1441. Build an Array With Stack Operations
Problem Description
The task involves simulating a stack operation to construct a specific sequence of numbers, given as the target
array, using a stream of numbers from 1 to n
. We can only use two operations: "Push", to add the next number in the stream on top of the stack, and "Pop", to remove the top element of the stack.
The goal is to ensure that the numbers in the stack, from bottom to top, match the sequence described in the target
array. We need to do this by adhering to specific rules:
- We can only take the next integer from the stream if the stream isn't empty and push it onto the stack.
- We can pop the top integer from the stack if the stack isn't empty.
- We must stop taking new integers from the stream and stop performing operations on the stack once it matches the
target
.
The expected output is a list of the stack operations ("Push" and "Pop") needed to attain the target
sequence.
Intuition
To solve this problem, we need to simulate the process of either pushing or popping elements to match the target
sequence. We iterate over the target
array and compare its elements with the next expected integer from the stream, which we represent with a variable (let's say, cur
). For every number in the target
array, we do the following:
- If
cur
is less than the currenttarget
value, it implies that there are numbers in the stream that we need to skip to match thetarget
. For each of these numbers, we "Push" (to simulate taking the number from the stream) and then "Pop" (to remove it). - We increment
cur
until it matches the currenttarget
value, then perform one "Push" operation without a corresponding "Pop" since this number should stay in the stack.
We continue this process for each element in the target
array. The sequence of "Push" and "Pop" operations that we accumulate forms the solution to rebuild the target
array from the stream. It is worth mentioning that multiple valid sequences of operations can exist, and we can return any of them.
This approach works because it effectively simulates the steps needed to recreate the target
array using the given stack operations, mirroring the actual manual process one might use if doing this with a physical stack and stream of numbers.
Learn more about Stack patterns.
Solution Approach
The solution takes a simple iterative approach to build the required sequence of stack operations. Here is a step-by-step explanation of the implementation details:
-
Initialize two variables:
cur
, which represents the next expected number from the stream starting from 1, andans
, which holds the list of stack operations performed. -
Loop through each value
v
in thetarget
array:-
Inside this loop, increment
cur
by 1 - it means we are considering the push of the next number in the stream. -
While
cur
is less than the currenttarget
valuev
, we realize that we are not interested in havingcur
in our final stack. Therefore, we perform a "Push" followed immediately by a "Pop". Add these two operations to theans
list, and then incrementcur
to consider the next number in the stream. -
By the time
cur
matchesv
, the current value of thetarget
array, we only need to "Push" as we want to keep this number in our stack. Hence, we add a "Push" operation toans
.
-
-
After iterating through each element in the
target
array, we have a list of operations that leads us to a stack that, if evaluated from bottom to top, will matchtarget
. -
Return the
ans
list, which contains the sequence of "Push" and "Pop" operations.
This algorithm effectively uses a simulation pattern, where we simulate each step needed to construct the target sequence from the available input numbers (1 to n
). The while
loop inside the iteration is essential as it allows us to skip over the numbers that are not included in the target
sequence by simulating a push/pop pair for each.
Here is how the pattern works with the code:
class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
cur, ans = 0, []
for v in target:
cur += 1
while cur < v:
ans.extend(['Push', 'Pop'])
cur += 1
ans.append('Push')
return ans
In this code:
- The
for
loop iterates over thetarget
. - The
while
loop handles skipping numbers not intarget
. - The
extend(['Push', 'Pop'])
method is used to simulate the push/pop operations for numbers we wish to skip. - When the condition of the
while
loop is not met (meaningcur
equalsv
), it appends aPush
toans
.
By the end of the iteration, ans
represents the sequence of stack operations that would build target
from the stream of integers within the range [1, n]
.
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Start EvaluatorExample Walkthrough
Consider the target array [2, 4]
and n = 4
. We want to simulate stack operations to reconstruct this sequence where the numbers in the stream range from 1 to n
.
Step 1: We initialize cur
to 0 and ans
to an empty list. These will keep track of the next expected number from the stream and the list of operations, respectively.
Step 2: We iterate through each value v
in the target array. The first value is 2
.
- We increment
cur
to 1 and start a while loop that will run sincecur < v
. - While
cur < 2
, we do a "Push" (as if we took1
from the stream and added it to the stack) followed by a "Pop" (we remove1
since it's not in the target). We add these operations toans
, resulting in['Push', 'Pop']
. - We increment
cur
again, nowcur
is 2 which matchesv
. - We perform a "Push" as
cur
is equal tov
to keep2
on the stack, updatingans
to['Push', 'Pop', 'Push']
.
Step 3: We move on to the next value in the target array, which is 4
.
- We increment
cur
to 3 since it is still less thanv
and perform another "Push" followed by "Pop", updatingans
to['Push', 'Pop', 'Push', 'Push', 'Pop']
. - Incrementing
cur
once more, nowcur
is 4 which matchesv
. - We perform a "Push" to keep
4
on the stack, updatingans
to['Push', 'Pop', 'Push', 'Push', 'Pop', 'Push']
.
Step 4: After processing each element in the target array, we have a complete list of operations that forms the output. Thus, the array ['Push', 'Pop', 'Push', 'Push', 'Pop', 'Push']
is the sequence of operations needed to simulate the stack to achieve the target sequence [2, 4]
.
