Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].
Return any permutation of A that maximizes its advantage with respect to B.
Example 1:
Input: A = [2,7,11,15], B = [1,10,4,11] Output: [2,11,7,15]
Example 2:
Input: A = [12,24,8,32], B = [13,25,32,11] Output: [24,32,8,12]
Note:
1 <= A.length = B.length <= 100000 <= A[i] <= 10^90 <= B[i] <= 10^9struct Solution;
type Pair = (usize, i32);
impl Solution {
fn advantage_count(mut a: Vec<i32>, b: Vec<i32>) -> Vec<i32> {
let n = a.len();
let mut b: Vec<Pair> = b.into_iter().enumerate().collect();
let mut l = 0;
let mut r = n - 1;
a.sort_unstable();
b.sort_unstable_by_key(|p| p.1);
let mut res: Vec<i32> = vec![0; n];
for i in 0..n {
if a[i] <= b[l].1 {
res[b[r].0] = a[i];
r -= 1;
} else {
res[b[l].0] = a[i];
l += 1;
}
}
res
}
}
#[test]
fn test() {
let a = vec![2, 7, 11, 15];
let b = vec![1, 10, 4, 11];
let res = vec![2, 11, 7, 15];
assert_eq!(Solution::advantage_count(a, b), res);
let a = vec![12, 24, 8, 32];
let b = vec![13, 25, 32, 11];
let res = vec![24, 32, 8, 12];
assert_eq!(Solution::advantage_count(a, b), res);
}