Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].
Return the answer in an array.
Example 1:
Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums[1]=1 does not exist any smaller number than it. For nums[2]=2 there exist one smaller number than it (1). For nums[3]=2 there exist one smaller number than it (1). For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
Example 2:
Input: nums = [6,5,4,8] Output: [2,1,0,3]
Example 3:
Input: nums = [7,7,7,7] Output: [0,0,0,0]
Constraints:
2 <= nums.length <= 5000 <= nums[i] <= 100struct Solution;
impl Solution {
fn smaller_numbers_than_current(nums: Vec<i32>) -> Vec<i32> {
let mut count = vec![0; 101];
for &x in &nums {
count[x as usize] += 1;
}
for i in 0..100 {
count[i + 1] += count[i]
}
let mut res = vec![];
for &x in &nums {
let v = if x == 0 { 0 } else { count[(x - 1) as usize] };
res.push(v)
}
res
}
}
#[test]
fn test() {
let nums = vec![8, 1, 2, 2, 3];
let res = vec![4, 0, 1, 1, 3];
assert_eq!(Solution::smaller_numbers_than_current(nums), res);
let nums = vec![6, 5, 4, 8];
let res = vec![2, 1, 0, 3];
assert_eq!(Solution::smaller_numbers_than_current(nums), res);
let nums = vec![7, 7, 7, 7];
let res = vec![0, 0, 0, 0];
assert_eq!(Solution::smaller_numbers_than_current(nums), res);
}