Check if Binary Heap is completely filled

Given an integer N which denotes the number of elements in a binary heap. Check if the binary Heap is completely filled or not.

Examples:

Input: 7
Output: YES
Explanation: At height = 0, no. of elements = 1
At height = 1, no. of elements = 2
At height = 2, no. of elements = 4 
Last level is completely filled because at level 2 we need 2² = 4 
elements to fill completely and 4 elements are here.

Input: 16
Output: NO
Explanation: At height=0, no. of elements=1
At height = 1, no. of elements = 2
At height = 2, no. of elements = 4
At height = 3, no. of elements = 8
At height = 4, no. of elements = 1 
Last level is not completely filled because at level 4 
we need 2⁴ = 16 elements to fill completely but here is only 1 element

Input: 31
Output: YES

Approach: The idea here is 

In a binary heap, a parent has 2 children. For each level h, number of elements are 2^h

Follow the steps below to solve the given problem:

• First find the number of nodes till height h = 2⁰ + 2¹ + 2² + . . . + 2^h.
• Above written series is a geometric progression, so using sum of GP = ( a( r^k-1 )/(r-1) ), here a = 2⁰ = 1, r = 2, k = h.
• Therefore sum = 1 * (2^h – 1) / (2 – 1) = 2^h – 1.
• Height of a binary tree = log₂N+ 1.
• Now check if the sum is equal to N or not.

Below is the implementation of the above approach: 

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to check whether
// it is completely filled or not
string isCompletelyFilled(int n)
{
    // Height of the heap
    int height = log2(n) + 1;

    // Sum of number of elements
    int sum = pow(2, height) - 1;

    // If sum equals to n print yes
    if (sum == n)
        return "Yes";
    return "No";
}

// Driver code
int main()
{
    int N = 7;

    // Function call
    cout << isCompletelyFilled(N);
    return 0;
}

Java

// JAVA program of above approach.
import java.util.*;
class GFG
{

// Function to check whether
// it is completely filled or not
static String isCompletelyFilled(int n)
{
    // Height of the heap
    int height =(int)(Math.log(n) / Math.log(2)) + 1;

    // Sum of number of elements
    int sum = (int)Math.pow(2, height) - 1;

    // If sum equals to n print yes
    if (sum == n)
        return "Yes";
    return "No";
}

// Driver code
public static void main(String[] args)
{
    int N = 7;

    // Function call
    System.out.print(isCompletelyFilled(N));
}
}

// This code is contributed by sanjoy_62.

Python3

# Python code to implement the approach
import math

# Function to check whether
# it is completely filled or not
def isCompletelyFilled(n) :
    
    # Height of the heap
    height = int(math.log2(n)) + 1

    # Sum of number of elements
    sum = pow(2, height) - 1

    # If sum equals to n print yes
    if (sum == n) :
        return "Yes"
    return "No"


# Driver code

N = 7

# Function call
print(isCompletelyFilled(N))

# This code is contributed by splevel62.

C#

// C# program of above approach.

using System;

public class GFG {

    // Function to check whether
    // it is completely filled or not
    static String isCompletelyFilled(int n)
    {
        // Height of the heap
        int height = (int)(Math.Log(n) / Math.Log(2)) + 1;

        // Sum of number of elements
        int sum = (int)Math.Pow(2, height) - 1;

        // If sum equals to n print yes
        if (sum == n)
            return "Yes";
        return "No";
    }

    static public void Main()
    {

        // Code
        int N = 7;

        // Function call
        Console.Write(isCompletelyFilled(N));
    }
}

// This code is contributed by lokeshmvs21.

Javascript

<script>
    // JavaScript code for the above approach

// Function to check whether
// it is completely filled or not
function isCompletelyFilled(n)
{
    // Height of the heap
    let height = Math.floor(Math.log(n) / Math.log(2)) + 1;
 
    // Sum of number of elements
    let sum = Math.pow(2, height) - 1;
 
    // If sum equals to n print yes
    if (sum == n)
        return "Yes";
    return "No";
}

    // Driver code
    let N = 7;
 
    // Function call
    document.write(isCompletelyFilled(N));
    
    // This code is contributed by code_hunt.
</script>

Output

Yes

Time Complexity: O(logN), time taken by pow is logN
Auxiliary Space: O(1)

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