Question

503. Next Greater Element II

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Description

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return _the next greater number for every element in_ nums.

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, return -1 for this number.

 

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]

 

Constraints:

• 1 <= nums.length <= 104
• -109 <= nums[i] <= 109

Solutions

Solution 1

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 {
  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;
  }
}

class Solution {
public:
  vector<int> nextGreaterElements(vector<int>& nums) {
    int n = nums.size();
    vector<int> ans(n, -1);
    stack<int> stk;
    for (int i = 0; i < (n << 1); ++i) {
      while (!stk.empty() && nums[stk.top()] < nums[i % n]) {
        ans[stk.top()] = nums[i % n];
        stk.pop();
      }
      stk.push(i % n);
    }
    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;
};

Solution 2

class Solution:
  def nextGreaterElements(self, nums: List[int]) -> List[int]:
    n = len(nums)
    ans = [-1] * n
    stk = []
    for i in range(n * 2 - 1, -1, -1):
      i %= n
      while stk and stk[-1] <= nums[i]:
        stk.pop()
      if stk:
        ans[i] = stk[-1]
      stk.append(nums[i])
    return ans

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 = n * 2 - 1; i >= 0; --i) {
      int j = i % n;
      while (!stk.isEmpty() && stk.peek() <= nums[j]) {
        stk.pop();
      }
      if (!stk.isEmpty()) {
        ans[j] = stk.peek();
      }
      stk.push(nums[j]);
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> nextGreaterElements(vector<int>& nums) {
    int n = nums.size();
    vector<int> ans(n, -1);
    stack<int> stk;
    for (int i = n * 2 - 1; ~i; --i) {
      int j = i % n;
      while (!stk.empty() && stk.top() <= nums[j]) stk.pop();
      if (!stk.empty()) ans[j] = stk.top();
      stk.push(nums[j]);
    }
    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 := n*2 - 1; i >= 0; i-- {
    j := i % n
    for len(stk) > 0 && stk[len(stk)-1] <= nums[j] {
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      ans[j] = stk[len(stk)-1]
    }
    stk = append(stk, nums[j])
  }
  return ans
}

/**
 * @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 = n * 2 - 1; ~i; --i) {
    const j = i % n;
    while (stk.length && stk[stk.length - 1] <= nums[j]) {
      stk.pop();
    }
    if (stk.length) {
      ans[j] = stk[stk.length - 1];
    }
    stk.push(nums[j]);
  }
  return ans;
};