Lena Sort


Problem Statement :


Lena developed a sorting algorithm described by the following pseudocode:

lena_sort(array nums) {
    if (nums.size <= 1) {
        return nums;
    }
    pivot = nums[0];
    array less;
    array more;
    for (i = 1; i < nums.size; ++i) {
    	// Comparison
        if (nums[i] < pivot) {
            less.append(nums[i]);
        }
        else {
            more.append(nums[i]);
        }
    }
    sorted_less = lena_sort(less);
    sorted_more = lena_sort(more);
    ans = sorted_less + pivot + sorted_more;
    
    return ans;
}
We consider a comparison to be any time some nums[i] is compared with pivot.

You must solve q queries where each query i consists of some len(i) and ci. For each query, construct an array of len(i) distinct elements in the inclusive range between 1 and 10^9 that will be sorted by lena_sort in exactly ci comparisons, then print each respective element of the unsorted array as a single line of leni space-separated integers; if no such array exists, print -1 instead.

Input Format

The first line contains a single integer denoting q (the number of queries).
Each line i of the q subsequent lines contains two space-separated integers describing the respective values of len(i) (the length of the array) and ci (the number of comparisons) for query i.

Constraints

1 <= q <= 10^5
1 <= len(i) <= 10^5
0 <= ci <= 10^9
1 <=  the sum of len(i) over all queries <= 10^6

Output Format

Print the answer to each query on a new line. For each query i, print len(i) space-separated integers describing each respective element in an unsorted array that Lena's algorithm will sort in exactly ci comparisons; if no such array exists, print -1 instead.


Solution :



title-img 


                            Solution in C :

In C++ :






#include <bits/stdc++.h>

using namespace std;

#ifdef WIN32
	#define I64 "%I64d"
#else
	#define I64 "%lld"
#endif

typedef long long ll;

#define f first
#define s second
#define mp make_pair
#define pb push_back
#define all(s) s.begin(), s.end()
#define sz(s) (int(s.size()))
#define fname "a"
#define MAXN 200002

int n, k;
int a[MAXN];
ll d[MAXN];
ll f[MAXN];

void go(int l, int r, int k, int ff) {
	if (l > r) return;
	if (l == r) {
		a[l] = ff;
		return;
	}
	int n = r - l + 1;
	k -= n - 1;
	for (int i = 0; i < n; ++i) {
		if (d[i] + d[n - 1 - i] <= k && f[i] + f[n - 1 - i] >= k)
                   {
			a[l] = ff + i;
			int k1 = d[i];
			int k2 = k - d[i];
			if (k2 > f[n - 1 - i]) {
				k1 += k2 - f[n - 1 - i];
				k2 = f[n - 1 - i];
			}
			go(l + 1, l + i, k1, ff);
			go(l + i + 1, r, k2, ff + i + 1);
			return;
		}
	}
}

inline void solve()
{
	scanf("%d%d", &n, &k);
	if (k < d[n] || k > f[n]) {
		puts("-1");
		return;
	}
	go(0, n - 1, k, 0);
	for (int i = 0; i < n; ++i)
		printf("%d%c", a[i] + 1, " \n"[i + 1 == n]);
}

int main()
{
	#ifdef LOCAL
	freopen(fname".in", "r", stdin);
	freopen(fname".out", "w", stdout);
	#endif

	for (int i = 2; i < MAXN; ++i) {
		d[i] = i - 1 + d[i / 2] + d[(i - 1) / 2];
		f[i] = 1LL * (i - 1) * i / 2;
	}

	int tt;
	scanf("%d", &tt);
	for (int t = 0; t < tt; ++t)
		solve();

	return 0;
}





In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        long[] mins = new long[100001];
        long[] maxes = new long[100001];
        mins[2] = 1;
        maxes[2] = 1;
        for (int i = 3; i <= 100000; i++) {
            mins[i] = i-1+mins[(i-1)/2]+mins[(i-1)-(i-1)/2];
            maxes[i] = maxes[i-1]+i-1;
        }
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for(int a0 = 0; a0 < q; a0++){
            int len = in.nextInt();
            int c = in.nextInt();
            if (maxes[len]<c||mins[len]>c) {
                System.out.println(-1);
                continue;
            }
            System.out.println(portion(len, c, 1, mins, maxes, new StringBuilder()));
        }
    }
    
    public static StringBuilder portion(int len, long c, int offset, long[] mins, long[] maxes, StringBuilder ans) {
        if (len==0) {
            return ans;
        }
        if (len==1) {
            ans.append(offset+" ");
            return ans;
        }
        int pivot = 0;
        c -= len-1;
        while (mins[pivot]+mins[len-pivot-1]>c||maxes[pivot]+maxes[len-pivot-1]<c)
            pivot++;
        long newc = mins[pivot];
        while (mins[len-pivot-1]>c-newc||maxes[len-pivot-1]<c-newc)
            newc++;
        ans.append((pivot+offset)+" ");
        portion(pivot, newc, offset, mins, maxes, ans);
        portion(len-pivot-1, c-newc, offset+pivot+1, mins, maxes, ans);
        return ans;
    }
}








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