Costly Intervals
Problem Statement :
Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the elements of Al. . .r. Let AND( l , r ) be the bitwise AND of the elements of Al. . .r. The cost of Al . . .r , denoted cost( l, r ), is defined as The size of Al . . .r is defined as r - l +1. You are given the array and and an integer . For each index from to , your goal is to find the largest size of any subarray such that and . Complete the function costlyIntervals which takes two integers n and k as first line of input, and array A1, A2, . . . , An in the second line of input. Return an array of n integers, where the ith element contains the answer for index i of the input array, 1 <= i <= n. Every element of the output array denotes the largest size of a subarray containing i whose cost is at least k, or -1 if there is no such subarray.
Solution :
Solution in C :
In C++ :
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Inf = 1000000007;
const int Maxn = 100005;
const int Maxm = 20;
const int Maxb = 31;
int n, k;
int a[Maxn];
int mx[Maxn][Maxm], mn[Maxn][Maxm];
int nxt[Maxn][Maxb][2];
int res[Maxn];
vector <int> add[Maxn], rem[Maxn];
map <int, int> M;
int Get(int ind, int forb, int &Mn, int &Mx, int cur)
{
int pnt = ind;
for (int i = Maxm - 1; i >= 0; i--)
if (pnt + (1 << i) <= forb) {
int candmx = max(Mx, mx[pnt][i]);
int candmn = min(Mn, mn[pnt][i]);
if (ll(cur) - ll(candmx - candmn) >= k) {
Mx = candmx; Mn = candmn;
pnt += 1 << i;
}
}
int res = pnt;
for (int i = Maxm - 1; i >= 0; i--)
if (pnt + (1 << i) <= forb) {
Mx = max(Mx, mx[pnt][i]);
Mn = min(Mn, mn[pnt][i]);
pnt += 1 << i;
}
return res;
}
int main() {
scanf("%d %d", &n, &k);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
mx[i][0] = mn[i][0] = a[i];
}
for (int j = 1; j < Maxm; j++)
for (int i = 1; i + (1 << j) <= n + 1; i++) {
int nxt = i + (1 << j - 1);
mx[i][j] = max(mx[i][j - 1], mx[nxt][j - 1]);
mn[i][j] = min(mn[i][j - 1], mn[nxt][j - 1]);
}
for (int i = 0; i < Maxb; i++)
nxt[n + 1][i][0] = nxt[n + 1][i][1] = n + 1;
for (int i = n; i > 0; i--)
for (int j = 0; j < Maxb; j++)
for (int k = 0; k < 2; k++)
if (bool(a[i] & 1 << j) == k) nxt[i][j][k] = i;
else nxt[i][j][k] = nxt[i + 1][j][k];
for (int i = 1; i <= n; i++) {
int Or = a[i], And = a[i];
int Mn = Inf, Mx = -Inf;
int st = i;
while (st <= n) {
int lim = n + 1;
for (int j = 0; j < Maxb; j++) {
if (!(Or & 1 << j)) lim = min(lim, nxt[st + 1][j][1]);
if (And & 1 << j) lim = min(lim, nxt[st + 1][j][0]);
}
int got = Get(st, lim, Mn, Mx, Or - And);
if (got > st) {
int cand = got - i;
add[i].push_back(cand); rem[got].push_back(cand);
}
for (int j = 0; j < Maxb; j++) {
if (!(Or & 1 << j) && lim == nxt[st + 1][j][1]) Or |= 1 << j;
if (bool(And & 1 << j) && lim == nxt[st + 1][j][0]) And ^= 1 << j;
}
st = lim;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j < add[i].size(); j++)
M[add[i][j]]++;
for (int j = 0; j < rem[i].size(); j++)
if (--M[rem[i][j]] == 0) M.erase(rem[i][j]);
if (!M.empty()) {
map <int, int>::iterator it = M.end(); it--;
printf("%d\n", it->first);
} else printf("-1\n");
}
return 0;
}
In Java :
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.InputMismatchException;
import java.util.TreeSet;
public class D2 {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
int n = ni(), K = ni();
int[] a = na(n);
int[] ra = new int[n];
for(int i = 0;i < n;i++)ra[i] = -a[i];
