Maximum of Minimum Values in All Subarrays - Problem

Given an integer array nums of size n, you need to solve n different queries. Each query asks: "What's the maximum possible minimum value among all subarrays of a specific length?"

For the i-th query (where 0 ≤ i < n):

  1. Consider all possible subarrays of length i + 1
  2. Find the minimum value in each subarray
  3. Return the maximum of all those minimum values

Your task is to return an array ans where ans[i] contains the answer to the i-th query.

Example: If nums = [0,1,2,4] and we want subarrays of length 2:

  • Subarray [0,1] → minimum = 0
  • Subarray [1,2] → minimum = 1
  • Subarray [2,4] → minimum = 2
  • Maximum of minimums = max(0,1,2) = 2

Input & Output

example_1.py — Basic Case
$ Input: [0,1,2,4]
› Output: [4,2,1,0]
💡 Note: Length 1: subarrays [0],[1],[2],[4] → minimums 0,1,2,4 → max = 4. Length 2: subarrays [0,1],[1,2],[2,4] → minimums 0,1,2 → max = 2. Length 3: subarrays [0,1,2],[1,2,4] → minimums 0,1 → max = 1. Length 4: subarray [0,1,2,4] → minimum 0 → max = 0.
example_2.py — Decreasing Array
$ Input: [10,20,50,1]
› Output: [50,20,10,1]
💡 Note: Length 1: max(10,20,50,1) = 50. Length 2: subarrays [10,20],[20,50],[50,1] → minimums 10,20,1 → max = 20. Length 3: subarrays [10,20,50],[20,50,1] → minimums 10,1 → max = 10. Length 4: minimum of entire array = 1.
example_3.py — Single Element
$ Input: [7]
› Output: [7]
💡 Note: Only one subarray [7] of length 1, its minimum is 7, so answer is [7].

Visualization

Tap to expand
🏔️ Mountain Range Analysis0124Observation Windows:Window size 2: shortest = 0, best = 1Window size 3: shortest = 1, best = 2Key Insight:Each mountain controls theanswer for its maximumpossible window size!
Understanding the Visualization
1
Identify Each Mountain's Domain
For each peak, find how far left and right you can look before seeing a shorter mountain
2
Calculate Maximum Influence
Each mountain determines the maximum window size where it could be the shortest
3
Fill the Observatory Log
Record the best shortest mountain for each window size, filling gaps by inheritance
Key Takeaway
🎯 Key Insight: Instead of checking every subarray, determine for each element the maximum subarray length where it could be the minimum value. This transforms an O(n³) problem into an elegant O(n) solution using monotonic stacks!

Time & Space Complexity

Time Complexity
⏱️
O(n)

Single pass with monotonic stack + single pass to process ranges + single pass to fill gaps

n
2n
✓ Linear Growth
Space Complexity
O(n)

Stack space for monotonic stack plus arrays for left/right boundaries and results

n
2n
⚡ Linearithmic Space

Related Problems

Constraints

  • 1 ≤ nums.length ≤ 105
  • 0 ≤ nums[i] ≤ 109
  • Follow-up: Can you solve this in O(n) time?
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