Leetcode #1950: Maximum of Minimum Values in All Subarrays
In this guide, we solve Leetcode #1950 Maximum of Minimum Values in All Subarrays in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer array nums of size n. You are asked to solve n queries for each integer i in the range 0 <= i < n.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Stack, Array, Monotonic Stack
Intuition
We need the next greater or smaller element efficiently, which is exactly what a monotonic stack offers.
Each element is pushed and popped at most once, yielding a linear-time scan.
Approach
Maintain a stack that is either increasing or decreasing, depending on the query.
When the invariant is broken, pop and resolve answers for those indices.
Steps:
- Scan elements once.
- Pop while the monotonic condition is violated.
- Use stack indices to update answers.
Example
Input: nums = [0,1,2,4]
Output: [4,2,1,0]
Explanation:
i=0:
- The subarrays of size 1 are [0], [1], [2], [4]. The minimum values are 0, 1, 2, 4.
- The maximum of the minimum values is 4.
i=1:
- The subarrays of size 2 are [0,1], [1,2], [2,4]. The minimum values are 0, 1, 2.
- The maximum of the minimum values is 2.
i=2:
- The subarrays of size 3 are [0,1,2], [1,2,4]. The minimum values are 0, 1.
- The maximum of the minimum values is 1.
i=3:
- There is one subarray of size 4, which is [0,1,2,4]. The minimum value is 0.
- There is only one value, so the maximum is 0.
Python Solution
class Solution:
def findMaximums(self, nums: List[int]) -> List[int]:
n = len(nums)
left = [-1] * n
right = [n] * n
stk = []
for i, x in enumerate(nums):
while stk and nums[stk[-1]] >= x:
stk.pop()
if stk:
left[i] = stk[-1]
stk.append(i)
stk = []
for i in range(n - 1, -1, -1):
while stk and nums[stk[-1]] >= nums[i]:
stk.pop()
if stk:
right[i] = stk[-1]
stk.append(i)
ans = [0] * n
for i in range(n):
m = right[i] - left[i] - 1
ans[m - 1] = max(ans[m - 1], nums[i])
for i in range(n - 2, -1, -1):
ans[i] = max(ans[i], ans[i + 1])
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.