Painter's partition problem

/*If all the buildings were painted by only one painter then it will take maximum time to paint all the house ie. sum(arr)
If each buildings were painted by one painter parallely then it will take minimum time to paint all the house ie. max(arr)
So our answer ranges between min to max so we will use binary search and we will verify mid time is possible to paint with given number of painters or not,if possible then we will try to reduce time else we will increase time.

In ispossible function:
we will allocated building to one labour while interating the houses,once it's time exceeds our given time,we will allocate another painter and we will in final,how many painter we used to paint the houses in given time ie.mid if exceeds the given painter in problem,return false else true.
*/
MOD = 10000003
def ispossible(arr,limit,maxm_allowed_painters):
    cur_time=0
    cur_painters=1
    for i in arr:
        cur_time+=i
        if cur_time>limit:
            cur_painters+=1
            cur_time=i
    if cur_painters>maxm_allowed_painters:
        return False
    return True
class Solution:
    def paint(self, A, B, C):
        ans=math.inf
        l=max(C)
        r=sum(C)
        while l<=r:
            mid=(l+r)//2
            temp=ispossible(C,mid,A)
            if temp:
                ans=min(ans,mid)
                r=mid-1
            else:
                l=mid+1
        return (ans*B)%MOD