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 :
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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