Next Largest Number to the Right

• Given an array of int nums
• Return an output array res where res[i] is the first number larger than nums[i] that can be found to the right in nums
• If no larger number exists to the right => -1

The point:

• Start on the right
• Check if the value is the next largest number for any value(s) to its left

Complexity :

Time	Space
O(n)	O(n)

• O(n) because each values from nums is push/pop once
• O(n) because the stack can store up to n values

About Rust :

• for i in (0..nums.len()).rev()
• *stack.last().unwrap()
• stack.is_empty()
• YES : tested on the Rust Playground

fn next_largest_number_to_the_right(nums: &[i32]) -> Vec<i32> {
    let mut res = vec![0; nums.len()];
    let mut stack = Vec::new();
    
    // start from the right
    for i in (0..nums.len()).rev() {
        // remove values smaller (equal) to current value
        while !stack.is_empty() && *stack.last().unwrap() <= nums[i] { // .last() returns Option<&T>
             stack.pop();
        }
        
        // record current value's next largest number or -1 if the stack is empty
        res[i] = if !stack.is_empty() {
            *stack.last().unwrap()
        } else {
            -1
        };
        
        stack.push(nums[i]);
    }
    
    res
}

fn main(){     // no main() if this code runs in a Jupyter cell 
    println!("{:?}", next_largest_number_to_the_right(&[5, 2, 4, 6, 1]));
    println!("{:?}", next_largest_number_to_the_right(&vec![5, 2, 4, 6, 1]));
    println!("{:?}", next_largest_number_to_the_right(&[]));
} // end of local scope OR end of main()