Kadane's Algorithm — (Dynamic Programming)

Don't be afraid this is not very difficult....

For new programmer, this is very essential... We know a little bit for that we don't answer many problem as like "largest sum contiguous sub array" that's really very easy to determine largest sum if we know the following algorithm.....

Initialize: max_so_far = -inf;

max_ending_here = 0;

Loop for each element of the array

(a) max_ending_here = max_ending_here + a[i]

(b) if(max_ending_here < 0) max_ending_here = 0

(c) if(max_so_far < max_ending_here) max_so_far = max_ending_here

return max_so_far

Explanation: Simple idea of the 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 = 4

Code :

for(int i = 0; i < N; i++){

max_ending_here += Number[i];

        if(max_ending_here > max_so_far)max_so_far = max_ending_here;

        if(max_ending_here < 0)max_ending_here = 0;

    }

So the complexity for this code is O(n)...

Problem Link:

10684 - The jackpot

507 - Jill Rides Again

787 - Maximum Sub-sequence Product

Happy coding...:)