1346. Check If N and Its Double Exist
Difficulty: Easy
Topics: Array, Hash Table, Two Pointers, Binary Search, Sorting
Given an array arr of integers, check if there exist two indices i and j such that :
- i != j
- 0
- arr[i] == 2 * arr[j]
Example 1:
- Input: arr = [10,2,5,3]
- Output: true
- Explanation: For i = 0 and j = 2, arr[i] == 10 == 2 * 5 == 2 * arr[j]
Example 2:
- Input: arr = [3,1,7,11]
- Output: false
- Explanation: There is no i and j that satisfy the conditions.
Constraints:
- 2
- -103 3
Hint:
- Loop from i = 0 to arr.length, maintaining in a hashTable the array elements from [0, i - 1].
- On each step of the loop check if we have seen the element 2 * arr[i] so far.
- Also check if we have seen arr[i] / 2 in case arr[i] % 2 == 0.
Solution:
We can use a hash table (associative array) to track the elements we have already encountered while iterating through the array. The idea is to check for each element arr[i] if its double (i.e., 2 * arr[i]) or half (i.e., arr[i] / 2 if it's an even number) has already been encountered.
Here’s a step-by-step solution:
Plan:
- Iterate through the array.
- For each element arr[i], check if we have seen 2 * arr[i] or arr[i] / 2 (if arr[i] is even) in the hash table.
- If any condition is satisfied, return true.
- Otherwise, add arr[i] to the hash table and continue to the next element.
- If no match is found by the end, return false.
Let's implement this solution in PHP: 1346. Check If N and Its Double Exist
<?php /** * @param Integer[] $arr * @return Boolean */ function checkIfExist($arr) { ... ... ... /** * go to ./solution.php */ } // Example usage $arr1 = [10, 2, 5, 3]; $arr2 = [3, 1, 7, 11]; echo checkIfExist($arr1) ? 'true' : 'false'; // Output: true echo "\n"; echo checkIfExist($arr2) ? 'true' : 'false'; // Output: false ?>
Explanation:
- Hash Table: We use the $hashTable associative array to store the elements we've encountered so far.
- First Condition: For each element arr[i], we check if arr[i] * 2 exists in the hash table.
- Second Condition: If the element is even, we check if arr[i] / 2 exists in the hash table.
- Adding to Hash Table: After checking, we add arr[i] to the hash table for future reference.
- Return: If we find a match, we immediately return true. If no match is found after the loop, we return false.
Time Complexity:
- The time complexity is O(n), where n is the length of the array. This is because each element is processed once and checking or adding elements in the hash table takes constant time on average.
Space Complexity:
- The space complexity is O(n) due to the storage required for the hash table.
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