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1441. Build an Array With Stack Operations

Description

You are given an integer array target and an integer n.

You have an empty stack with the two following operations:

  • "Push": pushes an integer to the top of the stack.
  • "Pop": removes the integer on the top of the stack.

You also have a stream of the integers in the range [1, n].

Use the two stack operations to make the numbers in the stack (from the bottom to the top) equal to target. You should follow the following rules:

  • If the stream of the integers is not empty, pick the next integer from the stream and push it to the top of the stack.
  • If the stack is not empty, pop the integer at the top of the stack.
  • If, at any moment, the elements in the stack (from the bottom to the top) are equal to target, do not read new integers from the stream and do not do more operations on the stack.

Return the stack operations needed to build target following the mentioned rules. If there are multiple valid answers, return any of them.

 

Example 1:

Input: target = [1,3], n = 3
Output: ["Push","Push","Pop","Push"]
Explanation: Initially the stack s is empty. The last element is the top of the stack.
Read 1 from the stream and push it to the stack. s = [1].
Read 2 from the stream and push it to the stack. s = [1,2].
Pop the integer on the top of the stack. s = [1].
Read 3 from the stream and push it to the stack. s = [1,3].

Example 2:

Input: target = [1,2,3], n = 3
Output: ["Push","Push","Push"]
Explanation: Initially the stack s is empty. The last element is the top of the stack.
Read 1 from the stream and push it to the stack. s = [1].
Read 2 from the stream and push it to the stack. s = [1,2].
Read 3 from the stream and push it to the stack. s = [1,2,3].

Example 3:

Input: target = [1,2], n = 4
Output: ["Push","Push"]
Explanation: Initially the stack s is empty. The last element is the top of the stack.
Read 1 from the stream and push it to the stack. s = [1].
Read 2 from the stream and push it to the stack. s = [1,2].
Since the stack (from the bottom to the top) is equal to target, we stop the stack operations.
The answers that read integer 3 from the stream are not accepted.

 

Constraints:

  • 1 <= target.length <= 100
  • 1 <= n <= 100
  • 1 <= target[i] <= n
  • target is strictly increasing.

Solutions

Solution 1: Simulation

We define a variable \(\textit{cur}\) to represent the current number to be read, initially set to \(\textit{cur} = 1\), and use an array \(\textit{ans}\) to store the answer.

Next, we iterate through each number \(x\) in the array \(\textit{target}\):

  • If \(\textit{cur} < x\), we add \(\textit{Push}\) and \(\textit{Pop}\) to the answer alternately until \(\textit{cur} = x\);
  • Then we add \(\textit{Push}\) to the answer, representing reading the number \(x\);
  • After that, we increment \(\textit{cur}\) and continue to process the next number.

After the iteration, we return the answer array.

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{target}\). Ignoring the space consumption of the answer array, the space complexity is \(O(1)\).

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class Solution:
    def buildArray(self, target: List[int], n: int) -> List[str]:
        ans = []
        cur = 1
        for x in target:
            while cur < x:
                ans.extend(["Push", "Pop"])
                cur += 1
            ans.append("Push")
            cur += 1
        return ans

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class Solution {
    public List<String> buildArray(int[] target, int n) {
        List<String> ans = new ArrayList<>();
        int cur = 1;
        for (int x : target) {
            while (cur < x) {
                ans.addAll(List.of("Push", "Pop"));
                ++cur;
            }
            ans.add("Push");
            ++cur;
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> buildArray(vector<int>& target, int n) {
        vector<string> ans;
        int cur = 1;
        for (int x : target) {
            while (cur < x) {
                ans.push_back("Push");
                ans.push_back("Pop");
                ++cur;
            }
            ans.push_back("Push");
            ++cur;
        }
        return ans;
    }
};

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func buildArray(target []int, n int) (ans []string) {
    cur := 1
    for _, x := range target {
        for ; cur < x; cur++ {
            ans = append(ans, "Push", "Pop")
        }
        ans = append(ans, "Push")
        cur++
    }
    return
}

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function buildArray(target: number[], n: number): string[] {
    const ans: string[] = [];
    let cur: number = 1;
    for (const x of target) {
        for (; cur < x; ++cur) {
            ans.push('Push', 'Pop');
        }
        ans.push('Push');
        ++cur;
    }
    return ans;
}

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impl Solution {
    pub fn build_array(target: Vec<i32>, n: i32) -> Vec<String> {
        let mut ans = Vec::new();
        let mut cur = 1;
        for &x in &target {
            while cur < x {
                ans.push("Push".to_string());
                ans.push("Pop".to_string());
                cur += 1;
            }
            ans.push("Push".to_string());
            cur += 1;
        }
        ans
    }
}

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
char** buildArray(int* target, int targetSize, int n, int* returnSize) {
    char** ans = (char**) malloc(sizeof(char*) * (2 * n));
    *returnSize = 0;
    int cur = 1;
    for (int i = 0; i < targetSize; i++) {
        while (cur < target[i]) {
            ans[(*returnSize)++] = "Push";
            ans[(*returnSize)++] = "Pop";
            cur++;
        }
        ans[(*returnSize)++] = "Push";
        cur++;
    }
    return ans;
}

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