Skip to content
Related Articles

Related Articles

Check if a triplet of buildings can be selected such that the third building is taller than the first building and smaller than the second building

View Discussion
Improve Article
Save Article
  • Difficulty Level : Expert
  • Last Updated : 04 Sep, 2021
View Discussion
Improve Article
Save Article

Given an array arr[] consisting of N integers, where each array element represents the height of a building situated on the X co-ordinates, the task is to check if it is possible to select 3 buildings, such that the third selected building is taller than the first selected building and shorter than the second selected building.

Examples:

Input: arr[] = {4, 7, 11, 5, 13, 2}
Output: Yes
Explanation:
One possible way is to select the building at indices [0, 1, 3] with heights 4, 7 and 5 respectively.

Input: arr[] = {11, 11, 12, 9}
Output: No

Recommended Practice

Approach: The given problem can be solved using the Stack data structure. Follow the steps below to solve the problem:

  • If N is less than 3, then print “No“.
  • Initialize an array, say preMin[], to store the prefix minimum array of array arr[].
  • Traverse the array arr[] and update preMin[i] as preMin[i] = min(preMin[i-1], arr[i]).
  • Now, initialize a Stack, say stack, to store the elements from the ending in ascending order.
  • Traverse the array arr[] in reverse using a variable, say i, and perform the following steps:
  • After completing the above steps, if none of the above cases satisfy, then print “No“.

Below is the implementation of the above approach:

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible
// to select three buildings that
// satisfy the given condition
string recreationalSpot(int arr[], int N)
{
 
    if (N < 3) {
        return "No";
    }
 
    // Stores prefix min array
    int preMin[N];
    preMin[0] = arr[0];
 
    // Iterate over the range [1, N-1]
    for (int i = 1; i < N; i++) {
        preMin[i] = min(preMin[i - 1], arr[i]);
    }
 
    // Stores the element from the
    // ending in increasing order
    stack<int> stack;
 
    // Iterate until j is greater than
    // or equal to 0
    for (int j = N - 1; j >= 0; j--) {
 
        // If current array element is
        // greater than the prefix min
        // upto j
        if (arr[j] > preMin[j]) {
 
            // Iterate while stack is not
            // empty and top element is
            // less than or equal to preMin[j]
 
            while (!stack.empty()
                   && stack.top() <= preMin[j]) {
                // Remove the top element
                stack.pop();
            }
 
            // If stack is not empty and top
            // element of the stack is less
            // than the current element
            if (!stack.empty() && stack.top() < arr[j]) {
                return "Yes";
            }
            // Push the arr[j] in stack
            stack.push(arr[j]);
        }
    }
    // If none of the above case
    // satisfy then return "No"
    return "No";
}
 
// Driver code
int main()
{
    // Input
    int arr[] = { 4, 7, 11, 5, 13, 2 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    cout << recreationalSpot(arr, size);
}

Java




// Java implementation of the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to check if it is possible
// to select three buildings that
// satisfy the given condition
static String recreationalSpot(int arr[], int N)
{
    if (N < 3)
    {
        return "No";
    }
 
    // Stores prefix min array
    int preMin[] = new int[N];
    preMin[0] = arr[0];
 
    // Iterate over the range [1, N-1]
    for(int i = 1; i < N; i++)
    {
        preMin[i] = Math.min(preMin[i - 1], arr[i]);
    }
 
    // Stores the element from the
    // ending in increasing order
    Stack<Integer> stack = new Stack<Integer>();
 
    // Iterate until j is greater than
    // or equal to 0
    for(int j = N - 1; j >= 0; j--)
    {
         
        // If current array element is
        // greater than the prefix min
        // upto j
        if (arr[j] > preMin[j])
        {
             
            // Iterate while stack is not
            // empty and top element is
            // less than or equal to preMin[j]
            while (stack.empty() == false &&
                   stack.peek() <= preMin[j])
            {
                 
                // Remove the top element
                stack.pop();
            }
 
            // If stack is not empty and top
            // element of the stack is less
            // than the current element
            if (stack.empty() == false &&
                stack.peek() < arr[j])
            {
                return "Yes";
            }
             
            // Push the arr[j] in stack
            stack.push(arr[j]);
        }
    }
     
    // If none of the above case
    // satisfy then return "No"
    return "No";
}
 
// Driver code
public static void main(String[] args)
{
     
    // Input
    int arr[] = { 4, 7, 11, 5, 13, 2 };
    int size = arr.length;
 
    System.out.println(recreationalSpot(arr, size));
}
}
 
// This code is contributed by Dharanendra L V.