int[][] stmin = build(a);
int[][] stmax = build(ra);
int[][] efs = new int[80*n][];
int efp = 0;
int esp = 0;
int[][] oas = new int[0][];
for(int i = n-1;i >= 0;i--){
int[][] noas = new int[40][];
int p = 0;
for(int j = 0;j < oas.length;j++){
oas[j][0] |= a[i];
oas[j][1] &= a[i];
if(p > 0 && noas[p-1][0] == oas[j][0] &&
noas[p-1][1] == oas[j][1]){
noas[p-1][2] = oas[j][2];
}else{
noas[p++] = oas[j];
}
}
if(!(p > 0 && noas[p-1][0] == a[i] &&
noas[p-1][1] == a[i])){
noas[p++] = new int[]{a[i], a[i], i};
}else{
noas[p-1][2] = i;
}
oas = Arrays.copyOf(noas, p);
// tr(i, oas);
int to = n;
for(int[] oa : oas){
// [oa[2], to)
int cha = oa[0] - oa[1];
int low = oa[2]-1, high = to;
while(high - low > 1){
int h = high+low>>1;
// [i,h]
// tr(h, oa, to, cha, -rmq(stmax, i, h+1) - rmq(stmin, i, h+1));
if(cha - (-rmq(stmax, i, h+1) - rmq(stmin, i, h+1)) >= K){
low = h;
}else{
high = h;
}
}
if(low >= oa[2]){
// tr(i, oa, to, low);
efs[efp++] = new int[]{i, low - i + 1, i};
efs[efp++] = new int[]{low+1, low - i + 1, i};
}
to = oa[2];
}
}
int I = -1;
int[] anss = new int[n];
Arrays.fill(anss, I);
Arrays.sort(efs, 0, efp, new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
return a[0] - b[0];
}
});
TreeSet<Long> set = new TreeSet<Long>();
int q = 0;
for(int i = 0;i < n;i++){
while(q < efp && efs[q][0] <= i){
long code = (long)efs[q][1]<<32|efs[q][2];
if(set.contains(code)){
set.remove(code);
}else{
set.add(code);
}
q++;
}
if(!set.isEmpty()){
Long first = set.last();
anss[i] = Math.max(anss[i], (int)(first>>>32));
}
}
for(int v : anss){
out.println(v);
}
}
public static int[][] build(int[] a)
{
int n = a.length;
int b = 32-Integer.numberOfLeadingZeros(n);
int[][] ret = new int[b][];
for(int i = 0, l = 1;i < b;i++, l*=2) {
if(i == 0) {
ret[i] = a;
}else {
ret[i] = new int[n-l+1];
for(int j = 0;j < n-l+1;j++) {
ret[i][j] = Math.min(ret[i-1][j], ret[i-1][j+l/2]);
}
}
}
return ret;
}
// [a,b)
public static int rmq(int[][] or, int l, int r)
{
assert l <= r;
if(l >= r)return Integer.MAX_VALUE;
// 1:0, 2:1, 3:1, 4:2, 5:2, 6:2, 7:2, 8:3
int t = 31-Integer.numberOfLeadingZeros(r-l);
return Math.min(or[t][l], or[t][r-(1<<t)]);
}
void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
public static void main(String[] args) throws Exception { new D2().run(); }
private byte[] inbuf = new byte[1024];
public int lenbuf = 0, ptrbuf = 0;
private int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }
private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}
In C :
#include <stdio.h>
#include <stdlib.h>
#define INF 200000
int get(int l,int r);
int max(int x,int y);
int min(int x,int y);
void init( int n );
void range_increment( int i, int j, int val );
int query( int i );
void sort_a(int*a,int size,int*new_size);
void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size);
int N,a[100000],b[100000],map[100001],dp[4][100000][18],dp1[30][100000],dp2[30][100000],tree[400000];
int main(){
int n,k,s,ns,h,l,m,i,j;
scanf("%d%d",&n,&k);
for(i=j=1,map[0]=0;i<=100000;i++)
if(j*2<=i){
j*=2;
map[i]=map[i-1]+1;
}
else
map[i]=map[i-1];
for(i=0;i<n;i++)
scanf("%d",a+i);
for(i=0;i<n;i++)
dp[0][i][0]=dp[1][i][0]=dp[2][i][0]=dp[3][i][0]=a[i];
for(j=1;1<<j<=n;j++)
for(i=0;i+(1<<j)-1<n;i++){
dp[0][i][j]=max(dp[0][i][j-1],dp[0][i+(1<<(j-1))][j-1]);
dp[1][i][j]=min(dp[1][i][j-1],dp[1][i+(1<<(j-1))][j-1]);
dp[2][i][j]=dp[2][i][j-1]|dp[2][i+(1<<(j-1))][j-1];
