S

```
/*
Moore's voting algorithm finds out the strictly majority-occurring element
without using extra space
and O(n) + O(n) time complexity
It is built on the intuition that a strictly major element will always have a net occurrence as 1.
Say, array given: 9 1 8 1 1
Here, the algorithm will work as:
(for finding element present >(n/2) times)
(assumed: all elements are >0)
Initialisation: ele=0, cnt=0
Loop beings.
loop 1: arr[0]=9
ele = 9
cnt=1 (since cnt = 0, cnt increments to 1 and ele = 9)
loop 2: arr[1]=1
ele = 9
cnt= 0 (since in this turn of the loop, the array[i] != ele, cnt decrements by 1)
loop 3: arr[2]=8
ele = 8
cnt=1 (since cnt = 0, cnt increments to 1 and ele = 8)
loop 4: arr[3]=1
ele = 8
cnt= 0 (since in this turn of the loop, the array[i] != ele, cnt decrements by 1)
loop 5: arr[4]=1
ele = 9
cnt=1 (since cnt = 0, cnt increments to 1 and ele = 1)
Now, this ele should be the majority element if there's any
To check, a quick O(n) loop is run to check if the count of ele is >(n/2), n being the length of the array
-1 is returned when no such element is found.
*/
pub fn moore_voting(arr: &[i32]) -> i32 {
let n = arr.len();
let mut cnt = 0; // initializing cnt
let mut ele = 0; // initializing ele
arr.iter().for_each(|&item| {
if cnt == 0 {
cnt = 1;
ele = item;
} else if item == ele {
cnt += 1;
} else {
cnt -= 1;
}
});
let cnt_check = arr.iter().filter(|&&x| x == ele).count();
if cnt_check > (n / 2) {
ele
} else {
-1
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_moore_voting() {
let arr1: Vec<i32> = vec![9, 1, 8, 1, 1];
assert!(moore_voting(&arr1) == 1);
let arr2: Vec<i32> = vec![1, 2, 3, 4];
assert!(moore_voting(&arr2) == -1);
}
}
```