Number of Visible People in a Queue - Problem

There are n people standing in a queue, 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 i^th 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 i^th person can see the j^th 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 i^th person can see to their right in the queue.

Input & Output

Example 1 — Basic Queue

$ Input: heights = [10,6,8,5,11,9]

› Output: [3,1,2,1,1,0]

💡 Note: Person 0 (height 10) can see persons 1, 2, and 4. Person 1 (height 6) can see person 2. Person 2 (height 8) can see persons 3 and 4. Person 3 (height 5) can see person 4. Person 4 (height 11) can see person 5. Person 5 sees no one.

Example 2 — Increasing Heights

$ Input: heights = [5,1,2,3,10]

› Output: [4,1,1,1,0]

💡 Note: Person 0 can see everyone to the right since they're all shorter until person 4. Each person can see the next taller person.

Example 3 — Decreasing Heights

$ Input: heights = [10,8,6,4,2]

› Output: [4,3,2,1,0]

💡 Note: In decreasing order, each person can see all people to their right since no one blocks the view.

Constraints

1 ≤ heights.length ≤ 10⁵
1 ≤ heights[i] ≤ 10⁹
All values in heights are distinct

Visualization

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Asked in

G Google 15 a Amazon 12 f Facebook 8 M Microsoft 6

The key insight is to use a monotonic decreasing stack to efficiently track which people are visible from each position. The optimal approach processes people from right to left, maintaining a stack of indices in decreasing height order. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n³) Space: O(1)

For each person i, iterate through all people j where j > i. For each pair (i,j), check if all people between them are shorter than both i and j.

Monotonic Stack (Optimal)

⏱️ Time: O(n) Space: O(n)

Process people from right to left, maintaining a stack of indices in decreasing order of heights. For each person, count visible people by popping from stack.

Brute Force - Check All Pairs — Algorithm Steps

For each person i from 0 to n-1
For each person j from i+1 to n-1
Check if min(heights[i], heights[j]) > max(heights[i+1...j-1])
If yes, increment visible count for person i

Visualization

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Step-by-Step Walkthrough

For Each Person

Start from person i and check all j > i

Check Visibility

For each pair (i,j), check if all between are shorter

Count Visible

Increment count if person j is visible from i

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* heights, int n, int* returnSize) {
    *returnSize = n;
    int* result = (int*)calloc(n, sizeof(int));
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int minHeight = (heights[i] < heights[j]) ? heights[i] : heights[j];
            int canSee = 1;
            
            for (int k = i + 1; k < j; k++) {
                if (heights[k] >= minHeight) {
                    canSee = 0;
                    break;
                }
            }
            
            if (canSee) {
                result[i]++;
            }
        }
    }
    
    return result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int heights[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        heights[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int returnSize;
    int* result = solution(heights, n, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Triple nested loop: n people × n people to check × n people between them

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

⚡ Linearithmic Space

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Number of Visible People in a Queue - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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