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Formatted question description: https://leetcode.ca/all/503.html

# 503. Next Greater Element II (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.

Related Topics:
Stack

Similar Questions:

## Solution 1. Monotonic stack

// OJ: https://leetcode.com/problems/next-greater-element-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> nextGreaterElements(vector<int>& A) {
stack<int> s;
int N = A.size();
vector<int> ans(N, -1);
for (int i = 0; i < 2 * N; ++i) {
int n = A[i % N];
while (s.size() && A[s.top()] < n) {
ans[s.top()] = n;
s.pop();
}
s.push(i % N);
}
return ans;
}
};

• class Solution {
public int[] nextGreaterElements(int[] nums) {
int length = nums.length;
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < length; i++)
int[] nextGreaterElements = new int[length];
for (int i = 0; i < length; i++)
nextGreaterElements[i] = -1;
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> indicesStack = new Stack<Integer>();
for (int i = 0; i < length; i++) {
int num = nums[i];
while (!stack.isEmpty() && stack.peek() < num) {
int prevNum = stack.pop();
int index = indicesStack.pop();
nextGreaterElements[index] = num;
}
stack.push(num);
indicesStack.push(i);
}
for (int i = 0; i < length; i++) {
int num = nums[i];
while (!stack.isEmpty() && stack.peek() < num) {
int prevNum = stack.pop();
int index = indicesStack.pop();
nextGreaterElements[index] = num;
}
stack.push(num);
indicesStack.push(i);
}
return nextGreaterElements;
}
}

############

class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
Arrays.fill(ans, -1);
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < (n << 1); ++i) {
while (!stk.isEmpty() && nums[stk.peek()] < nums[i % n]) {
ans[stk.pop()] = nums[i % n];
}
stk.push(i % n);
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/next-greater-element-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> nextGreaterElements(vector<int>& A) {
stack<int> s;
int N = A.size();
vector<int> ans(N, -1);
for (int i = 0; i < 2 * N; ++i) {
int n = A[i % N];
while (s.size() && A[s.top()] < n) {
ans[s.top()] = n;
s.pop();
}
s.push(i % N);
}
return ans;
}
};

• class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [-1] * n
stk = []
for i in range(n << 1):
while stk and nums[stk[-1]] < nums[i % n]:
ans[stk.pop()] = nums[i % n]
stk.append(i % n)
return ans

############

class Solution(object):
def nextGreaterElements(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
ans = [-1] * len(nums)
n = len(nums)
stack = []
nums *= 2
for i, num in enumerate(nums):
while stack and nums[stack[-1]] < num:
top = stack.pop()
if top < n:
ans[top] = num
stack.append(i)
return ans


• func nextGreaterElements(nums []int) []int {
n := len(nums)
ans := make([]int, n)
for i := range ans {
ans[i] = -1
}
var stk []int
for i := 0; i < (n << 1); i++ {
for len(stk) > 0 && nums[stk[len(stk)-1]] < nums[i%n] {
ans[stk[len(stk)-1]] = nums[i%n]
stk = stk[:len(stk)-1]
}
stk = append(stk, i%n)
}
return ans
}

• function nextGreaterElements(nums: number[]): number[] {
const stack: number[] = [],
len = nums.length;
const res: number[] = new Array(len).fill(-1);
for (let i = 0; i < 2 * len - 1; i++) {
const j = i % len;
while (stack.length !== 0 && nums[stack[stack.length - 1]] < nums[j]) {
res[stack[stack.length - 1]] = nums[j];
stack.pop();
}
stack.push(j);
}
return res;
}


• /**
* @param {number[]} nums
* @return {number[]}
*/
var nextGreaterElements = function (nums) {
const n = nums.length;
let stk = [];
let ans = new Array(n).fill(-1);
for (let i = 0; i < n << 1; i++) {
const j = i % n;
while (stk.length && nums[stk[stk.length - 1]] < nums[j]) {
ans[stk.pop()] = nums[j];
}
stk.push(j);
}
return ans;
};