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.


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

Example 1


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

k = 3


[0, 2, 3]

Example 2


nums = [10, 1, 0]

k = 2


[1, 0]

Solution :


                        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())

        else if (st.top() > nums[i]) {
            if (n - i >= k) {
                while (!st.empty() and st.top() > nums[i]) st.pop();
            } else if (st.size() + n - i >= k) {
                while (!st.empty() and st.size() + n - i > k and st.top() > nums[i]) st.pop();
            } else {
        } else {
    while (!st.empty() and st.size() > k) st.pop();
    vector<int> ans;
    while (!st.empty()) {
    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())
        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:

        return stack[:k]

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