1 2 1 2 3 2 3 4 5 4 6 7 6 7 8 ....10000
Given for all in i
Data Structure Used: Array
Algorithm:(Done Will Post Soon)
Working Code:
Time Complexity O(logn) yes its logn ??
Space Complexity O(1)
Run Here
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This is same as sorted array 1,2,3............. n and just do binary search.
ReplyDeletemid=(start+last)/2
while(start<last)
{
if(current = arr[mid])
return mid;
if(current < arr[mid])
last=mid;
else
start=mid;
}
@hemant yes it can be done in logn thanks for code :)
ReplyDeleteShashank