Lexicographically Largest Mountain List - Amazon Top Interview Questions


Problem Statement :


You are given three positive integers n, lower, and upper. You want to create a list of length n that is strictly increasing and then strictly decreasing and all the numbers are between [lower, upper], inclusive. Each of the increasing and decreasing parts should be non-empty.

Return the lexicographically largest list possible, or the empty list if it's not possible.

Constraints

3 ≤ n ≤ 100,000
1 ≤ lower ≤ upper < 2 ** 31


Example 1

Input

n = 5
lower = 2
upper = 6

Output

[5, 6, 5, 4, 3]


Explanation

Note that [6, 5, 4, 3, 2] is not valid since the strictly increasing part has to be non-empty.



Example 2

Input

n = 5
lower = 90
upper = 92

Output
[90, 91, 92, 91, 90]



Example 3

Input

n = 6
lower = 3
upper = 5


Output
[]


Explanation

It's impossible to make a strictly increasing then decreasing list of size 6 here.


Solution :



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                        Solution in C++ :

vector<int> solve(int n, int lower, int upper) {
    deque<int> d;
    int num = upper;
    for (int i = 0; i < n - 1; ++i) {
        if (num < lower) break;
        d.push_back(num--);
    }
    int right = d.size();
    num = upper - 1;
    for (int i = 0; i < n - right; ++i) {
        if (num < lower) break;
        d.push_front(num--);
    }
    if (d.size() != n) return {};
    vector<int> res;
    for (int i = 0; i < d.size(); ++i) res.push_back(d[i]);
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int N, int lower, int upper) {
        int[] ans = new int[N];
        if (N < 3 || lower == upper) {
            ans = new int[] {};
        } else if (N <= upper - lower + 2) {
            ans[0] = upper - 1;
            for (int i = 1; i < N; i++) ans[i] = upper - i + 1;
        } else if (N <= 2 * (upper - lower) + 1) {
            int cur = lower;
            int dir = 1;
            for (int i = N - 1; i >= 0; i--) {
                ans[i] = cur;
                if (cur == upper)
                    dir = -1;
                cur += dir;
            }
        } else {
            ans = new int[] {};
        }
        return ans;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, n, lower, upper):
        i = j = upper - 1
        n -= 3
        while j > lower and n:
            j -= 1
            n -= 1
        while i > lower and n:
            i -= 1
            n -= 1

        if n or lower == upper:
            return []
        return list(range(i, upper)) + list(range(upper, j - 1, -1))


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