1950. Maximum of Minimum Values in All Subarrays

Description

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.

To solve the ith query:

1. Find the minimum value in each possible subarray of size i + 1 of the array nums.
2. Find the maximum of those minimum values. This maximum is the answer to the query.

Return a 0-indexed integer array ans of size n such that ans[i] is the answer to the ith query.

A subarray is a contiguous sequence of elements in an array.

Example 1:

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.


Example 2:

Input: nums = [10,20,50,10]
Output: [50,20,10,10]
Explanation:
i=0:
- The subarrays of size 1 are [10], [20], [50], [10]. The minimum values are 10, 20, 50, 10.
- The maximum of the minimum values is 50.
i=1:
- The subarrays of size 2 are [10,20], [20,50], [50,10]. The minimum values are 10, 20, 10.
- The maximum of the minimum values is 20.
i=2:
- The subarrays of size 3 are [10,20,50], [20,50,10]. The minimum values are 10, 10.
- The maximum of the minimum values is 10.
i=3:
- There is one subarray of size 4, which is [10,20,50,10]. The minimum value is 10.
- There is only one value, so the maximum is 10.


Constraints:

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i] <= 109

Solutions

• class Solution {
public int[] findMaximums(int[] nums) {
int n = nums.length;
int[] left = new int[n];
int[] right = new int[n];
Arrays.fill(left, -1);
Arrays.fill(right, n);
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
while (!stk.isEmpty() && nums[stk.peek()] >= nums[i]) {
stk.pop();
}
if (!stk.isEmpty()) {
left[i] = stk.peek();
}
stk.push(i);
}
stk.clear();
for (int i = n - 1; i >= 0; --i) {
while (!stk.isEmpty() && nums[stk.peek()] >= nums[i]) {
stk.pop();
}
if (!stk.isEmpty()) {
right[i] = stk.peek();
}
stk.push(i);
}
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int m = right[i] - left[i] - 1;
ans[m - 1] = Math.max(ans[m - 1], nums[i]);
}
for (int i = n - 2; i >= 0; --i) {
ans[i] = Math.max(ans[i], ans[i + 1]);
}
return ans;
}
}

• class Solution {
public:
vector<int> findMaximums(vector<int>& nums) {
int n = nums.size();
vector<int> left(n, -1);
vector<int> right(n, n);
stack<int> stk;
for (int i = 0; i < n; ++i) {
while (!stk.empty() && nums[stk.top()] >= nums[i]) {
stk.pop();
}
if (!stk.empty()) {
left[i] = stk.top();
}
stk.push(i);
}
stk = stack<int>();
for (int i = n - 1; i >= 0; --i) {
while (!stk.empty() && nums[stk.top()] >= nums[i]) {
stk.pop();
}
if (!stk.empty()) {
right[i] = stk.top();
}
stk.push(i);
}
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
int m = right[i] - left[i] - 1;
ans[m - 1] = max(ans[m - 1], nums[i]);
}
for (int i = n - 2; i >= 0; --i) {
ans[i] = max(ans[i], ans[i + 1]);
}
return ans;
}
};

• 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


• func findMaximums(nums []int) []int {
n := len(nums)
left := make([]int, n)
right := make([]int, n)
for i := range left {
left[i], right[i] = -1, n
}
stk := []int{}
for i, x := range nums {
for len(stk) > 0 && nums[stk[len(stk)-1]] >= x {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
left[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
stk = []int{}
for i := n - 1; i >= 0; i-- {
x := nums[i]
for len(stk) > 0 && nums[stk[len(stk)-1]] >= x {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
right[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
ans := make([]int, n)
for i := range ans {
m := right[i] - left[i] - 1
ans[m-1] = max(ans[m-1], nums[i])
}
for i := n - 2; i >= 0; i-- {
ans[i] = max(ans[i], ans[i+1])
}
return ans
}