input : -2 -1 -3 -5 -6
in this case. the o/p would be -1 as the largest sum subarray. Isn’t it? But Kadane will give 0 as max_val would be 0 always
input : -2 -1 -3 -5 -6
in this case. the o/p would be -1 as the largest sum subarray. Isn’t it? But Kadane will give 0 as max_val would be 0 always
hello @amandeepdogra65
initilaise max_val with any value that is present in ur array .
and then it will work fine for every cases.
I didn’t get.
Suppose I initilise max_val=arr[0] and arr[0] is -2, in the first iteration itself max_val will become 0 when we will choose max among max_val and curr_sum. Isn’t it?
how u r implemnting kadane?
try this->
max_val=a[0];
cur=0;
for i -> 0 ....n-1:
cur+=a[i]
if cur > max_val
max_val=cur
if cur<0
cur=0
i got it. Thanks man!