Following the above steps ensures that we only push numbers onto the stack that are present in the target
and pop those that are not, using the simulation approach described in the solution.
This example clearly demonstrates the iterative process of the algorithm, incrementing the cur
variable, and conditionally adding 'Push' and 'Pop' operations to the ans
list to match the target
array.
Solution Implementation
1class Solution:
2 def buildArray(self, target: List[int], n: int) -> List[str]:
3 current_value = 1 # Start with the first value in the stack sequence
4 operations = [] # Initialize list to keep track of the operations
5
6 # Iterate over each value that needs to be in the target stack
7 for target_value in target:
8 # Keep pushing and popping until the current value matches the target value
9 while current_value < target_value:
10 operations.append('Push') # Add the value to the stack
11 operations.append('Pop') # Immediately remove it since it's not in target
12 current_value += 1 # Move to the next value in the sequence
13
14 # Once the current value matches the target value, add it to the stack
15 operations.append('Push')
16 # Move to the next value after successfully pushing the current target value
17 current_value += 1
18
19 # Return the list of operations that builds the target stack
20 return operations
21
1class Solution {
2 public List<String> buildArray(int[] target, int n) {
3 // Initialize the current number we are at, starting from 0 before the sequence begins.
4 int current = 0;
5
6 // Initialize the answer list to store the output sequence.
7 List<String> operations = new ArrayList<>();
8
9 // Iterate over the elements of the target array.
10 for (int targetValue : target) {
11 // Read the next number and compare it to the target value.
12 // If the current number is less than the target value, we "Push" and then "Pop".
13 while (++current < targetValue) {
14 operations.add("Push"); // Append "Push" to signify we pushed the current number.
15 operations.add("Pop"); // Append "Pop" to signify we removed the number we just added.
16 }
17 // After reaching the target value, perform a "Push" to add the target number to stack.
18 operations.add("Push");
19 }
20
21 // Return the list of operations needed to get the target sequence.
22 return operations;
23 }
24}
25
1#include <vector>
2#include <string>
3
4// The Solution class contains a method buildArray to generate a sequence of "Push" and "Pop"
5// operations to build a specific target array from another array that consists of the first 'n' integers.
6class Solution {
7public:
8 // This method takes in a target vector and an integer n and returns a vector of strings representing the required operations.
9 vector<string> buildArray(vector<int>& target, int n) {
10 // Initialize the current number to track the numbers we've reached.
11 int currentNumber = 0;
12
13 // Initialize an answer vector to hold the sequence of operations.
14 vector<string> operations;
15
16 // Iterate over each number in the target array.
17 for (int targetValue : target) {
18
19 // While the next number to push is less than our target value, emulate a push and a pop (i.e., skip the number).
20 while (++currentNumber < targetValue) {
21 operations.emplace_back("Push"); // Emulate adding currentNumber to the stack
22 operations.emplace_back("Pop"); // Emulate removing it since it's not our target
23 }
24
25 // Once we've reached the targetValue, add it to the stack with one "Push".
26 operations.emplace_back("Push"); // Add targetValue to the stack.
27 }
28
29 // Return the list of operations required to build the target array.
30 return operations;
31 }
32};
33
1// Function to simulate stack operations to build a target array
2function buildArray(target: number[], n: number): string[] {
3 // Initialize the result array to hold the sequence of operations
4 const operations = [];
5
6 // Track the current number to compare with target array
7 let currentNumber = 0;
8
9 // Iterate over the target array to determine operations
10 for (const targetNumber of target) {
11 // Increment current number and append 'Push' and 'Pop' until it matches the target number
12 while (++currentNumber < targetNumber) {
13 operations.push('Push'); // Simulate pushing the next number onto the stack
14 operations.push('Pop'); // Immediately remove it as it's not in the target
15 }
16 // When current number matches the target number, just perform the 'Push' operation
17 operations.push('Push');
18 }
19
20 // Return the array of operations
21 return operations;
22}
23
Time and Space Complexity
The time complexity of the given code can be analyzed based on the number of operations performed relative to the length of target
list and n
. Assume the length of target
is m
.
For every element in target
, the code performs a series of "Push" and "Pop" operations until it reaches the current target value. In the worst case, this could happen for each target value (i.e., if the target
list has all consecutive numbers starting from 1 up to m
). Since at each step of iterating through the target list you can have at most two operations (one "Push", and if the number is not in the target one "Pop") per number until you reach the target value, the number of these operations is linear with respect to the last value in target
, denoted as target[-1]
.
Therefore, the worst-case time complexity is O(target[-1])
, which in the worst case is O(n)
if the last value in target
is equal to n
.
The space complexity of the code is determined by the size of the ans
list, which holds a sequence of strings representing the operations. In the worst-case scenario as described above, the ans
list will contain a number of elements equal to twice the value of target[-1]
minus the length of target
(since each number not in target
adds two operations, and each number in target
adds one operation). Thus, the space complexity is also O(target[-1])
, which in the worst case is O(n)
.
Learn more about how to find time and space complexity quickly using problem constraints.
How would you design a stack which has a function min
that returns the minimum element in the stack, in addition to push
and pop
? All push
, pop
, min
should have running time O(1)
.
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