Dada una array binaria mat[][] de tamaño M * N y Q consultas de la forma {pi, pj, qi, qj} , la tarea para cada consulta es contar el número de 0 s en la subarray de la celda ( pi, pj) a (qi, qj) .
Ejemplos:
Entrada: mat[][] = {{0, 1, 0, 1, 1, 1, 0}, {1, 0, 1, 1, 1, 0, 1}, {1, 1, 0, 0, 1, 1, 0}, {1, 1, 1, 1, 1, 0, 1}, {0, 0, 1, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 0, 1 }}, Q[] = {{0, 1, 3, 2}, {2, 2, 4, 5}, {4, 3, 5, 6}}
Salida: 3 4 2
Explicación:
La subarray de los índices (0, 1) a (3, 2) contiene 3 0s.
La subarray de los índices (2, 2) a (4, 5) contiene 4 0s.
La subarray de los índices (4, 3) a (5, 6) contiene 2 0s.Entrada: mat[][] = {{0, 1, 0}, {1, 0, 1}}, Q[] = {{0, 0, 2, 2}}
Salida: 3
Enfoque ingenuo: el enfoque más simple para resolver el problema es contar el número de 0 para cada consulta iterando la subarray e imprimir el recuento para cada consulta.
Complejidad temporal: O(M*N*Q)
Espacio auxiliar: O(1)
Enfoque eficiente: para optimizar el enfoque anterior, la idea es preprocesar la array dada mat[][] . Cree una array de conteo, digamos prefixCnt[M][N] , tal que prefix_cnt[i][j] almacene el conteo de 0 s en la subarray dentro de los índices (0, 0) a (i, j) . Siga los pasos a continuación para calcular prefixCnt[][] :
- Inicialice prefixCnt[i][j] con 1 si mat[i][j] es 0 . De lo contrario, inicialice con 0 .
- Encuentre la suma del prefijo para cada fila y columna y actualice la array en consecuencia.
Una vez que se calcula el prefijo de array Cnt [] [] , el recuento de 0 en cualquier subarray se puede evaluar en tiempo O (1) . A continuación se muestra la expresión para contar los 0 en cada subarray desde (pi, pj) hasta (qi, qj) se puede calcular mediante:
cnt = prefijoCnt[qi][qj] – prefijoCnt[pi-1][qj] – prefijoCnt[qi][pj – 1] + prefijoCnt[pi – 1][pj – 1]
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define M 6
#define N 7
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
void preCompute(int mat[M][N],
int prefixCnt[M][N])
{
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// Initialize prefixCnt[i][j]
// with 1 if mat[i][j] is 0
if (mat[i][j] == 0) {
prefixCnt[i][j] = 1;
}
// Otherwise, assign with 0
else {
prefixCnt[i][j] = 0;
}
}
}
// Calculate prefix sum for each row
for (int i = 0; i < M; i++)
for (int j = 1; j < N; j++)
prefixCnt[i][j] += prefixCnt[i][j - 1];
// Calculate prefix sum for each column
for (int i = 1; i < M; i++)
for (int j = 0; j < N; j++)
prefixCnt[i][j] += prefixCnt[i - 1][j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
int countQuery(int prefixCnt[M][N],
int pi, int pj,
int qi, int qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
int cnt = prefixCnt[qi][qj];
// Subtract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1][qj];
// Subtract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi][pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1][pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
void count0s(int mat[M][N], int Q[][4],
int sizeQ)
{
// Stores the prefix sum of each
// row and column
int prefixCnt[M][N];
// Compute matrix prefixCnt[][]
preCompute(mat, prefixCnt);
for (int i = 0; i < sizeQ; i++) {
// Function Call for each query
cout << countQuery(prefixCnt, Q[i][0],
Q[i][1], Q[i][2],
Q[i][3])
<< ' ';
}
}
// Driver Code
int main()
{
// Given matrix
int mat[M][N] = {
{ 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1 },
{ 1, 1, 0, 0, 1, 1, 0 },
{ 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1 }
};
int Q[][4] = { { 0, 1, 3, 2 },
{ 2, 2, 4, 5 },
{ 4, 3, 5, 6 } };
int sizeQ = sizeof(Q) / sizeof(Q[0]);
// Function Call
count0s(mat, Q, sizeQ);
return 0;
}
Java
// Java program for the above approach
class GFG{
final static int M = 6;
final static int N = 7;
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
static void preCompute(int mat[][], int prefixCnt[][])
{
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// Initialize prefixCnt[i][j]
// with 1 if mat[i][j] is 0
if (mat[i][j] == 0)
{
prefixCnt[i][j] = 1;
}
// Otherwise, assign with 0
else
{
prefixCnt[i][j] = 0;
}
}
}
// Calculate prefix sum for each row
for(int i = 0; i < M; i++)
for(int j = 1; j < N; j++)
prefixCnt[i][j] += prefixCnt[i][j - 1];
// Calculate prefix sum for each column
for(int i = 1; i < M; i++)
for(int j = 0; j < N; j++)
prefixCnt[i][j] += prefixCnt[i - 1][j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
static int countQuery(int prefixCnt[][],
int pi, int pj,
int qi, int qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
int cnt = prefixCnt[qi][qj];
// Subtract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1][qj];
// Subtract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi][pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added;
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1][pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
static void count0s(int mat[][], int Q[][],
int sizeQ)
{
// Stores the prefix sum of each
// row and column
int prefixCnt[][] = new int[M][N];
// Compute matrix prefixCnt[][]
preCompute(mat, prefixCnt);
for(int i = 0; i < sizeQ; i++)
{
// Function Call for each query
System.out.print(countQuery(prefixCnt, Q[i][0],
Q[i][1],
Q[i][2],
Q[i][3]) + " ");
}
}
// Driver Code
public static void main (String[] args)
{
// Given matrix
int mat[][] = { { 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1 },
{ 1, 1, 0, 0, 1, 1, 0 },
{ 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1 } };
int Q[][] = { { 0, 1, 3, 2 },
{ 2, 2, 4, 5 },
{ 4, 3, 5, 6 } };
int sizeQ = Q.length;
// Function Call
count0s(mat, Q, sizeQ);
}
}
// This code is contributed by AnkThon
Python3
# Python3 program for the above approach M = 6 N = 7 # Function to compute the matrix # prefixCnt[M][N] from mat[M][N] such # that prefixCnt[i][j] stores the # count of 0's from (0, 0) to (i, j) def preCompute(mat, prefixCnt): for i in range(M): for j in range(N): # Initialize prefixCnt[i][j] # with 1 if mat[i][j] is 0 if (mat[i][j] == 0): prefixCnt[i][j] = 1 # Otherwise, assign with 0 else: prefixCnt[i][j] = 0 # Calculate prefix sum for each row for i in range(M): for j in range(1, N): prefixCnt[i][j] += prefixCnt[i][j - 1] # Calculate prefix sum for each column for i in range(1, M): for j in range(N): prefixCnt[i][j] += prefixCnt[i - 1][j] return prefixCnt # Function to compute count of 0's # in submatrix from (pi, pj) to # (qi, qj) from prefixCnt[M][N] def countQuery(prefixCnt, pi, pj, qi, qj): # Initialize that count of 0's # in the sub-matrix within # indices (0, 0) to (qi, qj) cnt = prefixCnt[qi][qj] # Subtract count of 0's within # indices (0, 0) and (pi-1, qj) if (pi > 0): cnt -= prefixCnt[pi - 1][qj] # Subtract count of 0's within # indices (0, 0) and (qi, pj-1) if (pj > 0): cnt -= prefixCnt[qi][pj - 1] # Add prefixCnt[pi - 1][pj - 1] # because its value has been added # once but subtracted twice if (pi > 0 and pj > 0): cnt += prefixCnt[pi - 1][pj - 1] return cnt # Function to count the 0s in the # each given submatrix def count0s(mat, Q, sizeQ): # Stores the prefix sum of each # row and column prefixCnt = [[ 0 for i in range(N)] for i in range(M)] # Compute matrix prefixCnt[][] prefixCnt = preCompute(mat, prefixCnt) for i in range(sizeQ): # Function Call for each query print(countQuery(prefixCnt, Q[i][0], Q[i][1], Q[i][2], Q[i][3]), end=" ") # Driver Code if __name__ == '__main__': # Given matrix mat = [[ 0, 1, 0, 1, 1, 1, 0 ], [ 1, 0, 1, 1, 1, 0, 1 ], [ 1, 1, 0, 0, 1, 1, 0 ], [ 1, 1, 1, 1, 1, 0, 1 ], [ 0, 0, 1, 0, 1, 1, 1 ], [ 1, 1, 0, 1, 1, 0, 1 ] ] Q= [ [ 0, 1, 3, 2 ], [ 2, 2, 4, 5 ], [ 4, 3, 5, 6 ] ] sizeQ = len(Q) # Function Call count0s(mat, Q, sizeQ) # This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG
{
