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:
<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> |
Output:
Maximum contiguous sum is 7
Another approach:
<script> // JavaScript Program to implement
// the above approach
function maxSubarraySum(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;
}
}
return max_so_far;
}
// This code is contributed by Potta Lokesh </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.
<script> // C# program to print largest // contiguous array sum 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;
} // 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
To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.
<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> |
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 refer complete article on Largest Sum Contiguous Subarray for more details!