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 :



title-img


                            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

Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →

Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →

Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →

Lazy White Falcon

White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi

View Solution →

Ticket to Ride

Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o

View Solution →

Heavy Light White Falcon

Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem. You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries: "1 u x" assign x to the value of the node . "2 u v" print the maxim

View Solution →