Max Transform


Problem Statement :


Transforming data into some other data is typical of a programming job. This problem is about a particular kind of transformation which we'll call the max transform.

Let  be a zero-indexed array of integers. For , let  denote the subarray of  from index  to index , inclusive.

Let's define the max transform of  as the array obtained by the following procedure:

Let  be a list, initially empty.
For  from  to :
For  from  to :
Let .
Append  to the end of .
Return .
The returned array is defined as the max transform of . We denote it by .

Complete the function solve that takes an integer array  as input.

Given an array , find the sum of the elements of , i.e., the max transform of the max transform of . Since the answer may be very large, only find it modulo .

Input Format

The first line of input contains a single integer  denoting the length of .

The second line contains  space-separated integers  denoting the elements of .

Constraints

1  <=  n  <=  2x 10^5
1  <=  Ai  <= 10^6



Output Format

Print a single line containing a single integer denoting the answer.


Solution :



title-img 


                            Solution in C :

In   C++  :





#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ii, ii> iii;
ll MOD = 1e9 + 7;
const ld E = 1e-9;
#define null NULL
#define ms(x) memset(x, 0, sizeof(x))
#ifndef LOCAL
#define endl "\n"
#endif
#ifndef LONG_LONG_MAX
#define LONG_LONG_MAX LLONG_MAX
#endif
#define sync ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define _read(x) freopen(x, "r", stdin)
#define _write(x) freopen(x, "w", stdout)
#define files(x) _read(x ".in"); _write(x ".out")
#define filesdatsol(x) _read(x ".DAT"); _write(x ".SOL")
#define output _write("output.txt")
#define input _read("input.txt")
#define prev time_prev
#ifndef M_PI
#define M_PI acos(-1)
#endif
#define remove tipa_remove
#define next tipa_next
#define left tipa_left
#define right tipa_right
#define mod % MOD
#define y1 hello_world
unsigned char ccc;
bool _minus = false;
template<typename T>
inline T sqr(T t) {
    return (t * t);
}
inline void read(ll &n) {
    n = 0;
    _minus = false;
    while (true) {
        ccc = getchar();
        if (ccc == ' ' || ccc == '\n')
            break;
        if (ccc == '-') {
            _minus = true;
            continue;
        }
        n = n * 10 + ccc - '0';
    }
    if (_minus)
        n *= -1;
}
inline bool read(int &n) {
    n = 0;
    _minus = false;
    while (true) {
        ccc = getchar();
        if (ccc == ' ' || ccc == '\n') {
            if (ccc == '\n')
                return true;
            break;
        }
        if (ccc == '-') {
            _minus = true;
            continue;
        }
        n = n * 10 + ccc - '0';
    }
    if (_minus)
        n *= -1;
    return false;
}
char wwww[19];
int kkkk;
inline void write(ll y) {
    long long x = y;
    kkkk = 0;
    if (x < 0) {
        putchar('-');
        x *= -1;
    }
    if (!x)
        wwww[++kkkk] = '0';
    else
        while (x) {
            ++kkkk;
            wwww[kkkk] = char(x % 10 + '0');
            x /= 10;
        }
    for (int i = kkkk; i >= 1; --i)
        putchar(wwww[i]);
}
#ifdef LOCAL
//#define __DEBUG
#endif
#ifdef __DEBUG
#define dbg if(1)
#else
#define dbg if(0)
#endif

inline ll sum(ll n){
    return (n * (n + 1)) / 2;
}



__int128 ans;

const int MAX = 4e5 + 10;

int ar[MAX];
ll tt[MAX];
ll ttt[MAX];
int n;

ll culc(int a){
    return tt[a] % MOD;
}

int t[MAX << 2];

void build(int v, int l, int r){
    if(l == r){
        t[v] = l;
        return;
    }
    int x = (l + r) >> 1;
    build(v << 1, l, x);
    build(v << 1 | 1, x + 1, r);
    t[v] = (ar[t[v << 1]] > ar[t[v << 1 | 1]] ? t[v << 1] : t[v << 1 | 1]);
}

int get_max(int v, int tl, int tr, int l, int r){
    if(l <= tl && tr <= r){
        return t[v];
    }
    if(tr < l || r < tl){
        return 0;
    }
    int x = (tl + tr) >> 1;
    int a = get_max(v << 1, tl, x, l, r);
    int b = get_max(v << 1 | 1, x + 1, tr, l, r);
    return (ar[a] > ar[b] ? a : b);
}

