Maximum people a person can see while standing in a line in both direction
Last Updated :
23 Jul, 2025
Given an array height[] which represents the height of n people standing in a line. A person i can see a person j if height[j] < height[i] and there is no person k standing in between them such that height[k] >= height[i]. Find the maximum number of people a person can see.
Note: A person can see any other person irrespective of whether they are standing in his front or back.
Examples:
Input: heights[] = {6, 2, 5, 4, 5, 1, 6}.
Output: 6
Explanation:
Person 1 (height = 6) can see five other people at following positions (2, 3, 4, 5. 6) in addition to himself, i.e. total 6.
Person 2 (height: 2) can see only himself.
Person 3 (height = 5) is able to see people 2nd, 3rd, and 4th person.
Only Person 4 (height = 4) can see himself.
Person 5 (height = 5) can see people 4th, 5th, and 6th.
Person 6 (height =1) can only see himself.
Person 7 (height = 6) can see 2nd, 3rd, 4th, 5th, 6th, and 7th people.
A maximum of six people can be seen by Person 1, 7th
Input: heights[] = {1, 3, 6, 4}
Output: 3
[Naive Approach] Using Nested Loops - O(n^2) Time and O(1) Space
The idea is to determine how many people a person can see by checking both left and right directions separately. For each person, we iterate backward to count visible people on the left until we hit a taller height, and forward to count on the right similarly. The key observation is that a person can only see others shorter than them until a taller person blocks the view.
C++
// C++ Code to find the maximum number of people
// a person can see in both directions
// using a naive approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum number of people
// a person can see in both directions
int maxPeople(vector<int>& height) {
int n = height.size();
int maxCount = 0;
// Iterate through each person
for (int i = 0; i < n; i++) {
// Count the person itself
int count = 1;
int maxLeft = 0, maxRight = 0;
// Check visibility towards the left
for (int j = i - 1; j >= 0; j--) {
if (height[j] > maxLeft) {
count++;
maxLeft = height[j];
}
}
// Check visibility towards the right
for (int j = i + 1; j < n; j++) {
if (height[j] > maxRight) {
count++;
maxRight = height[j];
}
}
// Update the maximum count
maxCount = max(maxCount, count);
}
return maxCount;
}
int main() {
vector<int> height = {6, 2, 5, 4, 5, 1, 6};
cout<< maxPeople(height);
return 0;
}
// Java Code to find the maximum number of people
// a person can see in both directions
// using a naive approach
import java.util.*;
class GfG {
// Function to find the maximum number of people
// a person can see in both directions
static int maxPeople(int[] height) {
int n = height.length;
int maxCount = 0;
// Iterate through each person
for (int i = 0; i < n; i++) {
// Count the person itself
int count = 1;
int maxLeft = 0, maxRight = 0;
// Check visibility towards the left
for (int j = i - 1; j >= 0; j--) {
if (height[j] > maxLeft) {
count++;
maxLeft = height[j];
}
}
// Check visibility towards the right
for (int j = i + 1; j < n; j++) {
if (height[j] > maxRight) {
count++;
maxRight = height[j];
}
}
// Update the maximum count
maxCount = Math.max(maxCount, count);
}
return maxCount;
}
public static void main(String[] args) {
int[] height = {6, 2, 5, 4, 5, 1, 6};
System.out.println(maxPeople(height));
}
}
# Python Code to find the maximum number of people
# a person can see in both directions
# using a naive approach
# Function to find the maximum number of people
# a person can see in both directions
def maxPeople(height):
n = len(height)
max_count = 0
# Iterate through each person
for i in range(n):
# Count the person itself
count = 1
max_left = 0
max_right = 0
# Check visibility towards the left
for j in range(i - 1, -1, -1):
if height[j] > max_left:
count += 1
max_left = height[j]
# Check visibility towards the right
for j in range(i + 1, n):
