for 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_farfor i=7, a[7] = -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:
- 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++
// C++ program to print largest contiguous array sum #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; } // Driver Code int main() { int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 }; int n = sizeof (a) / sizeof (a[0]); // Function Call int max_sum = maxSubArraySum(a, n); cout << "Maximum contiguous sum is " << max_sum; return 0; } |
Java
// Java program to print largest contiguous array sum import java.io.*; import java.util.*; class Kadane { // Driver Code 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)); } // Function Call 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
# Python program to find maximum contiguous subarray # Function to find the maximum contiguous subarray 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 # 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 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; } // Driver code public static void Main() { 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 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 ; } // 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 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 } // Driver code var a = [ -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> |
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:
- 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++
// C++ program to print largest contiguous array sum #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; } /*Driver program to test maxSubArraySum*/ 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
// Java program to print largest // contiguous array sum 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); } // 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 |
Python3
# Python program to print largest contiguous array sum from sys import maxsize # Function to find the maximum contiguous subarray # and print its starting and end index 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)) # 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 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); } // Driver code public static void Main() { int [] a = { -2, -3, 4, -1, -2, 1, 5, -3 }; int n = a.Length; maxSubArraySum(a, n); } } // This code is contributed // by anuj_67. |
PHP
<?php // PHP program to print largest // contiguous array sum 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" ; } // Driver Code $a = array (-2, -3, 4, -1, -2, 1, 5, -3); $n = sizeof( $a ); maxSubArraySum( $a , $n ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // javascript program to print largest // contiguous array sum 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); } // Driver code var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ]; var n = a.length; maxSubArraySum(a, n); // This code is contributed by Rajput-Ji </script> |
Maximum contiguous sum is 7 Starting index 2 Ending index 6
Time Complexity: O(n)
Auxiliary Space: O(1)
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.
Recursive implementation of Kadane’s algorithm to find the maximum contiguous sum of an integer array.
To determine the maximum subarray sum of an integer array, Kadane’s Algorithm uses a Divide and Conquer strategy. This algorithm’s fundamental concept is to break the given array into smaller subarrays, recursively solve the issue for each subarray, and then combine the solutions to find the overall solution.
Recursive algorithm to find the maximum contiguous sum of an integer array:
- The input array “arr” and its length “n” are the two parameters for the function “maxSubArraySum”.
- When there is only one element in the array, the base case of the recursion occurs. In this situation, the function merely returns that element.
- Finding the midpoint of the array “m” in the recursive case splits the problem into two smaller subproblems. (i.e., the index of the middle element).
- To find the largest subarray sum in those subarrays, the function recursively calls itself on the left half of the array and the right half of the array (starting from the midpoint). “left_max” and “right_max” contain these values.
- The function loops through the right half of the array, keeping track of the maximum subarray sum ending at each index, to determine the maximum subarray sum that crosses the middle element of the array. This value is stored in “right_sum”.
- The function then performs a loop through the array’s left half, recording the highest subarray sum beginning at each index. “left_sum” is where this value is kept.
- The sum of the maximum subarray sum beginning at the left half and ending at the right half (i.e., “left_sum + right_sum”) crosses the middle element. “cross_max” is where this value is kept.
- The “left_max”, “right_max”, and “cross_max” subarray sums are returned by the function as their respective maximum values.
Below is the Implementation of above algorithm:
C++
#include <climits> #include <iostream> using namespace std; int maxSubArraySum( int arr[], int n) { // Base case: when there is only one element in the // array if (n == 1) { return arr[0]; } // Recursive case: divide the problem into smaller // sub-problems int m = n / 2; // Find the maximum subarray sum in the left half int left_max = maxSubArraySum(arr, m); // Find the maximum subarray sum in the right half int right_max = maxSubArraySum(arr + m, n - m); // Find the maximum subarray sum that crosses the middle // element int left_sum = INT_MIN, right_sum = INT_MIN, sum = 0; for ( int i = m; i < n; i++) { sum += arr[i]; right_sum = max(right_sum, sum); } sum = 0; for ( int i = m - 1; i >= 0; i--) { sum += arr[i]; left_sum = max(left_sum, sum); } int cross_max = left_sum + right_sum; // Return the maximum of the three subarray sums return max(cross_max, max(left_max, right_max)); } int main() { int arr[] = { -2, -3, 4, -1, -2, 1, 5, -3 }; int n = sizeof (arr) / sizeof (arr[0]); int max_sum = maxSubArraySum(arr, n); cout << "Maximum contiguous sum is " << max_sum << endl; return 0; } |