ll culc(int a, int b){
    ll ans = sum(a);
    a--;
    int r = a + b;
    int l = a + b - 2 * min(a, b);
    l = max(l, 0);
    ans += ttt[r] - ttt[l];
    int e = min(a, b);
    a -= e;
    b -= e;
    ans += culc(a) + culc(b);
    return ans;
}

int get_max(int l, int r){
    return get_max(1, 1, n, l, r);
}

int get_max(int l1, int r1, int l2, int r2){
    int a = get_max(l1, r1);
    int b = get_max(l2, r2);
    return (ar[a] > ar[b] ? a : b);
}

ll solve_1(int l, int r){
    if(l > r)
        return 0;
    if(l == r){
        ans += ar[l];
        return 1;
    }
    int pos = get_max(l, r);
    ll has = culc(r - l + 1);
    has -= solve_1(l, pos - 1);
    has -= solve_1(pos + 1, r);
    has %= MOD;
    ans += has * ar[pos];
    return culc(r - l + 1);
}

ll solve_2(int r, int l){
    if(r == 0){
        return solve_1(l, n);
    }
    if(l == n + 1){
        return solve_1(1, r);
    }
    int pos = get_max(1, r, l, n);
    ll has = culc(r, n - l + 1);
    if(pos <= r){
        has -= solve_2(pos - 1, l);
        has -= solve_1(pos + 1, r);
    }else{
        has -= solve_2(r, pos + 1);
        has -= solve_1(l, pos - 1);
    }
    has %= MOD;
    ans += has * ar[pos];
    return culc(r, n - l + 1);
}

namespace solve_long {
    vector<int> get(vector<int> v){
        vector<int> t;
        int n = (int) v.size();
        for(int k = 0; k < n; k++){
            for(int i = 0; i < n - k; i++){
                int ans = 0;
                for(int j = i; j <= i + k; j++){
                    ans = max(ans, v[j]);
                }
                t.push_back(ans);
            }
        }
        return t;
    }
    int max_val = 0;
    int get_max(vector<int> &v, int l, int r){
        int pos = l;
        for(int i = l + 1; i <= r && v[pos] != max_val; i++){
            if(v[i] > v[pos]){
                pos = i;
            }
        }
        return pos;
    }
    ll sum(vector<int> &v, int l, int r){
        if(l > r)
            return 0;
        int pos = get_max(v, l, r);
        ll ans = (pos - l + 1) * 1LL * (r - pos + 1) * v[pos];
        return ans + sum(v, l, pos - 1) + sum(v, pos + 1, r);
    }
    
    ll sum(vector<int> v){
        max_val = 0;
        for(int a : v){
            max_val = max(max_val, a);
        }
        return sum(v, 0, (int) v.size() - 1);
    }
    __int128 solve(vector<int> v){
        return sum((get(v))) % MOD;
    }
}



ostream& operator<<(ostream &cout, __int128 a){
    string s = "";
    while(a){
        s += char(a % 10 + '0');
        a /= 10;
    }
    reverse(s.begin(), s.end());
    cout << s;
    return cout;
}



__int128 solve_ok(vector<int> v){
    n = (int) v.size();
    for(int i = 1; i <= n; i++){
        ar[i] = v[i - 1];
    }
    build(1, 1, n);
    ans = 0;
    __int128 g = n;
    g = (g * (g + 1)) / 2;
    g = (g * (g + 1)) / 2;
    
    int pos = get_max(1, n);
    g -= solve_2(pos - 1, pos + 1);
    
    ans += g * ar[pos];
    return ans % MOD;
}

int main() {
    sync;
    srand((unsigned int) time(NULL));
    cout.precision(10);
    cout << fixed;
#ifdef LOCAL
    input;
    output;
    ll start = (ll) clock();
#else
    
#endif
  
    for(int i = 1; i < MAX; i++){
        tt[i] = sum(i) + tt[i - 1];
    }
    for(int i = 1; i < MAX; i++){
        ttt[i] = sum(i);
        if(i > 1){
            ttt[i] += ttt[i - 2];
        }
    }
    
    int n;
    cin >> n;
  
    vector<int> v(n);
    for(int i = 0; i < n; i++){
        cin >> v[i];
    }
    cout << solve_ok(v) << endl;
    
