Question

1944. Number of Visible People in a Queue

中文文档

Description

There are n people standing in a queue, and they numbered from 0 to n - 1 in left to right order. You are given an array heights of distinct integers where heights[i] represents the height of the ith person.

A person can see another person to their right in the queue if everybody in between is shorter than both of them. More formally, the ith person can see the jth person if i < j and min(heights[i], heights[j]) > max(heights[i+1], heights[i+2], ..., heights[j-1]).

Return _an array _answer_ of length _n_ where _answer[i]_ is the number of people the _ith_ person can see to their right in the queue_.

 

Example 1:

Input: heights = [10,6,8,5,11,9]
Output: [3,1,2,1,1,0]
Explanation:
Person 0 can see person 1, 2, and 4.
Person 1 can see person 2.
Person 2 can see person 3 and 4.
Person 3 can see person 4.
Person 4 can see person 5.
Person 5 can see no one since nobody is to the right of them.

Example 2:

Input: heights = [5,1,2,3,10]
Output: [4,1,1,1,0]

 

Constraints:

• n == heights.length
• 1 <= n <= 105
• 1 <= heights[i] <= 105
• All the values of heights are unique.

Solutions

Solution 1

class Solution:
  def canSeePersonsCount(self, heights: List[int]) -> List[int]:
    n = len(heights)
    ans = [0] * n
    stk = []
    for i in range(n - 1, -1, -1):
      while stk and stk[-1] < heights[i]:
        ans[i] += 1
        stk.pop()
      if stk:
        ans[i] += 1
      stk.append(heights[i])
    return ans

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

class Solution {
public:
  vector<int> canSeePersonsCount(vector<int>& heights) {
    int n = heights.size();
    vector<int> ans(n);
    stack<int> stk;
    for (int i = n - 1; ~i; --i) {
      while (stk.size() && stk.top() < heights[i]) {
        ++ans[i];
        stk.pop();
      }
      if (stk.size()) {
        ++ans[i];
      }
      stk.push(heights[i]);
    }
    return ans;
  }
};

func canSeePersonsCount(heights []int) []int {
  n := len(heights)
  ans := make([]int, n)
  stk := []int{}
  for i := n - 1; i >= 0; i-- {
    for len(stk) > 0 && stk[len(stk)-1] < heights[i] {
      ans[i]++
      stk = stk[:len(stk)-1]
    }
    if len(stk) > 0 {
      ans[i]++
    }
    stk = append(stk, heights[i])
  }
  return ans
}

function canSeePersonsCount(heights: number[]): number[] {
  const n = heights.length;
  const ans: number[] = new Array(n).fill(0);
  const stk: number[] = [];
  for (let i = n - 1; ~i; --i) {
    while (stk.length && stk.at(-1) < heights[i]) {
      ++ans[i];
      stk.pop();
    }
    if (stk.length) {
      ++ans[i];
    }
    stk.push(heights[i]);
  }
  return ans;
}

impl Solution {
  pub fn can_see_persons_count(heights: Vec<i32>) -> Vec<i32> {
    let n = heights.len();
    let mut ans = vec![0; n];
    let mut stack = Vec::new();
    for i in (0..n).rev() {
      while !stack.is_empty() {
        ans[i] += 1;
        if heights[i] <= heights[*stack.last().unwrap()] {
          break;
        }
        stack.pop();
      }
      stack.push(i);
    }
    ans
  }
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* canSeePersonsCount(int* heights, int heightsSize, int* returnSize) {
  int* ans = malloc(sizeof(int) * heightsSize);
  memset(ans, 0, sizeof(int) * heightsSize);
  int stack[heightsSize];
  int i = 0;
  for (int j = heightsSize - 1; j >= 0; j--) {
    while (i) {
      ans[j]++;
      if (heights[j] <= heights[stack[i - 1]]) {
        break;
      }
      i--;
    }
    stack[i++] = j;
  }
  *returnSize = heightsSize;
  return ans;
}