Dados tres números N , A y X , la tarea es construir la array binaria lexicográficamente más pequeña de tamaño N , que contenga A 0 s y tenga un recuento de inversión de X .
Ejemplos:
Entrada: N=5, A=2, X=1
Salida: 0 1 0 1 1
Explicación:
El número de inversiones en esta array es 1 (segundo y tercer índice).Entrada: N=5, A=2, X=3
Salida: 0 1 1 1 0
Enfoque: El problema dado se puede resolver usando la técnica de dos punteros basada en las siguientes observaciones:
- La array con A 0 s que tiene inversión 0 es la array con todos los 0 al principio y luego todos los 1 s.
- Si se intercambia un elemento 0 en el índice i y un elemento 1 en el índice j , entonces el conteo de inversión aumenta en 1 s en el rango [i, j].
- La cuenta de inversión máxima posible es A*(NA).
Siga los pasos a continuación para resolver el problema:
- Si X es mayor que A*(NA) , imprima -1 y luego regrese.
- Inicialice una array, digamos arr[] de tamaño N y complete los primeros índices A con 0 s y el resto con 1 s.
- Inicialice dos variables curr como A-1 y prev como N-1 para iterar sobre la array .
- Itere hasta que X sea mayor que 0 y curr , no sea menor que 0 , y realice los siguientes pasos:
- Si X es mayor o igual que prev-cur , haga lo siguiente:
- Intercambie los dos elementos en arr[prev] y arr[curr].
- Reste prev-cur de X .
- Decrementar prev y curr en 1.
- De lo contrario, haga lo siguiente:
- Intercambia los dos elementos arr[curr] y arr[cur+1] .
- Incremente curr en 1 y disminuya X en 1.
- Si X es mayor o igual que prev-cur , haga lo siguiente:
- Imprima la array arr.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct lexicographically
// smallest binary string of length N, having
// A 0s and X inversions
void binaryArrayInversions(int N, int A, int X)
{
// If X inversions are not possible
if (A * (N - A) < X) {
cout << "-1";
return;
}
// Initialize array and fill with 0
int Arr[N] = { 0 };
// Fill last N-A indices with 1
fill(Arr + A, Arr + N, 1);
// Stores the index of current 0
int cur = A - 1;
// Stores the index of current 1
int prev = N - 1;
// Iterate until X is greater than
// 0 and cur is greater than equal
// to 0
while (X && cur >= 0) {
// If X is greater than or
// equal to the prev-cur
if (X >= prev - cur) {
// Swap current 0 and current 1
swap(Arr[prev], Arr[cur]);
// Update X
X -= prev - cur;
// Decrement prev and cur by 1
prev--;
cur--;
}
// Otherwise
else {
// Swap current 0 with the next index
swap(Arr[cur], Arr[cur + 1]);
// Increment cur by 1
cur++;
// Decrement X by 1
X--;
}
}
// Print the array
for (auto u : Arr)
cout << u << " ";
}
// Driver code
int main()
{
// Input
int N = 5;
int A = 2;
int X = 1;
// Function call
binaryArrayInversions(N, A, X);
return 0;
}
Java
// Java program for the above approach
import java.util.Arrays;
class GFG{
// Function to construct lexicographically
// smallest binary string of length N, having
// A 0s and X inversions
static void binaryArrayInversions(int N, int A, int X)
{
// If X inversions are not possible
if (A * (N - A) < X)
{
System.out.println("-1");
return;
}
// Initialize array and fill with 0
int []Arr = new int[N];
// Fill last N-A indices with 1
Arrays.fill(Arr, 0);
for(int i = A; i < N; i++)
Arr[i] = 1;
// Stores the index of current 0
int cur = A - 1;
// Stores the index of current 1
int prev = N - 1;
// Iterate until X is greater than
// 0 and cur is greater than equal
// to 0
while (X != 0 && cur >= 0)
{
// If X is greater than or
// equal to the prev-cur
if (X >= prev - cur)
{
// Swap current 0 and current 1
int temp = Arr[prev];
Arr[prev] = Arr[cur];
Arr[cur] = temp;
// Update X
X -= prev - cur;
// Decrement prev and cur by 1
prev--;
cur--;
}
// Otherwise
else
{
// Swap current 0 with the next index
int temp = Arr[cur];
Arr[cur] = Arr[cur + 1];
Arr[cur + 1] = temp;
// Increment cur by 1
cur++;
// Decrement X by 1
X--;
}
}
// Print the array
for(int i = 0; i < Arr.length; i++)
System.out.print(Arr[i] + " ");
}
// Driver code
public static void main(String args[])
{
// Input
int N = 5;
int A = 2;
int X = 1;
// Function call
binaryArrayInversions(N, A, X);
}
}
// This code is contributed by gfgking
Python3
# Python3 program for the above approach
# Function to construct lexicographically
# smallest binary string of length N, having
# A 0s and X inversions
def binaryArrayInversions(N, A, X):
# If X inversions are not possible
if (A * (N - A) < X):
print("-1")
return
# Initialize array and fill with 0
Arr = [0]*N
for i in range(A,N):
Arr[i]=1
# Stores the index of current 0
cur = A - 1
# Stores the index of current 1
prev = N - 1
# Iterate until X is greater than
# 0 and cur is greater than equal
# to 0
while (X and cur >= 0):
# If X is greater than or
# equal to the prev-cur
if (X >= prev - cur):
# Swap current 0 and current 1
Arr[prev], Arr[cur] = Arr[cur],Arr[prev]
# Update X
X -= prev - cur
# Decrement prev and cur by 1
prev -= 1
cur -= 1
# Otherwise
else:
# Swap current 0 with the next index
Arr[cur], Arr[cur + 1] = Arr[cur + 1], Arr[cur]
# Increment cur by 1
cur += 1
# Decrement X by 1
X -= 1
# Print the array
for u in Arr:
print(u, end = " ")
# Driver code
if __name__ == '__main__':
# Input
N = 5
A = 2
X = 1
# Function call
binaryArrayInversions(N, A, X)
# This code is contributed by mohit kumar 29.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to construct lexicographically
// smallest binary string of length N, having
// A 0s and X inversions
static void binaryArrayInversions(int N, int A, int X)
{
// If X inversions are not possible
if (A * (N - A) < X) {
Console.Write("-1");
return;
}
// Initialize array and fill with 0
int []Arr = new int[N];
// Fill last N-A indices with 1
Array.Clear(Arr, 0, N);
for(int i=A;i<N;i++)
Arr[i] = 1;
// Stores the index of current 0
int cur = A - 1;
// Stores the index of current 1
int prev = N - 1;
// Iterate until X is greater than
// 0 and cur is greater than equal
// to 0
while (X!=0 && cur >= 0)
{
// If X is greater than or
// equal to the prev-cur
if (X >= prev - cur)
{
// Swap current 0 and current 1
int temp = Arr[prev];
Arr[prev] = Arr[cur];
Arr[cur] = temp;
// Update X
X -= prev - cur;
// Decrement prev and cur by 1
prev--;
cur--;
}
// Otherwise
else {
// Swap current 0 with the next index
int temp = Arr[cur];
Arr[cur] = Arr[cur + 1];
Arr[cur + 1] = temp;
// Increment cur by 1
cur++;
// Decrement X by 1
X--;
}
}
// Print the array
for(int i = 0; i < Arr.Length; i++)
Console.Write(Arr[i] +" ");
}
// Driver code
public static void Main()
{
// Input
int N = 5;
int A = 2;
int X = 1;
// Function call
binaryArrayInversions(N, A, X);
}
}
// This code is contributed by SURENDRA_GANGWAR.
Javascript
<script>
// JavaScript program for the above approach
// Function to construct lexicographically
// smallest binary string of length N, having
// A 0s and X inversions
function binaryArrayInversions(N, A, X) {
// If X inversions are not possible
if (A * (N - A) < X) {
document.write("-1");
return;
}
// Initialize array and fill with 0
let Arr = new Array(N).fill(0);
// Fill last N-A indices with 1
Arr.forEach((item, i) => {
if (i >= Arr.length - (N - A)) {
Arr[i] = 1
}
})
// Stores the index of current 0
let cur = A - 1;
// Stores the index of current 1
let prev = N - 1;
// Iterate until X is greater than
// 0 and cur is greater than equal
// to 0
while (X && cur >= 0) {
// If X is greater than or
// equal to the prev-cur
if (X >= prev - cur) {
// Swap current 0 and current 1
let temp = Arr[prev];
Arr[prev] = Arr[cur];
Arr[cur] = temp;
// Update X
X -= prev - cur;
// Decrement prev and cur by 1
prev--;
cur--;
}
// Otherwise
else {
// Swap current 0 with the next index
let temp = Arr[cur + 1];
Arr[cur + 1] = Arr[cur];
Arr[cur] = temp;
// Increment cur by 1
cur++;
// Decrement X by 1
X--;
}
}
// Print the array
document.write(Arr);
}
// Driver code
// Input
let N = 5;
let A = 2;
let X = 1;
// Function call
binaryArrayInversions(N, A, X);
</script>
Producción
0 1 0 1 1
Complejidad temporal: O(N)
Espacio auxiliar: O(1)