LeetCode 1966: Binary Searchable Numbers in an Unsorted Array Solution

Master LeetCode problem 1966 (Binary Searchable Numbers in an Unsorted Array), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1966. Binary Searchable Numbers in an Unsorted Array

Medium

Array Binary Search

Problem Explanation

Explanation:

To solve this problem, we need to find the count of binary searchable numbers in an unsorted array. A number is considered binary searchable if it is greater than all the elements to its left and less than all the elements to its right.

We can iterate through each element in the array and for each element, we can check if it is greater than all elements to its left and less than all elements to its right. If it satisfies this condition, we increment a counter to keep track of the number of binary searchable numbers.

Algorithm:

Initialize a counter to keep track of the number of binary searchable numbers.
Iterate through each element in the array.
For each element, check if it is greater than all elements to its left and less than all elements to its right.
If the element satisfies the condition, increment the counter.
Return the counter as the result.

Time Complexity:

The time complexity of this approach is O(n^2), where n is the number of elements in the array.

Space Complexity:

The space complexity of this approach is O(1) as we are not using any extra space apart from some variables.:

Solution Code

class Solution {
    public int numBinarySearchableNumbers(int[] nums) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            boolean isBinarySearchable = true;
            for (int j = 0; j < i; j++) {
                if (nums[j] >= nums[i]) {
                    isBinarySearchable = false;
                    break;
                }
            }
            if (!isBinarySearchable) continue;
            for (int k = i + 1; k < nums.length; k++) {
                if (nums[k] <= nums[i]) {
                    isBinarySearchable = false;
                    break;
                }
            }
            if (isBinarySearchable) {
                count++;
            }
        }
        return count;
    }
}

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