20250415 finished.
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@ -599,3 +599,5 @@ mod p3272_find_the_count_of_good_integers;
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mod p1922_count_good_numbers;
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mod p1534_count_good_triplets;
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mod p2179_count_good_triplets_in_an_array;
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99
src/problem/p2179_count_good_triplets_in_an_array.rs
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99
src/problem/p2179_count_good_triplets_in_an_array.rs
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@ -0,0 +1,99 @@
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/**
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* [2179] Count Good Triplets in an Array
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*/
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pub struct Solution {}
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// submission codes start here
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struct FenwickTree {
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tree: Vec<i64>,
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}
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impl FenwickTree {
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fn new(size: usize) -> Self {
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Self {
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tree: vec![0; size + 1],
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}
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}
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fn update(&mut self, mut index: usize, delta: i64) {
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index += 1;
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while index < self.tree.len() {
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self.tree[index] += delta;
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index += Self::low_bit(index)
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}
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}
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fn query(&self, mut index: usize) -> i64 {
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index += 1;
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let mut result = 0;
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while index > 0 {
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result += self.tree[index];
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index -= Self::low_bit(index)
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}
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result
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}
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// 二进制最低位1以及后面的0组成的数
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fn low_bit(i: usize) -> usize {
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i & (!i + 1)
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}
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}
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impl Solution {
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pub fn good_triplets(nums1: Vec<i32>, nums2: Vec<i32>) -> i64 {
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let n = nums1.len();
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// nums2中各个值到下标的映射数组
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let mut pos2 = vec![0; n];
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// index_mapping 是nums1中一个数在nums2中的下标
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// reverse_index_mapping是该下标同nums1中数字下标的对应关系
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let mut reversed_index_mapping = vec![0; n];
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for (i, v) in nums2.iter().enumerate() {
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pos2[*v as usize] = i;
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}
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for i in 0..n {
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// nums1中的一个值在nums2中的下标到该值在nums1中的下标的映射
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reversed_index_mapping[pos2[nums1[i] as usize]] = i;
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}
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let mut tree = FenwickTree::new(n);
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let mut result = 0;
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// 枚举num1s中的一个值在nums2中的下标
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for j in 0..n {
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let pos = reversed_index_mapping[j];
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let left = tree.query(pos);
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tree.update(pos, 1);
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let right = ((n - 1 - pos) - (j - left as usize)) as i64;
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result += left * right;
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}
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result
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}
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}
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// submission codes end
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn test_2179() {
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assert_eq!(
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1,
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Solution::good_triplets(vec![2, 0, 1, 3], vec![0, 1, 2, 3])
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);
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assert_eq!(
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4,
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Solution::good_triplets(vec![4, 0, 1, 3, 2], vec![4, 1, 0, 2, 3])
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);
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}
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}
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