Requisito previo: árboles de segmentos
Dada una array binaria arr[] que consta solo de 0 y 1 y una array 2D Q[][] que consta de K consultas, la tarea es encontrar la distancia mínima entre dos 0 en el rango [L, R] de la array para cada consulta {L, R}.
Ejemplos:
Entrada: arr[] = {1, 0, 0, 1}, Q[][] = {{0, 2}}
Salida: 1
Explicación:
Claramente, en el rango [0, 2], el primer 0 se encuentra en índice 1 y último en el índice 2.
Distancia mínima = 2 – 1 = 1.Entrada: arr[] = {1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0}, Q[][] = {{3, 9}, { 10, 13}}
Salida: 2 3
Explicación:
En el rango [3, 9], la distancia mínima entre 0 es 2 (Índice 4 y 6).
En el rango [10, 13], la distancia mínima entre 0 es 3 (Índice 10 y 13).
Enfoque: La idea es usar un árbol de segmentos para resolver este problema:
- Cada Node en el árbol de segmentos tendrá el índice del 0 más a la izquierda así como el 0 más a la derecha y un número entero que contiene la distancia mínima entre los 0 en el subarreglo {L, R}.
- Sea min la distancia mínima entre dos ceros. Luego, el valor de min se puede encontrar después de formar el árbol de segmentos como:
min = mínimo (valor de min en el Node izquierdo, el valor de min en el Node derecho y la diferencia entre el índice más a la izquierda de 0 en el Node derecho y índice más a la derecha de 0 en el Node izquierdo). - Después de calcular y almacenar la distancia mínima para cada Node, todas las consultas se pueden responder en tiempo logarítmico.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the minimum
// distance between two elements
// with value 0 within a subarray (l, r)
#include <bits/stdc++.h>
using namespace std;
// Structure for each node
// in the segment tree
struct node {
int l0, r0;
int min0;
} seg[100001];
// A utility function for
// merging two nodes
node task(node l, node r)
{
node x;
x.l0 = (l.l0 != -1) ? l.l0 : r.l0;
x.r0 = (r.r0 != -1) ? r.r0 : l.r0;
x.min0 = min(l.min0, r.min0);
// If both the nodes are valid
if (l.r0 != -1 && r.l0 != -1)
// Computing the minimum distance to store
// in the segment tree
x.min0 = min(x.min0, r.l0 - l.r0);
return x;
}
// A recursive function that constructs
// Segment Tree for given string
void build(int qs, int qe, int ind, int arr[])
{
// If start is equal to end then
// insert the array element
if (qs == qe) {
if (arr[qs] == 0) {
seg[ind].l0 = seg[ind].r0 = qs;
seg[ind].min0 = INT_MAX;
}
else {
seg[ind].l0 = seg[ind].r0 = -1;
seg[ind].min0 = INT_MAX;
}
return;
}
int mid = (qs + qe) >> 1;
// Build the segment tree
// for range qs to mid
build(qs, mid, ind << 1, arr);
// Build the segment tree
// for range mid+1 to qe
build(mid + 1, qe, ind << 1 | 1, arr);
// Merge the two child nodes
// to obtain the parent node
seg[ind] = task(seg[ind << 1],
seg[ind << 1 | 1]);
}
// Query in a range qs to qe
node query(int qs, int qe, int ns, int ne, int ind)
{
node x;
x.l0 = x.r0 = -1;
x.min0 = INT_MAX;
// If the range lies in this segment
if (qs <= ns && qe >= ne)
return seg[ind];
// If the range is out of the bounds
// of this segment
if (ne < qs || ns > qe || ns > ne)
return x;
// Else query for the right and left
// child node of this subtree
// and merge them
int mid = (ns + ne) >> 1;
node l = query(qs, qe, ns, mid, ind << 1);
node r = query(qs, qe, mid + 1, ne, ind << 1 | 1);
x = task(l, r);
return x;
}
// Driver code
int main()
{
int arr[] = { 1, 1, 0, 1, 0, 1,
0, 1, 0, 1, 0, 1, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
// Build the segment tree
build(0, n - 1, 1, arr);
// Queries
int Q[][2] = { { 3, 9 }, { 10, 13 } };
for (int i = 0; i < 2; i++) {
// Finding the answer for every query
// and printing it
node ans = query(Q[i][0], Q[i][1],
0, n - 1, 1);
cout << ans.min0 << endl;
}
return 0;
}
Java
// Java program to find the minimum
// distance between two elements
// with value 0 within a subarray (l, r)
class GFG{
// Structure for each Node
// in the segment tree
static class Node
{
int l0, r0;
int min0;
};
static Node[] seg = new Node[100001];
// A utility function for
// merging two Nodes
static Node task(Node l, Node r)
{
Node x = new Node();
x.l0 = (l.l0 != -1) ? l.l0 : r.l0;
x.r0 = (r.r0 != -1) ? r.r0 : l.r0;
x.min0 = Math.min(l.min0, r.min0);
// If both the Nodes are valid