#ifdef LOCAL
    cout << "=====" << endl;
    cout << (clock() - start) * 1. / CLOCKS_PER_SEC << endl;
    cout << clock() << endl;
    cout << "=====" << endl;
#endif
    
   // assert(false);
    
}







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.InputMismatchException;

public class MaxTransform {
InputStream is;
PrintWriter out;
String INPUT = "";
int mod = 1000000007;

void solve()
{
int n = ni();
int[] a = na(n);
if(n == 1){
out.println(a[0]);
return;
}
long all = (long)n*(n+1)/2%mod;
all = all*(all+1)/2%mod;
for(int i = 1;i <= n-1;i++){
all -= (long)(n-i+1+n-i)*(n-i+1+n-i+1)/2;
if(i < n-1)all += (long)(n-i)*(n-i+1)/2;
all %= mod;
}
int amax = 0;
for(int v : a)amax = Math.max(amax, v);
long ans = all*amax;
ans %= mod;
int[] b = new int[n];
for(int i = 0;i < n;i++)b[i] = -a[i];
SegmentTreeRMQPos st = new SegmentTreeRMQPos(b);
imos = new long[n+3];
dfs(0, n, a, st);

for(int i = 0;i <= n+2;i++){
imos[i] %= mod;
}
for(int i = 0;i <= n+1;i++){
imos[i+1] += imos[i];
imos[i+1] %= mod;
}
for(int i = 0;i <= n+1;i++){
imos[i+1] += imos[i];
imos[i+1] %= mod;
}
for(int i = 1;i <= n;i++){
ans += (long)i*imos[i];
ans %= mod;
}


int[] sufs = new int[n];
for(int i = 0;i < n;i++){
sufs[i] = a[n-1-i];
if(i > 0)sufs[i] = Math.max(sufs[i], sufs[i-1]);
}
int[] pres = new int[n];
for(int i = 0;i < n;i++){
pres[i] = a[i];
if(i > 0)pres[i] = Math.max(pres[i], pres[i-1]);
}
long[] cpres = new long[n+1];
for(int i = 0;i < n;i++){
cpres[i+1] = cpres[i] + pres[i];
}
long[] csufs = new long[n+1];
for(int i = 0;i < n;i++){
csufs[i+1] = csufs[i] + sufs[i];
}

long temp = 0;
for(int i = n-1;i >= 1;i--){
if(i < n-1)temp += maxsum(sufs[i-1], pres, i+1, n, cpres);
temp += maxsum(pres[i], sufs, i-1, n, csufs);
ans += temp;
ans %= mod;
}
ans %= mod;
if(ans < 0)ans += mod;
out.println(ans);
}

long maxsum(int v, int[] a, int l, int r, long[] cum)
{
int ind = Arrays.binarySearch(a, l, r, v);
if(ind < 0)ind = -ind-1;
long ret = cum[r] - cum[ind] + (long)(ind-l) * v;
ret %= mod;
return ret;
}

long[] imos;

void dfs(int l, int r, int[] a, SegmentTreeRMQPos st)
{
if(l >= r)return;
st.minx(l, r);
int arg = st.minpos;
imos[1] += a[arg];
imos[arg-l+2] -= a[arg];
imos[r-arg+1] -= a[arg];
imos[r-l+2] += a[arg];
dfs(l, arg, a, st);
dfs(arg+1, r, a, st);
}

public static class SegmentTreeRMQPos {
public int M, H, N;
public int[] st;
public int[] pos;

public SegmentTreeRMQPos(int n)
{
N = n;
M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
H = M>>>1;
st = new int[M];
pos = new int[M];
for(int i = 0;i < N;i++)pos[H+i] = i;
Arrays.fill(st, 0, M, Integer.MAX_VALUE);
for(int i = H-1;i >= 1;i--)propagate(i);
}

public SegmentTreeRMQPos(int[] a)
{
N = a.length;
M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
H = M>>>1;
st = new int[M];
pos = new int[M];
for(int i = 0;i < N;i++){
st[H+i] = a[i];
pos[H+i] = i;
}
Arrays.fill(st, H+N, M, Integer.MAX_VALUE);
for(int i = H-1;i >= 1;i--)propagate(i);
}

public void update(int pos, int x)
{
st[H+pos] = x;
for(int i = (H+pos)>>>1;i >= 1;i >>>= 1)propagate(i);
}