dp[3][i][j]=dp[3][i][j-1]&dp[3][i+(1<<(j-1))][j-1];
}
for(i=0;i<n;i++)
for(j=0;j<30;j++)
if(a[i]&(1<<j)){
dp1[j][i]=i;
dp2[j][i]=INF;
}
else{
dp1[j][i]=INF;
dp2[j][i]=i;
}
for(i=n-2;i>=0;i--)
for(j=0;j<30;j++){
dp1[j][i]=min(dp1[j][i],dp1[j][i+1]);
dp2[j][i]=min(dp2[j][i],dp2[j][i+1]);
}
init(n);
for(i=0;i<n;i++){
for(j=s=0;j<30;j++){
if(dp1[j][i]!=INF)
b[s++]=dp1[j][i];
if(dp2[j][i]!=INF)
b[s++]=dp2[j][i];
}
sort_a(b,s,&ns);
for(j=ns-1;j>=0;j--)
if(get(i,b[j])>=k){
if(j==ns-1)
h=n-1;
else
h=b[j+1]-1;
l=s=b[j];
while(l<=h){
m=(h+l)/2;
if(get(i,m)>=k){
s=m;
l=m+1;
}
else
h=m-1;
}
range_increment(i,s,s-i+1);
break;
}
}
for(i=0;i<n;i++)
printf("%d\n",query(i));
return 0;
}
int get(int l,int r){
int a,b,c,d,len;
len=map[r-l+1];
a=max(dp[0][l][len],dp[0][r-(1<<len)+1][len]);
b=min(dp[1][l][len],dp[1][r-(1<<len)+1][len]);
c=dp[2][l][len]|dp[2][r-(1<<len)+1][len];
d=dp[3][l][len]&dp[3][r-(1<<len)+1][len];
return c-d-a+b;
}
int max(int x,int y){
return (x>y)?x:y;
}
int min(int x,int y){
return (x<y)?x:y;
}
void init( int n )
{
N = 1;
while( N < n ) N *= 2;
int i;
for( i = 1; i < N + n; i++ ) tree[i] = -1;
}
void range_increment( int i, int j, int val )
{
for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
{
if( i % 2 == 1 ) tree[i] = max(val,tree[i]);
if( j % 2 == 0 ) tree[j] = max(val,tree[j]);
}
}
int query( int i )
{
int ans = -1,j;
for( j = i + N; j; j /= 2 ) ans = max(tree[j],ans);
return ans;
}
void sort_a(int*a,int size,int*new_size){
if (size < 2){
(*new_size)=size;
return;
}
int m = (size+1)/2,i;
int *left,*right;
left=(int*)malloc(m*sizeof(int));
right=(int*)malloc((size-m)*sizeof(int));
for(i=0;i<m;i++)
left[i]=a[i];
for(i=0;i<size-m;i++)
right[i]=a[i+m];
int new_l_size=0,new_r_size=0;
sort_a(left,m,&new_l_size);
sort_a(right,size-m,&new_r_size);
merge(a,left,right,new_l_size,new_r_size,new_size);
free(left);
free(right);
return;
}
void merge(int*a,int*left,int*right,int left_size, int right_size,int*new_size){
int i = 0, j = 0,index=0;
while (i < left_size|| j < right_size) {
if (i == left_size) {
a[index++] = right[j];
j++;
} else if (j == right_size) {
a[index++] = left[i];
i++;
} else if (left[i] <= right[j]) {
a[index++] = left[i];
i++;
} else {
a[index++] = right[j];
j++;
}
if(index>1&&a[index-2]==a[index-1])
index--;
}
(*new_size)=index;
return;
}
In Python3 :
import sys
def cost(a):
x = 0
y = 1
for i in a:
x |= i
y &= i
return((x-y)-(max(a)-min(a)))
def costlyIntervals(n, k, A):
ans = []
for m in range(n):
cs = -1
for i in range(0,n-1):
for j in range(i,n):
l = A[i:j+1]
if A[m] in l:
x = cost(l)
if x >= k:
cs = max(cs,len(l))
ans.append(cs)
return(ans)
if __name__ == "__main__":
n, k = input().strip().split(' ')
n, k = [int(n), int(k)]
A = list(map(int, input().strip().split(' ')))
result = costlyIntervals(n, k, A)
print ("\n".join(map(str, result)))
View More Similar Problems
Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →Find Merge Point of Two Lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share
View Solution →Inserting a Node Into a Sorted Doubly Linked List
Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function
View Solution →Reverse a doubly linked list
This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.
View Solution →Tree: Preorder Traversal
Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's
View Solution →Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →