Next Greater Element II

Stack

Medium

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

Example 1:

Input:
[1,2,1]

Output:
[2,-1,2]

Explanation:
The first 1's next greater number is 2;

The number 2 can't find next greater number;

The second 1's next greater number needs to search circularly, which is also 2.

Note:The length of given array won't exceed 10000.

Solution

Monotonous Stack + Circular Array Search (21 ms, faster than 94.64%)

class Solution {
public int[] nextGreaterElements(int[] nums) {
int len = nums.length;
int[] res = new int[len];
Deque<Integer> stack = new ArrayDeque<>();

Arrays.fill(res, -1);

for (int i = 0; i < 2 * len; i++) {
while (!stack.isEmpty() && nums[i % len] > nums[stack.peek()]) {
int idx = stack.pop();
res[idx] = nums[i % len];
}
stack.push(i % len);
}

return res;
}
}
• Time complexity : O(n). Only two traversals of the nums array are done. Further, at most 2n elements are pushed and popped from the stack.

• Space complexity : O(n). A stack of size n is used. res array of size n is used.

A little optimization with i < len and checking if stack.isEmpty() then break

class Solution {
public int[] nextGreaterElements(int[] nums) {
int len = nums.length;
int[] res = new int[len];
Deque<Integer> stack = new ArrayDeque<>();

Arrays.fill(res, -1);

for (int i = 0; i < 2 * len; i++) {
while (!stack.isEmpty() && nums[i % len] > nums[stack.peek()]) {
int idx = stack.pop();
res[idx] = nums[i % len];
}
if (i < len) {
stack.push(i);
}
if (stack.isEmpty()) {
break;
}
}

return res;
}
}