problem

You are given a 0-indexed array nums that consists of n distinct positive integers. Apply m operations to this array, where in the ith operation you replace the number operations[i][0] with operations[i][1].

It is guaranteed that in the ith operation:

  • operations[i][0] exists in nums.
  • operations[i][1] does not exist in nums.

Return the array obtained after applying all the operations.

 

Example 1:

Input: nums = [1,2,4,6], operations = [[1,3],[4,7],[6,1]]
Output: [3,2,7,1]
Explanation: We perform the following operations on nums:
- Replace the number 1 with 3. nums becomes [3,2,4,6].
- Replace the number 4 with 7. nums becomes [3,2,7,6].
- Replace the number 6 with 1. nums becomes [3,2,7,1].
We return the final array [3,2,7,1].

Example 2:

Input: nums = [1,2], operations = [[1,3],[2,1],[3,2]]
Output: [2,1]
Explanation: We perform the following operations to nums:
- Replace the number 1 with 3. nums becomes [3,2].
- Replace the number 2 with 1. nums becomes [3,1].
- Replace the number 3 with 2. nums becomes [2,1].
We return the array [2,1].

 

Constraints:

  • n == nums.length
  • m == operations.length
  • 1 <= n, m <= 105
  • All the values of nums are distinct.
  • operations[i].length == 2
  • 1 <= nums[i], operations[i][0], operations[i][1] <= 106
  • operations[i][0] will exist in nums when applying the ith operation.
  • operations[i][1] will not exist in nums when applying the ith operation.

submission

// use a hashmap to keep track: value -> index
impl Solution {
    pub fn array_change(
        mut nums: Vec<i32>,
        operations: Vec<Vec<i32>>,
    ) -> Vec<i32> {
        let mut indices: std::collections::HashMap<i32, usize> = nums
            .iter()
            .enumerate()
            .map(|(idx, &val)| (val, idx))
            .collect();
        for op in operations {
            let (old, new) = (op[0], op[1]);
            // test data guaranteed old exist
            let idx = indices[&old];
            // update nums and indices
            // we don't actually need to remove index
            nums[idx] = new;
            indices.insert(new, idx);
        }
        nums
    }
}