Question

That given an array a consisting of n integers returns the maximuim sum of any slice of a

Answer 1

Step-by-step explanation:

Initialize:

max_so_far = 0

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

Program:

// C++ program to print largest contiguous array sum

#include<iostream>

#include<climits>

using namespace std;