Dada una array arr[] de N elementos y Q consultas de la forma [X] . Para cada consulta, la tarea es encontrar el intervalo más grande [L, R] de la array tal que el elemento más grande en el intervalo sea arr[X] , tal que 1 ≤ L ≤ X ≤ R .
Nota: La array tiene una indexación basada en 1.
Ejemplos:
Entrada: N = 5, arr[] = {2, 1, 2, 3, 2}, Q = 3, query[] = {1, 2, 4} Salida: [1, 3] [2
,
2
]
[ 1, 5]
Explicación:
en la primera consulta, x = 1, entonces arr[x] = 2 y la respuesta es L = 1 y R = 3. aquí, podemos ver que max(arr[1], arr[2], arr[3]) = arr[x], que son los intervalos máximos.
En la segunda consulta, x = 2, entonces arr[x] = 1 y dado que es el elemento más pequeño de la array, el intervalo contiene solo un elemento, por lo que el rango es [2, 2].
En la tercera consulta, x = 4, por lo que arr[x] = 4, que es el elemento máximo de arr[], por lo que la respuesta es una array completa, L = 1 y R = N.Entrada: N = 4, arr[] = { 1, 2, 2, 4}, Q = 2, query[] = {1, 2} Salida: [1, 1] [1, 3] Explicación
:
En
la
primera
consulta , x = 1, entonces arr[x] = 1 y dado que es el elemento más pequeño de la array, el intervalo contiene solo un elemento, por lo que el rango es [1, 1].
En la segunda consulta, x = 2, entonces arr[x] = 2 y la respuesta es L = 1 y R = 3. aquí, podemos ver que max(arr[1], arr[2], arr[3]) = arr[x] = arr[2] = 2, que son los intervalos máximos.
Enfoque: la idea es precalcular el intervalo más grande para cada valor K en arr[] de 1 a N . A continuación se muestran los pasos:
- Para cada elemento K en arr[] , fije el índice del elemento K , luego encuentre cuánto podemos extender el intervalo hacia la izquierda y hacia la derecha.
- Disminuya el iterador izquierdo hasta que arr[left] ≤ K y, de manera similar, incremente el iterador derecho hasta que arr[right] ≤ K .
- El valor final de izquierda y derecha representa el índice inicial y final del intervalo, que se almacena en arrL[] y arrR[] respectivamente.
- Después de haber calculado previamente el rango de intervalo para cada valor. Luego, para cada consulta, necesitamos imprimir el rango de intervalo para arr[x] , es decir, arrL[arr[x]] y arrR[arr[x]] .
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 precompute the interval
// for each query
void utilLargestInterval(int arr[],
int arrL[],
int arrR[],
int N)
{
// For every values [1, N] find
// the longest intervals
for (int maxValue = 1;
maxValue <= N; maxValue++) {
int lastIndex = 0;
// Iterate the array arr[]
for (int i = 1; i <= N; i++) {
if (lastIndex >= i
|| arr[i] != maxValue)
continue;
int left = i, right = i;
// Shift the left pointers
while (left > 0
&& arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N
&& arr[right] <= maxValue)
right++;
left++, right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for (int j = left; j <= right; j++) {
if (arr[j] == maxValue) {
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
void largestInterval(
int arr[], int query[], int N, int Q)
{
// To store the L and R of X
int arrL[N + 1], arrR[N + 1];
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for each query
for (int i = 0; i < Q; i++) {
cout << "[" << arrL[query[i]]
<< ", " << arrR[query[i]]
<< "]\n";
}
}
// Driver Code
int main()
{
int N = 5, Q = 3;
// Given array arr[]
int arr[N + 1] = { 0, 2, 1, 2, 3, 2 };
// Given Queries
int query[Q] = { 1, 2, 4 };
// Function Call
largestInterval(arr, query, N, Q);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to precompute the interval
// for each query
static void utilLargestInterval(int arr[],
int arrL[],
int arrR[],
int N)
{
// For every values [1, N] find
// the longest intervals
for(int maxValue = 1;
maxValue <= N; maxValue++)
{
int lastIndex = 0;
// Iterate the array arr[]
for(int i = 1; i <= N; i++)
{
if (lastIndex >= i ||
arr[i] != maxValue)
continue;
int left = i, right = i;
// Shift the left pointers
while (left > 0 &&
arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N &&
arr[right] <= maxValue)
right++;
left++;
right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for(int j = left; j <= right; j++)
{
if (arr[j] == maxValue)
{
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
static void largestInterval(int arr[],
int query[],
int N, int Q)
{
// To store the L and R of X
int []arrL = new int[N + 1];
int []arrR = new int[N + 1];
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for
// each query
for(int i = 0; i < Q; i++)
{
System.out.print("[" + arrL[query[i]] +
", " + arrR[query[i]] + "]\n");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 5, Q = 3;
// Given array arr[]
int arr[] = { 0, 2, 1, 2, 3, 2 };
// Given queries
int query[] = { 1, 2, 4 };
// Function call
largestInterval(arr, query, N, Q);
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the above approach
# Function to precompute the interval
# for each query
def utilLargestInterval(arr, arrL, arrR, N):
# For every values [1, N] find
# the longest intervals
for maxValue in range(1, N + 1):
lastIndex = 0
# Iterate the array arr[]
for i in range(N + 1):
if (lastIndex >= i or
arr[i] != maxValue):
continue
left = i
right = i
# Shift the left pointers
while (left > 0 and
arr[left] <= maxValue):
left -= 1
# Shift the right pointers
while (right <= N and
arr[right] <= maxValue):
