Suppose we have an array nums, we have to find the maximum min-product of each non-empty subarray of nums. Since the answer can be big enough, return it in modulo 10^9+7. The min-product of the array is equal to the minimum value in the array multiplied by the sum of the array. So if we have an array is like [4,3,6] (minimum value is 3) has a min-product of 3*(4+3+6) = 3*13 = 39.
So, if the input is like nums = [2,3,4,3], then the output will be 30 because we can get subarray [3,4,3] for maximizing the result, so 3*(3+4+3) = 3*10 = 30.
To solve this, we will follow these steps −
m := 10^9+7
stack := a new stack
rsum := 0, res := 0
insert 0 at the end of nums
for each index i and value v in nums, do
while stack is not empty and nums[index of top of stack] >= v, do
index, val) := top of stack and pop it from stack
arrSum:= rsum
if stack not empty, then
arrSum:= rsum - value of top of stack
res:= maximum of res and (nums[index]*arrSum)
rsum := rsum + v
push (i, rsum) at the end of stack
return res mod m
Example
Let us see the following implementation to get better understanding −
def solve(nums):
m = int(1e9+7)
stack = []
rsum = 0
res = 0
nums.append(0)
for i, v in enumerate(nums):
while stack and nums[stack[-1][0]] >= v:
index, _ = stack.pop()
arrSum=rsum
if stack:
arrSum=rsum-stack[-1][1]
res=max(res, nums[index]*arrSum)
rsum += v
stack.append((i, rsum))
return res % m
nums = [2,3,4,3]
print(solve(nums))Input
[2,3,4,3]
Output
30