Last Occurrence [Linear Search] easy

You have been given an array of size N consisting of integers. In addition you have been given an element M you need to find and print the index of the last occurrence of this element M in the array if it exists in it, otherwise print -1. Consider this array to be 1 indexed.

Input Format:

The first line consists of 2 integers N and M denoting the size of the array and the element to be searched for in the array respectively . The next line contains N space separated integers denoting the elements of of the array.

Output Format

Print a single integer denoting the index of the last occurrence of integer M in the array if it exists, otherwise print -1.

Constraints

SAMPLE INPUT

5 1
1 2 3 4 1

SAMPLE OUTPUT

5

Solution

In main.rs we can add main function that process in and out streams.

use linear_sort_reverse_search_rust_easy::reverse_search;
use std::io;

fn main() -> io::Result<()> {
    reverse_search(&mut io::stdin(), &mut io::stdout())
}

in lib.rs there is rest of our code

use std::io::{Error, Read, Write};

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn basic_test() {
        let mut output: Vec<u8> = Vec::new();

        reverse_search(&mut "5 1
1 2 3 4 1".as_bytes(), &mut output).unwrap();
        assert_eq!(&output, b"5\n");
    }


    #[test]
    fn not_found_test() {
        let mut output: Vec<u8> = Vec::new();

        reverse_search(&mut "5 7
1 2 3 4 1".as_bytes(), &mut output).unwrap();
        assert_eq!(&output, b"-1\n");
    }
}

pub fn reverse_search(
    handle: &mut impl Read ,
    output: &mut impl Write,
) -> Result<(), Error> {
    let mut buffer = "".to_string();
    let mut out = "".to_string();
    handle.read_to_string(&mut buffer)?;

    let mut lines = buffer.lines();

    let mut some_line = match lines.next() {
        Some(line) => line,
        _ => ""
    };

    let mut iterator = some_line.split_ascii_whitespace();

    let mut len: usize = match iterator.next() {
        Some(p) => p.trim().parse().expect("can't read"),
        None => 0
    };
    let  needle = match iterator.next() {
        Some(p) => p.trim().parse().expect("can't read"),
        None => 0
    };

    some_line = match lines.next() {
        Some(line) => line,
        _ => ""
    };

    let mut vec:Vec<i32> = vec![0; len];

    iterator = some_line.split_ascii_whitespace();

    for n in 0..len {
        vec[n] = match iterator.next() {
            Some(p) => p.trim().parse().expect("can't read"),
            None => 0
        };
    }

    let mut iter = vec.iter().rev();

    while let Some(num) = iter.next() {
        if *num == needle {
            out = format!("{}\n", len);
            break;
        }
        len -= 1;
    }

    if len == 0 {
        out = format!("-1\n");
    }

    output.write_all(out.to_uppercase().as_bytes())?;

    Ok(())
}

linear-sort-reverse-search-rust-easy