right += 1
left += 1
right -= 1
lastIndex = right
# Store the range of interval
# in arrL[] and arrR[].
for j in range(left, right + 1):
if (arr[j] == maxValue):
arrL[j] = left
arrR[j] = right
# Function to find the largest interval
# for each query in Q[]
def largestInterval(arr, query, N, Q):
# To store the L and R of X
arrL = [0 for i in range(N + 1)]
arrR = [0 for i in range(N + 1)]
# Function call
utilLargestInterval(arr, arrL, arrR, N);
# Iterate to find ranges for each query
for i in range(Q):
print('[' + str(arrL[query[i]]) +
', ' + str(arrR[query[i]]) + ']')
# Driver code
if __name__=="__main__":
N = 5
Q = 3
# Given array arr[]
arr = [ 0, 2, 1, 2, 3, 2 ]
# Given Queries
query = [ 1, 2, 4 ]
# Function call
largestInterval(arr, query, N, Q)
# This code is contributed by rutvik_56
C#
// C# program for the above approach
using System;
class GFG{
// Function to precompute the interval
// for each query
static void utilLargestInterval(int []arr,
int []arrL,
int []arrR,
int N)
{
// For every values [1, N] find
// the longest intervals
for(int maxValue = 1;
maxValue <= N; maxValue++)
{
int lastIndex = 0;
// Iterate the array []arr
for(int i = 1; i <= N; i++)
{
if (lastIndex >= i ||
arr[i] != maxValue)
continue;
int left = i, right = i;
// Shift the left pointers
while (left > 0 &&
arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N &&
arr[right] <= maxValue)
right++;
left++;
right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for(int j = left; j <= right; j++)
{
if (arr[j] == maxValue)
{
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
static void largestInterval(int []arr,
int []query,
int N, int Q)
{
// To store the L and R of X
int []arrL = new int[N + 1];
int []arrR = new int[N + 1];
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for
// each query
for(int i = 0; i < Q; i++)
{
Console.Write("[" + arrL[query[i]] +
", " + arrR[query[i]] + "]\n");
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 5, Q = 3;
// Given array []arr
int []arr = { 0, 2, 1, 2, 3, 2 };
// Given queries
int []query = { 1, 2, 4 };
// Function call
largestInterval(arr, query, N, Q);
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript program for the above approach
// Function to precompute the interval
// for each query
function utilLargestInterval(arr, arrL, arrR, N)
{
// For every values [1, N] find
// the longest intervals
for (var maxValue = 1;
maxValue <= N; maxValue++) {
var lastIndex = 0;
// Iterate the array arr[]
for (var i = 1; i <= N; i++) {
if (lastIndex >= i
|| arr[i] != maxValue)
continue;
var left = i, right = i;
// Shift the left pointers
while (left > 0
&& arr[left] <= maxValue)
left--;
// Shift the right pointers
while (right <= N
&& arr[right] <= maxValue)
right++;
left++, right--;
lastIndex = right;
// Store the range of interval
// in arrL[] and arrR[].
for (var j = left; j <= right; j++) {
if (arr[j] == maxValue) {
arrL[j] = left;
arrR[j] = right;
}
}
}
}
}
// Function to find the largest interval
// for each query in Q[]
function largestInterval( arr, query, N, Q)
{
// To store the L and R of X
var arrL = Array(N+1).fill(0),arrR = Array(N+1).fill(0);
// Function Call
utilLargestInterval(arr, arrL,
arrR, N);
// Iterate to find ranges for each query
for (var i = 0; i < Q; i++) {
document.write( "[" + arrL[query[i]]
+ ", " + arrR[query[i]]
+ "]<br>");
}
}
// Driver Code
var N = 5, Q = 3;
// Given array arr[]
var arr = [0, 2, 1, 2, 3, 2];
// Given Queries
var query = [1, 2, 4];
// Function Call
largestInterval(arr, query, N, Q);
// This code is contributed by itsok.
</script>
Producción:
[1, 3] [2, 2] [1, 5]
Complejidad de Tiempo: O(Q + N 2 )
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por grand_master y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA