1-answer.md

✅ Counts the Frequencies to solve the 🧑🏻‍💻 Binary Search Problem on 🦀 Rust language

🧑🏻‍💻 Approach

The function search takes in a vector of integers (nums) and a target integer (target). A copy of the nums vector is created and stored in nums2 to avoid modifying the original vector. The nums2 vector is sorted in ascending order using the sort() function. Pointers low and high are initialized to keep track of the search range in the sorted vector. The binary search algorithm is performed using a while loop, which continues until low is greater than high. Inside the loop, the middle index mid is calculated as the average of low and high. The element at the middle index (nums2[mid]) is compared to the target. If they are equal, it means the target has been found, and the index mid is returned as the result. If the element at the middle index is less than the target, it means the target is in the right half of the remaining range. Therefore, the low pointer is moved to mid + 1. If the element at the middle index is greater than the target, it means the target is in the left half of the remaining range. Therefore, the high pointer is moved to mid - 1. If the target is not found after the loop finishes, -1 is returned as the result.

🔐 Code

    impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let mut nums2 = nums.clone();
    nums2.sort();

    let mut low = 0;
    let mut high = nums2.len() as i32 - 1;

    while low <= high{
        let mid= (low+(high - low)/2) as usize;
        if nums2[mid] == target{
            return mid as i32;
        }
        else if nums2[mid]< target {
            low = mid as i32 +1;
        }
        else {
            high = mid as i32 -1
        }
    }
    -1
    }
}