Skip to content

1944. Number of Visible People in a Queue

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: Monotonic Stack

We observe that for the \(i\)-th person, the people he can see must be strictly increasing in height from left to right.

Therefore, we can traverse the array \(\textit{heights}\) in reverse order, using a stack \(\textit{stk}\) that is monotonically increasing from top to bottom to record the heights of the people we have traversed.

For the \(i\)-th person, if the stack is not empty and the top element of the stack is less than \(\textit{heights}[i]\), we increment the count of people the \(i\)-th person can see, then pop the top element of the stack, until the stack is empty or the top element of the stack is greater than or equal to \(\textit{heights}[i]\). If the stack is not empty at this point, it means the top element of the stack is greater than or equal to \(\textit{heights}[i]\), so we increment the count of people the \(i\)-th person can see by 1.

Next, we push \(\textit{heights}[i]\) onto the stack and continue to the next person.

After traversing, we return the answer array \(\textit{ans}\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the array \(\textit{heights}\).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * 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;
}

Comments