1389. Create Target Array in the Given Order
Problem Description
The problem gives us two arrays: nums
and index
. The goal is to construct a new array called target
. To create this target array, we need to follow specific rules:
- Start with an empty
target
array. - Read elements from
nums
andindex
arrays from left to right. For each pair(nums[i], index[i])
, insert the valuenums[i]
into thetarget
array at the position specified byindex[i]
. - Keep inserting elements into the
target
array until there are no more elements to read fromnums
andindex
.
The challenge requires us to return this target
array after all insertions are complete. It is guaranteed in the problem statement that the insertion operations, as described by the index
array, will be valid—which means they won't lead to any out-of-bounds errors.
Intuition
The intuition behind the solution is straightforward, as the problem specifies the exact steps needed to arrive at the target
array. With each pair of elements from nums
and index
, we directly follow the rule and use the insert operation.
Here's the intuitive approach step-by-step:
- Start with an empty list for the
target
. - Loop over
nums
andindex
using thezip()
function in Python, which conveniently gives us pairs of(nums[i], index[i])
. - For each pair, use the
insert()
method on thetarget
list, which allows us to place elements not just at the end of the list (likeappend
) but at any position we specify. - The iteration continues until every element from
nums
has been placed into thetarget
at the correct positions. - Finally, return the
target
array, which now contains all the elements fromnums
sorted by the rules of theindex
array.
This solution is elegant due to Python's list handling capabilities and delivers the expected target
array using a direct translation of the problem's rules into code.
Solution Approach
The implementation of the solution is straightforward and follows the problem description closely. It utilizes the built-in list data structure in Python and the insert()
method it provides. The solution does not rely on any sophisticated algorithms or complex patterns; rather, it uses a basic iterative approach that corresponds with the rules defined in the problem.
Here's the detailed implementation description:
- Start by initializing an empty list
target
that will eventually hold the final array. - Use the built-in Python function
zip()
to iterate over both thenums
andindex
arrays simultaneously.zip(nums, index)
creates an iterator of tuples where the first item in each passed iterator is paired together, then the second item, and so on. In this case, it pairs each element innums
with its corresponding element inindex
. - For each pair
(x, i)
obtained from zippingnums
andindex
, perform an insert operation:target.insert(i, x)
. This line is the core of the solution, wherei
is the position in the target array where we want to insert the element, andx
is the actual element fromnums
we want to insert. - The
insert()
method takes two arguments: the first argument is the index at which to insert the item, and the second argument is the item to insert. It modifies the list in place, which means no new list is created, the existingtarget
list is updated. - This process is repeated until there are no more elements to read, meaning every element from
nums
has been placed into thetarget
list in the order specified byindex
. - Return the
target
list.
By utilizing the insert()
method, we can insert elements at specific positions, which provides a seamless way to construct the target
array. The solution follows the insertion rules exactly as specified and the simplicity of the approach reflects the clarity of the problem’s instructions. It is worth noting that while insert()
operation is efficient for small to medium-sized lists, inserting elements into a list has a time complexity of O(n) per operation because it may require shifting over other elements. However, for the constraints of this problem, the approach is suitable and efficient.
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Start EvaluatorExample Walkthrough
Let's consider the following small example to illustrate the solution approach:
Suppose nums = [0, 1, 2, 3, 4]
and index = [0, 1, 2, 2, 1]
.
Following the algorithm:
-
Initialize the
target
list as empty:target = []
. -
Now begin iterating over the
nums
andindex
arrays simultaneously.- First pair:
(0, 0)
- insert0
at index0
oftarget
→target = [0]
. - Second pair:
(1, 1)
- insert1
at index1
oftarget
→target = [0, 1]
. - Third pair:
(2, 2)
- insert2
at index2
oftarget
→target = [0, 1, 2]
. - Fourth pair:
(3, 2)
- insert3
at index2
oftarget
. This will push the current element at index2
(which is2
) to the right →target = [0, 1, 3, 2]
. - Fifth pair:
(4, 1)
- insert4
at index1
oftarget
. This will push elements starting from index1
to the right →target = [0, 4, 1, 3, 2]
.
