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1950. Maximum of Minimum Values in All Subarrays

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  // Similar to 1950. Maximum of Minimum Values in All Subarrays
  vector<int> findMaximums(vector<int>& nums) {
    const int n = nums.size();
    vector<int> ans(n);
    // prevMin[i] := the index k s.t.
    // nums[k] is the previous minimum in nums[0..n)
    vector<int> prevMin(n, -1);
    // nextMin[i] := the index k s.t.
    // nums[k] is the next minimum innums[i + 1..n)
    vector<int> nextMin(n, n);
    stack<int> stack;

    for (int i = 0; i < n; ++i) {
      while (!stack.empty() && nums[stack.top()] > nums[i]) {
        const int index = stack.top();
        stack.pop();
        nextMin[index] = i;
      }
      if (!stack.empty())
        prevMin[i] = stack.top();
      stack.push(i);
    }

    // For each nums[i], let l = nextMin[i] + 1 and r = nextMin[i] - 1.
    // nums[i] is the minimum in nums[l..r].
    // So, the ans[r - l + 1] will be at least nums[i].
    for (int i = 0; i < n; ++i) {
      const int sz = nextMin[i] - prevMin[i] - 1;
      ans[sz - 1] = max(ans[sz - 1], nums[i]);
    }

    // ans[i] should always >= ans[i + 1..n).
    for (int i = n - 2; i >= 0; --i)
      ans[i] = max(ans[i], ans[i + 1]);

    return ans;
  }
};

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class Solution {
  // Similar to 1950. Maximum of Minimum Values in All Subarrays
  public int[] findMaximums(int[] nums) {
    final int n = nums.length;
    int[] ans = new int[n];
    // prevMin[i] := the index k s.t.
    // nums[k] is the previous minimum in nums[0..n)
    int[] prevMin = new int[n];
    // nextMin[i] := the index k s.t.
    // nums[k] is the next minimum innums[i + 1..n)
    int[] nextMin = new int[n];
    Deque<Integer> stack = new ArrayDeque<>();

    Arrays.fill(prevMin, -1);
    Arrays.fill(nextMin, n);

    for (int i = 0; i < n; ++i) {
      while (!stack.isEmpty() && nums[stack.peek()] > nums[i]) {
        final int index = stack.pop();
        nextMin[index] = i;
      }
      if (!stack.isEmpty())
        prevMin[i] = stack.peek();
      stack.push(i);
    }

    // For each nums[i], let l = nextMin[i] + 1 and r = nextMin[i] - 1.
    // nums[i] is the minimum in nums[l..r].
    // So, the ans[r - l + 1] will be at least nums[i].
    for (int i = 0; i < n; ++i) {
      final int sz = nextMin[i] - prevMin[i] - 1;
      ans[sz - 1] = Math.max(ans[sz - 1], nums[i]);
    }

    // ans[i] should always >= ans[i + 1..n).
    for (int i = n - 2; i >= 0; --i)
      ans[i] = Math.max(ans[i], ans[i + 1]);

    return ans;
  }
}

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class Solution:
  # Similar to 1950. Maximum of Minimum Values in All Subarrays
  def findMaximums(self, nums: list[int]) -> list[int]:
    n = len(nums)
    ans = [0] * n
    # prevMin[i] := the index k s.t.
    # nums[k] is the previous minimum in nums[0..n)
    prevMin = [-1] * n
    # nextMin[i] := the index k s.t.
    # nums[k] is the next minimum innums[i + 1..n)
    nextMin = [n] * n
    stack = []

    for i, num in enumerate(nums):
      while stack and nums[stack[-1]] > nums[i]:
        index = stack.pop()
        nextMin[index] = i
      if stack:
        prevMin[i] = stack[-1]
      stack.append(i)

    # For each nums[i], let l = nextMin[i] + 1 and r = nextMin[i] - 1.
    # nums[i] is the minimum in nums[l..r].
    # So, the ans[r - l + 1] will be at least nums[i].
    for num, l, r in zip(nums, prevMin, nextMin):
      sz = r - l - 1
      ans[sz - 1] = max(ans[sz - 1], num)

    # ans[i] should always >= ans[i + 1..n).
    for i in range(n - 2, -1, -1):
      ans[i] = max(ans[i], ans[i + 1])

    return ans