Check If N and Its Double Exist Leetcode solution in Java

Problem Description

Given an array arr of integers, check if there exists two integers N and M such that N is the double of M ( i.e N = 2* M).

Example 1 :-

Input: arr = [10,2,5,3]
Output: true
Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.

Example 2 :-

Input: arr = [7,1,14,11]
Output: true
Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.

Example 3 :-

Input: arr = [3,1,7,11]
Output: false
Explanation: In this case does not exist N and M, such that N = 2 * M.

See full problem description on Leetcode :-

• Check If N and Its Double Exist
  

Solution 1 :- Using For loop

class Solution {
    
    public boolean checkIfExist(int[] arr) {
        
        // Loop through array
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < arr.length; j++) {   

                // Check if i and j is not same and N * 2 = M         
                if (i != j && arr[i] * 2 == arr[j]) {
                    return true;
                }
            }
        }
        return false;    
    }
}

Explanation :-

• Both for loop goes until arr length.
• In if loop checking both i and j are not at same position and array's i element is double of array's j element. If condition becomes true then return true otherwise return false after both loop end.

Solution 2 :- Using ArrayList

public boolean checkIfExist(int[] arr) {    
    // Declare and Initialize arraylist
    ArrayList<Integer> list = new ArrayList<Integer>();
    
    // Adding all array elements into arraylist
    for(int i = 0; i < arr.length; i++) {
        list.add(arr[i]);
    }
    
    // Check if current element * 2 is exist or not. if exist then return true
    for(int i = 0; i < arr.length; i++) {
        if(list.contains(arr[i] * 2) && i != list.indexOf(arr[i] * 2)) {
            return true;
        }
    }
    return false;
}

Explanation :-

• Declare and initialize arraylist.
• After loop through array and add all array elements into arraylist one by one.
• Now again loop through array and check arraylist contains our element that matches our condition ( N and M such that N is the double of M).   
• list.indexOf(arr[i] * 2), this condition check first occurrence of given number. this condition is helpfull when there is only one 0 given in array. Becuse of this condition it does not count itself.

Hope you understand better.

Happy Coding.

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