private void propagate(int i)
{
if(st[2*i] <= st[2*i+1]){
st[i] = st[2*i];
pos[i] = pos[2*i];
}else{
st[i] = st[2*i+1];
pos[i] = pos[2*i+1];
}
}

public int minpos;
public int minval;

public int minx(int l, int r){
minval = Integer.MAX_VALUE;
minpos = -1;
if(l >= r)return minval;
while(l != 0){
int f = l&-l;
if(l+f > r)break;
int v = st[(H+l)/f];
if(v < minval){
    minval = v;
    minpos = pos[(H+l)/f];
}
l += f;
}

while(l < r){
int f = r&-r;
int v = st[(H+r)/f-1];
if(v < minval){
    minval = v;
    minpos = pos[(H+r)/f-1];
}
r -= f;
}
return minval;
}

public int min(int l, int r){ 
minpos = -1;
minval = Integer.MAX_VALUE;
min(l, r, 0, H, 1);
return minval;
}

private void min(int l, int r, int cl, int cr, int cur)
{
if(l <= cl && cr <= r){
if(st[cur] < minval){
    minval = st[cur];
    minpos = pos[cur];
}
}else{
int mid = cl+cr>>>1;
if(cl < r && l < mid)min(l, r, cl, mid, 2*cur);
if(mid < r && l < cr)min(l, r, mid, cr, 2*cur+1);
}
}
}


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 MaxTransform().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))){ 
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  :





#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse4")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int mod = 1000000007, _2 = 500000004;
int N, MX = 0, tp, a[200010], 
i_1[200010], st[200010], mxl[200010],
 mxr[200010], sxl[200010], sxr[200010];
long long M, CNT, ANS = 0;
void calc(int w, int x, int y)
{
if( x < y )
{
int temp = x;
x = y;
y = temp;
}
int k;
if( x == y )
{
k = ( ( (long long)( x + y ) * i_1[y] % mod - 
(long long)x * x % mod ) % mod + mod ) % mod;
}
else
{
k = ( ( (long long)y * ( i_1[x-1] - i_1[y] ) % 
mod + (long long)( x + y ) * i_1[y] % 
mod ) % mod + mod ) % mod;
}
ANS = ( ANS + (long long)w * k ) % mod;
CNT -= k;
if( CNT < 0 )
{
CNT += mod;
}
}
void calcl(int w, int x, int y)
{
if( x == 1 || y == 0 )
{
return;
}
int k;
if( y < x )
{
k = i_1[y];
}
else
{
k = ( i_1[x-1] + (long long)( 
    y - x + 1 ) * ( x - 1 ) ) % mod;
}
ANS = ( ANS + (long long)w * k ) % mod;
CNT -= k;
if( CNT < 0 )
{
CNT += mod;
}
}
void calcr(int w, int x, int y)
{
if( x == 0 || y == 1 )
{
return;
}
int k;
if( y + 1 <= x )
{
k = i_1[y-1];
}
else
{
k = ( i_1[x] + (long long)( y - x - 1 ) * x ) % mod;
}
ANS = ( ANS + (long long)w * k ) % mod;
CNT -= k;
if( CNT < 0 )
{
CNT += mod;
}
}
int main()
{
int p;
scanf("%d", &N);
for( int i = 1 ; i <= N ; i++ )
{
scanf("%d", &a[i]);
MX = MX > a[i] ? MX : a[i];
}
M = ( (long long)N * 
( N + 1 ) >> 1 ) % mod;
M = (long long)M * ( M + 1 ) % mod * _2 % mod;
CNT = M;
for( int i = 1 ; i <= N ; i++ )
{
i_1[i] = ( i_1[i-1] + i ) % mod;
}
for( int i = 1 ; i <= N ; i++ )
{
sxl[i] = sxl[i-1] > a[i] ? sxl[i-1] : a[i];
}
for( int i = N ; i ; i-- )