if height[j] > max_right:
count += 1
max_right = height[j]
# Update the maximum count
max_count = max(max_count, count)
return max_count
if __name__ == "__main__":
height = [6, 2, 5, 4, 5, 1, 6]
print(maxPeople(height))
// C# Code to find the maximum number of people
// a person can see in both directions
// using a naive approach
using System;
class GfG {
// Function to find the maximum number of people
// a person can see in both directions
public static int maxPeople(int[] height) {
int n = height.Length;
int maxCount = 0;
// Iterate through each person
for (int i = 0; i < n; i++) {
// Count the person itself
int count = 1;
int maxLeft = 0, maxRight = 0;
// Check visibility towards the left
for (int j = i - 1; j >= 0; j--) {
if (height[j] > maxLeft) {
count++;
maxLeft = height[j];
}
}
// Check visibility towards the right
for (int j = i + 1; j < n; j++) {
if (height[j] > maxRight) {
count++;
maxRight = height[j];
}
}
// Update the maximum count
maxCount = Math.Max(maxCount, count);
}
return maxCount;
}
public static void Main() {
int[] height = {6, 2, 5, 4, 5, 1, 6};
Console.WriteLine(maxPeople(height));
}
}
// JavaScript Code to find the maximum number of people
// a person can see in both directions
// using a naive approach
// Function to find the maximum number of people
// a person can see in both directions
function maxPeople(height) {
let n = height.length;
let maxCount = 0;
// Iterate through each person
for (let i = 0; i < n; i++) {
// Count the person itself
let count = 1;
let maxLeft = 0, maxRight = 0;
// Check visibility towards the left
for (let j = i - 1; j >= 0; j--) {
if (height[j] > maxLeft) {
count++;
maxLeft = height[j];
}
}
// Check visibility towards the right
for (let j = i + 1; j < n; j++) {
if (height[j] > maxRight) {
count++;
maxRight = height[j];
}
}
// Update the maximum count
maxCount = Math.max(maxCount, count);
}
return maxCount;
}
// Driver code
let height = [6, 2, 5, 4, 5, 1, 6];
console.log(maxPeople(height));
[Expected Approach] Using Monotonic Stack - O(n) Time and O(n) Space
If we take a closer look at this problem, we can notice that for every element, we need to find next greater (Closest greater element on right side) and previous greater (Closest greater on the left side). All the elements before next greater would be visible on the right side and all the elements after previous greater would be visible on the left side. We mainly implement the same stack idea as next greater and previous greater problems to find the next greater and previous greater for all elements in O(n) time. Finally, we return maximum of all values.
C++
// C++ Code to find max people visible in
// both directions using Monotonic Stack
#include <bits/stdc++.h>
using namespace std;
// Function to compute max people visible using stack
int maxPeople(vector<int>& height) {
int n = height.size();
// Initialize each person sees himself
vector<int> left(n, 1), right(n, 1);
stack<int> st;
// Compute left visibility using a decreasing stack
for (int i = 0; i < n; i++) {
while (!st.empty() && height[st.top()] < height[i]) {
left[i] += left[st.top()];
st.pop();
}
st.push(i);
}
// Clear stack for right computation
while (!st.empty()) st.pop();
// Compute right visibility using a decreasing stack
for (int i = n - 1; i >= 0; i--) {
while (!st.empty() && height[st.top()] < height[i]) {
right[i] += right[st.top()];
st.pop();
}
st.push(i);
}
// Find the max people visible
int maxCount = 0;
for (int i = 0; i < n; i++) {
maxCount = max(maxCount, left[i] + right[i] - 1);
}
return maxCount;
}
int main() {
vector<int> height = {6, 2, 5, 4, 5, 1, 6};
cout << maxPeople(height);
return 0;
}
// Java Code to find max people visible in
// both directions using Monotonic Stack
import java.util.Stack;
class GfG {
// Function to compute max people visible using stack
static int maxPeople(int[] height) {
int n = height.length;
// Initialize each person sees himself
int[] left = new int[n], right = new int[n];
for (int i = 0; i < n; i++) {
left[i] = right[i] = 1;
}
Stack<Integer> st = new Stack<>();
// Compute left visibility using a decreasing stack
for (int i = 0; i < n; i++) {
while (!st.isEmpty() && height[st.peek()] < height[i]) {
left[i] += left[st.pop()];
}
st.push(i);
}
// Clear stack for right computation
st.clear();
// Compute right visibility using a decreasing stack
for (int i = n - 1; i >= 0; i--) {
while (!st.isEmpty() && height[st.peek()] < height[i]) {
right[i] += right[st.pop()];
}
st.push(i);
}
// Find the max people visible
int maxCount = 0;
for (int i = 0; i < n; i++) {
maxCount = Math.max(maxCount, left[i] + right[i] - 1);
}
return maxCount;
}
public static void main(String[] args) {
int[] height = {6, 2, 5, 4, 5, 1, 6};
System.out.println(maxPeople(height));
}
}
# Python Code to find max people visible in
# both directions using Monotonic Stack
def maxPeople(height):
n = len(height)
# Initialize each person sees himself
left = [1] * n
right = [1] * n
st = []
# Compute left visibility using a decreasing stack
for i in range(n):
while st and height[st[-1]] < height[i]:
left[i] += left[st.pop()]
st.append(i)
# Clear stack for right computation
st.clear()
# Compute right visibility using a decreasing stack
for i in range(n - 1, -1, -1):
while st and height[st[-1]] < height[i]:
right[i] += right[st.pop()]
st.append(i)
# Find the max people visible
maxCount = 0
for i in range(n):
maxCount = max(maxCount, left[i] + right[i] - 1)
return maxCount
if __name__ == "__main__":
height = [6, 2, 5, 4, 5, 1, 6]
print(maxPeople(height))
// C# Code to find max people visible in
// both directions using Monotonic Stack
using System;
using System.Collections.Generic;
class GfG {
// Function to compute max people visible using stack
public static int maxPeople(int[] height) {
int n = height.Length;
// Initialize each person sees himself
int[] left = new int[n], right = new int[n];
for (int i = 0; i < n; i++) {
left[i] = right[i] = 1;
}
Stack<int> st = new Stack<int>();
// Compute left visibility using a decreasing stack
for (int i = 0; i < n; i++) {
while (st.Count > 0 && height[st.Peek()] < height[i]) {
left[i] += left[st.Pop()];
}
st.Push(i);
}
// Clear stack for right computation
st.Clear();
// Compute right visibility using a decreasing stack
for (int i = n - 1; i >= 0; i--) {
while (st.Count > 0 && height[st.Peek()] < height[i]) {
right[i] += right[st.Pop()];
}
st.Push(i);
}
// Find the max people visible
int maxCount = 0;
for (int i = 0; i < n; i++) {
maxCount = Math.Max(maxCount, left[i] + right[i] - 1);
}
return maxCount;
}
public static void Main() {
int[] height = {6, 2, 5, 4, 5, 1, 6};
Console.WriteLine(maxPeople(height));
}
}
// Javascript Code to find max people visible in
// both directions using Monotonic Stack
function maxPeople(height) {
let n = height.length;
// Initialize each person sees himself
let left = new Array(n).fill(1);
let right = new Array(n).fill(1);
let st = [];
// Compute left visibility using a decreasing stack
for (let i = 0; i < n; i++) {
while (st.length > 0 && height[st[st.length - 1]] < height[i]) {
left[i] += left[st.pop()];
}
st.push(i);
}
// Clear stack for right computation
st = [];
// Compute right visibility using a decreasing stack
for (let i = n - 1; i >= 0; i--) {
while (st.length > 0 && height[st[st.length - 1]] < height[i]) {
right[i] += right[st.pop()];
}
st.push(i);
}
// Find the max people visible
let maxCount = 0;
for (let i = 0; i < n; i++) {
maxCount = Math.max(maxCount, left[i] + right[i] - 1);
}
return maxCount;
}
let height = [6, 2, 5, 4, 5, 1, 6];
console.log(maxPeople(height));
Further Optimization
We can further optimize the above code to wok using single for loop only. Please refer Largest Area in a Histogram problem fore reference.
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