- First pair:
-
At this point, we have read all elements from
nums
andindex
, and thetarget
list is fully constructed. -
The final step is to return the
target
list which is[0, 4, 1, 3, 2]
.
This example clearly demonstrates how elements from the nums
array are inserted into the target
array at the positions dictated by the corresponding index
values. Each insert operation respects the current state of the target
array, potentially shifting elements to make room for the new ones. The final target
array reflects the ordered insertions as per the given nums
and index
arrays.
Solution Implementation
1from typing import List # Importing List from typing module for type hints
2
3class Solution:
4 def createTargetArray(self, nums: List[int], indices: List[int]) -> List[int]:
5 # Initialize an empty target array
6 target = []
7
8 # Loop over the pairs of elements from nums and their corresponding indices
9 for num, idx in zip(nums, indices):
10 # Insert the element 'num' at the index 'idx' of the target array
11 target.insert(idx, num)
12
13 # Return the final target array
14 return target
15
1class Solution {
2 public int[] createTargetArray(int[] nums, int[] index) {
3 // Get the length of the input array.
4 int n = nums.length;
5 // Initialize an ArrayList to hold the target elements.
6 List<Integer> targetList = new ArrayList<>();
7 // Iterate through each element of nums and index arrays.
8 for (int i = 0; i < n; ++i) {
9 // Add the current element from nums into the targetList at the position given by index[i].
10 targetList.add(index[i], nums[i]);
11 }
12
13 // Initialize the target array.
14 int[] targetArray = new int[n];
15 // Convert the ArrayList back into an array.
16 for (int i = 0; i < n; ++i) {
17 targetArray[i] = targetList.get(i);
18 }
19 // Return the resultant target array.
20 return targetArray;
21 }
22}
23
1#include <vector> // Include vector header for std::vector
2using namespace std;
3
4class Solution {
5public:
6 /**
7 * Create a target array by inserting elements from the 'nums' array into the
8 * 'target' array at positions specified by the 'index' array.
9 *
10 * @param nums A vector of integers to insert into the target array.
11 * @param index A vector of integers indicating the indices at which
12 * to insert the elements from the nums vector.
13 * @return The target vector after all insertions.
14 */
15 vector<int> createTargetArray(vector<int>& nums, vector<int>& index) {
16 vector<int> target; // The target vector that we'll return
17
18 // Iterate over all elements in 'nums' and 'index'
19 for (int i = 0; i < nums.size(); ++i) {
20 // At the position index[i], insert the value nums[i] into the 'target' vector
21 target.insert(target.begin() + index[i], nums[i]);
22 }
23
24 return target; // Return the completed target vector
25 }
26};
27
1// Function to create a target array according to the specified order
2function createTargetArray(nums: number[], indices: number[]): number[] {
3 // Initialize the target array to hold the result
4 const targetArray: number[] = [];
5
6 // Iterate over the nums array to populate the target array
7 for (let i = 0; i < nums.length; i++) {
8 // Insert the current number at the specified index in the target array
9 // The splice method modifies the array in place and can insert elements
10 targetArray.splice(indices[i], 0, nums[i]);
11 }
12
13 // Return the resulting target array after the loop completes
14 return targetArray;
15}
16
Time and Space Complexity
Time Complexity
The given code has a time complexity of O(n^2)
. This is because for each element x
in nums
, the method insert()
is called which can take O(n)
time in the worst case, as it requires shifting all the elements after the inserted element by one position to make space for the new element. Since there are n
insert operations, and each might take up to O(n)
time, the overall time complexity is O(n * n)
, which simplifies to O(n^2)
.
Space Complexity
The space complexity of the code is O(n)
. The target
list that is created, in the worst case, will contain all the elements from the nums
list, thus the amount of space used grows linearly with the input size n
. No additional space other than the target
list (which is the desired output) is used, hence the space complexity is O(n)
.
Learn more about how to find time and space complexity quickly using problem constraints.
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