#1966 - Binary Searchable Numbers in an Unsorted Array
Problem Description
Consider a function that implements an algorithm similar to Binary Search. The function has two input parameters: sequence is a sequence of integers, and target is an integer value.
The purpose of the function is to find if the target exists in the sequence.
The pseudocode of the function is as follows: func(sequence, target) while sequence is not empty randomly choose an element from sequence as the pivot if pivot = target, return true else if pivot < target, remove pivot and all elements to its left from the sequence else, remove pivot and all elements to its right from the sequence end while return false When the sequence is sorted, the function works correctly for all values. When the sequence is not sorted, the function does not work for all values, but may still work for some values.
Given an integer array nums, representing the sequence, that contains unique numbers and may or may not be sorted, return the number of values that are guaranteed to be found using the function, for every possible pivot selection.
Solution
/**
* @param {number[]} nums
* @return {number}
*/
var binarySearchableNumbers = function(nums) {
const n = nums.length;
const leftMax = new Array(n);
const rightMin = new Array(n);
leftMax[0] = nums[0];
for (let i = 1; i < n; i++) {
leftMax[i] = Math.max(leftMax[i - 1], nums[i]);
}
rightMin[n - 1] = nums[n - 1];
for (let i = n - 2; i >= 0; i--) {
rightMin[i] = Math.min(rightMin[i + 1], nums[i]);
}
let count = 0;
for (let i = 0; i < n; i++) {
const leftMaxVal = i > 0 ? leftMax[i - 1] : -Infinity;
const rightMinVal = i < n - 1 ? rightMin[i + 1] : Infinity;
if (leftMaxVal < nums[i] && nums[i] < rightMinVal) {
count++;
}
}
return count;
};