{
sxr[i] = sxr[i+1] > a[i] ? sxr[i+1] : a[i];
}
tp = 0;
for( int i = 1 ; i <= N ; i++ )
{
while( tp > 0 && a[st[tp]] <= a[i] )
{
tp--;
}
if(tp)
{
mxl[i] = st[tp] + 1;
}
else
{
mxl[i] = 1;
}
st[++tp] = i;
}
tp = 0;
for( int i = N ; i ; i-- )
{
while( tp > 0 && a[st[tp]] < a[i] )
{
tp--;
}
if(tp)
{
mxr[i] = st[tp] - 1;
}
else
{
mxr[i] = N;
}
st[++tp] = i;
}
for( int i = 1 ; i <= N ; i++ )
{
calc(a[i], i-mxl[i]+1, mxr[i]-i+1);
}
p = N;
for( int i = 1 ; i <= N ; i++ )
{
int g = sxl[i];
while( p > i && sxr[p] < g )
{
p--;
}
while( p < i )
{
p++;
}
calcl(g, i, N-p);
}
p = 1;
for( int i = N ; i ; i-- )
{
int g = sxr[i];
while( p < i && sxl[p] <= g )
{
p++;
}
while( p > i )
{
p--;
}
calcr(g, N-i+1, p-1);
}
CNT = ( CNT % mod + mod ) % mod;
ANS = ( ANS + (long long)CNT * MX ) % mod;
printf("%lld", ANS);
return 0;
}








In   Python 3 :






#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the solve function below.

import math
import os
import random
import re
import sys
sys.setrecursionlimit(9999999)
from decimal import Decimal
def t1(n):
    return Decimal(n * (n + 1) / 2)


def t2(n):
    return Decimal(n * (n + 1) * (n + 2) / 6)


def u2(n):
    return Decimal(n * (n + 2) * (2 * n + 5) / 24)


def countzip(a, b):
    return u2(a + b) - u2(abs(a - b)) + t2(abs(a - b))


def countends(x, n, ex):
    return countzip(n, ex) - countzip(x, ex) - countzip(n - 1 - x, 0)


def countsplit(x, n):
    return t1(t1(n)) - t1(x) - countzip(n - x - 1, x - 1)


K = 20
lg = [0] * (1 << K)
for i in range(K):
    lg[1 << i] = i
for i in range(1, 1 << K):
    lg[i] = max(lg[i], lg[i - 1])


def make_rangemax(A):
    n = len(A)
    assert 1 << K > n

    key = lambda x: A[x]
    mxk = []
    mxk.append(range(n))
    for k in range(K - 1):
        mxk.append(list(mxk[-1]))
        for i in range(n - (1 << k)):
            mxk[k + 1][i] = max(
            mxk[k][i], mxk[k][i + (1 << k)],
             key=key)

    def rangemax(i, j):
        k = lg[j - i]
        return max(mxk[k][i], mxk[k][j - (1 << k)], key=key)

    return rangemax


def brutesolo(A):
    rangemax = make_rangemax(A)
    stack = [(0, len(A))]
    ans = 0
    while stack:
        i, j = stack.pop()
        if i != j:
            x = rangemax(i, j)
            stack.append((i, x))
            stack.append((x + 1, j))
            ans += A[x] * (x - i + 1) * (j - x)
    return ans


def make_brute(A):
    rangemax = make_rangemax(A)

    def brute(i, j):
        stack = [(i, j)]
        ans = 0
        while stack:
            i, j = stack.pop()
            if i != j:
                x = rangemax(i, j)
                stack.append((i, x))
                stack.append((x + 1, j))
                ans += A[x] * countends(x - i, j - i, 0)
        return ans

    return brute, rangemax


def ends(A, B):
    brutea, rangemaxa = make_brute(A)
    bruteb, rangemaxb = make_brute(B)

    stack = [(len(A), len(B))]
    ans = 0
    while stack:
        i, j = stack.pop()
        if i == 0:
            ans += bruteb(0, j)
        elif j == 0:
            ans += brutea(0, i)
        else:
            x = rangemaxa(0, i)
            y = rangemaxb(0, j)
            if A[x] < B[y]:
                ans += bruteb(y + 1, j)
                ans += B[y] * countends(y, j, i)
                stack.append((i, y))
            else:
                ans += brutea(x + 1, i)
                ans += A[x] * countends(x, i, j)
                stack.append((x, j))

    return ans


def maxpairs(a):
    return [max(x, y) for x, y in zip(a, a[1:])]


def solve(A):
    n = len(A)
    x = max(range(n), key=lambda x: A[x])
    return (int((brutesolo(A[:x]) +
    ends(A[x + 1:][::-1], maxpairs(A[:x])) + 
    A[x] * countsplit(x, n))%(10**9+7)))



if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input())

    A = list(map(int, input().rstrip().split()))

    result = solve(A)

    fptr.write(str(result) + '\n')

    fptr.close()








View More Similar Problems

Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →

Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →

Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

View Solution →

Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

View Solution →

Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

View Solution →

Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

View Solution →