LeetCode 2832: Maximal Range That Each Element Is Maximum in It Solution
Master LeetCode problem 2832 (Maximal Range That Each Element Is Maximum in It), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2832. Maximal Range That Each Element Is Maximum in It
Problem Explanation
Explanation:
The problem asks us to find the maximal range for each element in an array, where the element is the maximum value in that range. We can achieve this by iterating through the array and for each element, finding the range of indices where it is the maximum value.
To solve this problem, we can use two arrays left and right to store the leftmost and rightmost indices where the current element is the maximum value respectively. Then, we can compute the range for each element by calculating right[i] - left[i] + 1 where i is the index of the element.
Solution:
Solution Code
class Solution {
public int[] maxRange(int[] nums) {
int n = nums.length;
int[] left = new int[n];
int[] right = new int[n];
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < n; i++) {
while (!stack.isEmpty() && nums[i] >= nums[stack.peek()]) {
stack.pop();
}
left[i] = stack.isEmpty() ? 0 : stack.peek() + 1;
stack.push(i);
}
stack.clear();
for (int i = n - 1; i >= 0; i--) {
while (!stack.isEmpty() && nums[i] > nums[stack.peek()]) {
stack.pop();
}
right[i] = stack.isEmpty() ? n - 1 : stack.peek() - 1;
stack.push(i);
}
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = right[i] - left[i] + 1;
}
return result;
}
}Try It Yourself
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