### prabhat7298's blog

By prabhat7298, history, 22 months ago, I want to find largest length subarray having sum greater than k.

One of the solution I found somewhere is binary searching on the length of subarray. Length of subarray can vary from 1 to n. We can binary search in range low=1 to high=n and for every mid=(low+high)/2 we can check for all subarray of length=mid if any subarray has sum greater than k in O(n). If any such subarray exist then we can search for higher length subarray i.e. low=mid+1 otherwise we decrease the search length i.e. high=mid-1.

int maxlen(vector<int> v)
{
int hi = n, lo = 1, ans = -1, mid, cnt = 0;
while(lo <= hi) {
mid = hi+lo>>1;
if(cnt = count(mid)) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
return ans;
}

int count(int len) {
int cnt = 0;
for(int i = len; i <= n; i++)
if(prefixsum[i] - prefixsum[i - len] > K)
cnt++;
return cnt;
}


What puzzles me is that if we get sum < k for present length subarray then increasing the search length of subarray will ensure that we'll get some subarray having subarray>k. And if we have subarray>k for any length then by decreasing search length we can get more such subarrays having greater than k sum. It could be that we could have decreased search length to get some subarray>k.

So, the question boils down to deciding the predicate of binary search i.e. at each step how we'll decide to change the search range?

PS: I know it can be easily solved by binary searching on prefix subarray rather than on length but I want to know the correctnes of the solution or how it works  Comments (6)
 » If there is subarray of length l with sum > k, it doesn't guarantee that there exists subarray with length l-1. For example, if k=1 and array is [2, -1, 2], [2, -1, 2] is ok, but [2, -1] and [-1, 2] is not.
•  » » So, according to you we can't decide the direction of search space depending upon the existence of a particular length subarray having sum>K or
 » It can be done using some data structure (e.g. segment tree or Fenwick (binary indexed) tree). Let's precalc each prefix sum of the array — $pref_i$. Now sum on some segment $[l, r]$ is equal to $pref_r - pref_{l - 1}$ ($pref_{-1} = 0$). $pref_r - pref_{l - 1} > k$ then $pref_r - k > pref_{l - 1}$. Now let's iterate over each element we should minimum $val$ on segment $(-\infty, pref_i - k)$ and update $val_{pref_i}$ with $i$.
•  » » Thank you very much! But could you plz also comment on the correctness of the above solution?
•  » » » Well, for each $r$ we choose only subarrays with $sum > k$ because of the equation and choose the minimum $l - 1$ so that $r - l + 1$ is maximal. If you don't understand it now you should try solving easier tasks.
•  » » » » I understood it thanks :)