2454. Next Greater Element IV
Explanation
- We can solve this problem using a stack to keep track of elements in decreasing order.
- We iterate through the array from right to left and maintain a stack of elements.
- For each element, we pop elements from the stack as long as they are less than the current element. The top of the stack will be the second greater element for the current element.
- If the stack becomes empty before finding the second greater element, we set it to -1.
- We store the result in an array and return it.
Time Complexity: O(n) Space Complexity: O(n)
import java.util.Stack;
class Solution {
public int[] nextGreaterElement(int[] nums) {
int n = nums.length;
int[] result = new int[n];
Stack<Integer> stack = new Stack<>();
for (int i = n - 1; i >= 0; i--) {
while (!stack.isEmpty() && nums[i] >= stack.peek()) {
stack.pop();
}
result[i] = stack.isEmpty() ? -1 : stack.peek();
stack.push(nums[i]);
}
return result;
}
}Code Editor (Testing phase)
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