if (l.r0 != -1 && r.l0 != -1)
// Computing the minimum distance to store
// in the segment tree
x.min0 = Math.min(x.min0, r.l0 - l.r0);
return x;
}
// A recursive function that constructs
// Segment Tree for given string
static void build(int qs, int qe,
int ind, int arr[])
{
// If start is equal to end then
// insert the array element
if (qs == qe)
{
if (arr[qs] == 0)
{
seg[ind].l0 = seg[ind].r0 = qs;
seg[ind].min0 = Integer.MAX_VALUE;
}
else
{
seg[ind].l0 = seg[ind].r0 = -1;
seg[ind].min0 = Integer.MAX_VALUE;
}
return;
}
int mid = (qs + qe) >> 1;
// Build the segment tree
// for range qs to mid
build(qs, mid, ind << 1, arr);
// Build the segment tree
// for range mid+1 to qe
build(mid + 1, qe, ind << 1 | 1, arr);
// Merge the two child Nodes
// to obtain the parent Node
seg[ind] = task(seg[ind << 1],
seg[ind << 1 | 1]);
}
// Query in a range qs to qe
static Node query(int qs, int qe, int ns,
int ne, int ind)
{
Node x = new Node();
x.l0 = x.r0 = -1;
x.min0 = Integer.MAX_VALUE;
// If the range lies in this segment
if (qs <= ns && qe >= ne)
return seg[ind];
// If the range is out of the bounds
// of this segment
if (ne < qs || ns > qe || ns > ne)
return x;
// Else query for the right and left
// child Node of this subtree
// and merge them
int mid = (ns + ne) >> 1;
Node l = query(qs, qe, ns, mid,
ind << 1);
Node r = query(qs, qe, mid + 1,
ne, ind << 1 | 1);
x = task(l, r);
return x;
}
// Driver code
public static void main(String[] args)
{
for(int i = 0; i < 100001; i++)
{
seg[i] = new Node();
}
int arr[] = { 1, 1, 0, 1, 0, 1, 0,
1, 0, 1, 0, 1, 1, 0 };
int n = arr.length;
// Build the segment tree
build(0, n - 1, 1, arr);
// Queries
int[][] Q = { { 3, 9 }, { 10, 13 } };
for(int i = 0; i < 2; i++)
{
// Finding the answer for every query
// and printing it
Node ans = query(Q[i][0], Q[i][1],
0, n - 1, 1);
System.out.println(ans.min0);
}
}
}
// This code is contributed by sanjeev2552
Python3
# Python3 program to find the minimum # distance between two elements with # value 0 within a subarray (l, r) import sys # Structure for each node # in the segment tree class node(): def __init__(self): self.l0 = 0 self.r0 = 0 min0 = 0 seg = [node() for i in range(100001)] # A utility function for # merging two nodes def task(l, r): x = node() x.l0 = l.l0 if (l.l0 != -1) else r.l0 x.r0 = r.r0 if (r.r0 != -1) else l.r0 x.min0 = min(l.min0, r.min0) # If both the nodes are valid if (l.r0 != -1 and r.l0 != -1): # Computing the minimum distance to # store in the segment tree x.min0 = min(x.min0, r.l0 - l.r0) return x # A recursive function that constructs # Segment Tree for given string def build(qs, qe, ind, arr): # If start is equal to end then # insert the array element if (qs == qe): if (arr[qs] == 0): seg[ind].l0 = seg[ind].r0 = qs seg[ind].min0 = sys.maxsize else: seg[ind].l0 = seg[ind].r0 = -1 seg[ind].min0 = sys.maxsize return mid = (qs + qe) >> 1 # Build the segment tree # for range qs to mid build(qs, mid, ind << 1, arr) # Build the segment tree # for range mid+1 to qe build(mid + 1, qe, ind << 1 | 1, arr) # Merge the two child nodes # to obtain the parent node seg[ind] = task(seg[ind << 1], seg[ind << 1 | 1]) # Query in a range qs to qe def query(qs, qe, ns, ne, ind): x = node() x.l0 = x.r0 = -1 x.min0 = sys.maxsize # If the range lies in this segment if (qs <= ns and qe >= ne): return seg[ind] # If the range is out of the bounds # of this segment if (ne < qs or ns > qe or ns > ne): return x # Else query for the right and left # child node of this subtree # and merge them mid = (ns + ne) >> 1 l = query(qs, qe, ns, mid, ind << 1) r = query(qs, qe, mid + 1, ne, ind << 1 | 1) x = task(l, r) return x # Driver code if __name__=="__main__": arr = [ 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0 ] n = len(arr) # Build the segment tree build(0, n - 1, 1, arr) # Queries Q = [ [ 3, 9 ], [ 10, 13 ] ] for i in range(2): # Finding the answer for every query # and printing it ans = query(Q[i][0], Q[i][1], 0, n - 1, 1) print(ans.min0) # This code is contributed by rutvik_56
Producción:
2 3
Tema relacionado: Árbol de segmentos
Publicación traducida automáticamente
Artículo escrito por muskan_garg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA