K Lexicographically Smallest Subsequence - Google Top Interview Questions


Problem Statement :


Given a list of integers nums and an integer k, return the lexicographically smallest subsequence of length k.

Constraints

k ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [1, 2, 0, 9, 2, 3]

k = 3

Output

[0, 2, 3]

Example 2

Input

nums = [10, 1, 0]

k = 2

Output

[1, 0]



Solution :



title-img




                        Solution in C++ :

vector<int> solve(vector<int>& nums, int k) {
    int n = nums.size();
    stack<int> st;
    for (int i = 0; i < n; i++) {
        if (st.empty())
            st.push(nums[i]);

        else if (st.top() > nums[i]) {
            if (n - i >= k) {
                while (!st.empty() and st.top() > nums[i]) st.pop();
                st.push(nums[i]);
            } else if (st.size() + n - i >= k) {
                while (!st.empty() and st.size() + n - i > k and st.top() > nums[i]) st.pop();
                st.push(nums[i]);
            } else {
                st.push(nums[i]);
            }
        } else {
            st.push(nums[i]);
        }
    }
    while (!st.empty() and st.size() > k) st.pop();
    vector<int> ans;
    while (!st.empty()) {
        ans.push_back(st.top());
        st.pop();
    }
    reverse(begin(ans), end(ans));
    return ans;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k > nums.length)
            return new int[0];
        Deque<Integer> deque = new ArrayDeque();
        for (int i = 0; i < nums.length; i++) {
            while (!deque.isEmpty() && nums[i] < nums[deque.peekLast()]
                && (nums.length - (i + 1)) >= k - deque.size())
                deque.pollLast();
            deque.addLast(i);
        }
        int[] res = new int[k];
        for (int i = 0; i < k; i++) res[i] = nums[deque.pollFirst()];
        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, k):
        if k <= 0:
            return []

        n = len(nums)
        stack = []

        for index, num in enumerate(nums):
            while stack and num < stack[-1] and (len(stack) + n - index - 1) >= k:
                stack.pop()
            stack.append(num)

        return stack[:k]
                    


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