Python3




# Python3 implementation of the above approach
 
# Function to check if it is possible
# to select three buildings that
# satisfy the given condition
def recreationalSpot(arr, N):
     
    if (N < 3):
        return "No"
 
    # Stores prefix min array
    preMin = [0] * N
    preMin[0] = arr[0]
 
    # Iterate over the range [1, N-1]
    for i in range(1, N):
        preMin[i] = min(preMin[i - 1], arr[i])
 
    # Stores the element from the
    # ending in increasing order
    stack = []
 
    # Iterate until j is greater than
    # or equal to 0
    for j in range(N - 1, -1, -1):
         
        # If current array element is
        # greater than the prefix min
        # upto j
        if (arr[j] > preMin[j]):
 
            # Iterate while stack is not
            # empty and top element is
            # less than or equal to preMin[j]
            while (len(stack) > 0 and
                   stack[-1] <= preMin[j]):
                        
                # Remove the top element
                del stack[-1]
 
            # If stack is not empty and top
            # element of the stack is less
            # than the current element
            if (len(stack) > 0 and stack[-1] < arr[j]):
                return "Yes"
 
            # Push the arr[j] in stack
            stack.append(arr[j])
 
    # If none of the above case
    # satisfy then return "No"
    return "No"
 
# Driver code
if __name__ == '__main__':
     
    # Input
    arr =  [ 4, 7, 11, 5, 13, 2 ]
    size = len(arr)
 
    print (recreationalSpot(arr, size))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
using System.Collections.Generic;
public class GFG{
 
// Function to check if it is possible
// to select three buildings that
// satisfy the given condition
static String recreationalSpot(int []arr, int N)
{
    if (N < 3)
    {
        return "No";
    }
 
    // Stores prefix min array  
    int []preMin = new int[N];
    preMin[0] = arr[0];
 
    // Iterate over the range [1, N-1]
    for(int i = 1; i < N; i++)
    {
        preMin[i] = Math.Min(preMin[i - 1], arr[i]);
    }
 
    // Stores the element from the
    // ending in increasing order
    Stack<int> stack = new Stack<int>();
 
    // Iterate until j is greater than
    // or equal to 0
    for(int j = N - 1; j >= 0; j--)
    {
         
        // If current array element is
        // greater than the prefix min
        // upto j
        if (arr[j] > preMin[j])
        {
             
            // Iterate while stack is not
            // empty and top element is
            // less than or equal to preMin[j]
            while (stack.Count!=0 &&
                   stack.Peek() <= preMin[j])
            {
                 
                // Remove the top element
                stack.Pop();
            }
 
            // If stack is not empty and top
            // element of the stack is less
            // than the current element
            if (stack.Count!=0 &&
                stack.Peek() < arr[j])
            {
                return "Yes";
            }
             
            // Push the arr[j] in stack
            stack.Push(arr[j]);
        }
    }
     
    // If none of the above case
    // satisfy then return "No"
    return "No";
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Input
    int []arr = { 4, 7, 11, 5, 13, 2 };
    int size = arr.Length;
 
    Console.WriteLine(recreationalSpot(arr, size));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript implementation of the above approach
 
// Function to check if it is possible
// to select three buildings that
// satisfy the given condition
function recreationalSpot(arr, N)
{
 
    if (N < 3) {
        return "No";
    }
 
    // Stores prefix min array
    var preMin = new Array(N);
    preMin[0] = arr[0];
    var i;
    // Iterate over the range [1, N-1]
    for (i = 1; i < N; i++) {
        preMin[i] = Math.min(preMin[i - 1], arr[i]);
    }
 
    // Stores the element from the
    // ending in increasing order
    var st = [];
    var j;
    // Iterate until j is greater than
    // or equal to 0
    for (j = N - 1; j >= 0; j--) {
 
        // If current array element is
        // greater than the prefix min
        // upto j
        if (arr[j] > preMin[j]) {
 
            // Iterate while st is not
            // empty and top element is
            // less than or equal to preMin[j]
 
            while (st.length>0
                   && st[st.length-1] <= preMin[j]) {
                // Remove the top element
                st.pop();
            }
 
            // If st is not empty and top
            // element of the st is less
            // than the current element
            if (st.length>0 && st[st.length-1] < arr[j]) {
                return "Yes";
            }
            // Push the arr[j] in st
            st.push(arr[j]);
        }
    }
    // If none of the above case
    // satisfy then return "No"
    return "No";
}
 
// Driver code
 
    // Input
    var arr = [4, 7, 11, 5, 13, 2];
    var size = arr.length;
 
    document.write(recreationalSpot(arr, size));
 
</script>

Output: 

Yes

 

Time Complexity: O(N)
Auxiliary Space: O(N)

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!