Java
import java.util.Arrays; public class Program { // Define a function to find the maximum subarray sum public static int maxSubArraySum( int [] arr) { // Base case: when there is only one element in the // array if (arr.length == 1 ) { return arr[ 0 ]; } // Recursive case: divide the problem into smaller // sub-problems int m = arr.length / 2 ; // Find the maximum subarray sum in the left half int leftMax = maxSubArraySum(Arrays.copyOfRange(arr, 0 , m)); // Find the maximum subarray sum in the right half int rightMax = maxSubArraySum( Arrays.copyOfRange(arr, m, arr.length)); // Find the maximum subarray sum that crosses the // middle element int leftSum = Integer.MIN_VALUE; int rightSum = Integer.MIN_VALUE; int sum = 0 ; // Traverse the array from the middle to the right for ( int i = m; i < arr.length; i++) { sum += arr[i]; rightSum = Math.max(rightSum, sum); } sum = 0 ; // Traverse the array from the middle to the left for ( int i = m - 1 ; i >= 0 ; i--) { sum += arr[i]; leftSum = Math.max(leftSum, sum); } int crossMax = leftSum + rightSum; // Return the maximum of the three subarray sums return Math.max(crossMax, Math.max(leftMax, rightMax)); } // Example usage public static void main(String[] args) { int [] arr = { - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 }; int maxSum = maxSubArraySum(arr); System.out.println( "Maximum contiguous sum is " + maxSum); } } |
Python
import sys # Define a function to find the maximum subarray sum def maxSubArraySum(arr): # Base case: when there is only one element in the array if len (arr) = = 1 : return arr[ 0 ] # Recursive case: divide the problem into smaller sub-problems m = len (arr) / / 2 # Find the maximum subarray sum in the left half left_max = maxSubArraySum(arr[:m]) # Find the maximum subarray sum in the right half right_max = maxSubArraySum(arr[m:]) # Find the maximum subarray sum that crosses the middle element left_sum = - sys.maxsize - 1 right_sum = - sys.maxsize - 1 sum = 0 # Traverse the array from the middle to the right for i in range (m, len (arr)): sum + = arr[i] right_sum = max (right_sum, sum ) sum = 0 # Traverse the array from the middle to the left for i in range (m - 1 , - 1 , - 1 ): sum + = arr[i] left_sum = max (left_sum, sum ) cross_max = left_sum + right_sum # Return the maximum of the three subarray sums return max (cross_max, max (left_max, right_max)) # Example usage arr = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ] max_sum = maxSubArraySum(arr) print ( "Maximum contiguous sum is" , max_sum) |
C#
using System; using System.Collections.Generic; using System.Linq; class Program { // Define a function to find the Maximum subarray sum public static int MaxSubArraySum( int [] arr) { // Base case: when there is only one element in the // array if (arr.Length == 1) { return arr[0]; } // Recursive case: divide the problem into smaller // sub-problems int m = arr.Length / 2; // Find the Maximum subarray sum in the left half int leftMax = MaxSubArraySum(arr.Take(m).ToArray()); // Find the Maximum subarray sum in the right half int rightMax = MaxSubArraySum(arr.Skip(m).ToArray()); // Find the Maximum subarray sum that crosses the // middle element int leftSum = int .MinValue; int rightSum = int .MinValue; int sum = 0; // Traverse the array from the middle to the right for ( int i = m; i < arr.Length; i++) { sum += arr[i]; rightSum = Math.Max(rightSum, sum); } sum = 0; // Traverse the array from the middle to the left for ( int i = m - 1; i >= 0; i--) { sum += arr[i]; leftSum = Math.Max(leftSum, sum); } int crossMax = leftSum + rightSum; // Return the Maximum of the three subarray sums return Math.Max(crossMax, Math.Max(leftMax, rightMax)); } // Example usage public static void Main( string [] args) { int [] arr = { -2, -3, 4, -1, -2, 1, 5, -3 }; int MaxSum = MaxSubArraySum(arr); Console.WriteLine( "Maximum contiguous sum is " + MaxSum); } } // This code is contributed by Tapesh(tapeshdua420) |
Javascript
function maxSubArraySum(arr, n) { // Base case: when there is only one element in the array if (n === 1) { return arr[0]; } // Recursive case: divide the problem into smaller sub-problems const m = Math.floor(n / 2); // Find the maximum subarray sum in the left half const leftMax = maxSubArraySum(arr.slice(0, m), m); // Find the maximum subarray sum in the right half const rightMax = maxSubArraySum(arr.slice(m), n - m); // Find the maximum subarray sum that crosses the middle element let leftSum = -Infinity, rightSum = -Infinity, sum = 0; for (let i = m; i < n; i++) { sum += arr[i]; rightSum = Math.max(rightSum, sum); } sum = 0; for (let i = m - 1; i >= 0; i--) { sum += arr[i]; leftSum = Math.max(leftSum, sum); } const crossMax = leftSum + rightSum; // Return the maximum of the three subarray sums return Math.max(crossMax, leftMax, rightMax); } const arr = [-2, -3, 4, -1, -2, 1, 5, -3]; const n = arr.length; const maxSum = maxSubArraySum(arr, n); console.log(`Maximum contiguous sum is ${maxSum}`); |
Maximum contiguous sum is 7
Practice Problem:
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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