static int M = 6;
static int N = 7;
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
static void preCompute(int [,]mat, int [,]prefixCnt)
{
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// Initialize prefixCnt[i][j]
// with 1 if mat[i,j] is 0
if (mat[i, j] == 0)
{
prefixCnt[i, j] = 1;
}
// Otherwise, assign with 0
else
{
prefixCnt[i, j] = 0;
}
}
}
// Calculate prefix sum for each row
for(int i = 0; i < M; i++)
for(int j = 1; j < N; j++)
prefixCnt[i, j] += prefixCnt[i, j - 1];
// Calculate prefix sum for each column
for(int i = 1; i < M; i++)
for(int j = 0; j < N; j++)
prefixCnt[i, j] += prefixCnt[i - 1, j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
static int countQuery(int [,]prefixCnt,
int pi, int pj,
int qi, int qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
int cnt = prefixCnt[qi, qj];
// Subtract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1, qj];
// Subtract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi, pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added;
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1, pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
static void count0s(int [,]mat, int [,]Q,
int sizeQ)
{
// Stores the prefix sum of each
// row and column
int [,]prefixCnt = new int[M, N];
// Compute matrix prefixCnt[,]
preCompute(mat, prefixCnt);
for(int i = 0; i < sizeQ; i++)
{
// Function Call for each query
Console.Write(countQuery(prefixCnt, Q[i, 0],
Q[i, 1],
Q[i, 2],
Q[i, 3]) + " ");
}
}
// Driver Code
public static void Main(string[] args)
{
// Given matrix
int [,]mat = { { 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1 },
{ 1, 1, 0, 0, 1, 1, 0 },
{ 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1 } };
int [,]Q = { { 0, 1, 3, 2 },
{ 2, 2, 4, 5 },
{ 4, 3, 5, 6 } };
int sizeQ = Q.GetLength(0);
// Function Call
count0s(mat, Q, sizeQ);
}
}
// This code is contributed by AnkThon
Javascript
<script>
// javascript program of the above approach
let M = 6;
let N = 7;
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
function preCompute(mat, prefixCnt)
{
for(let i = 0; i < M; i++)
{
for(let j = 0; j < N; j++)
{
// Initialize prefixCnt[i][j]
// with 1 if mat[i][j] is 0
if (mat[i][j] == 0)
{
prefixCnt[i][j] = 1;
}
// Otherwise, assign with 0
else
{
prefixCnt[i][j] = 0;
}
}
}
// Calculate prefix sum for each row
for(let i = 0; i < M; i++)
for(let j = 1; j < N; j++)
prefixCnt[i][j] += prefixCnt[i][j - 1];
// Calculate prefix sum for each column
for(let i = 1; i < M; i++)
for(let j = 0; j < N; j++)
prefixCnt[i][j] += prefixCnt[i - 1][j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
function countQuery(prefixCnt,
pi, pj,
qi, qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
let cnt = prefixCnt[qi][qj];
// Subtract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1][qj];
// Subtract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi][pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added;
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1][pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
function count0s(mat, Q,
sizeQ)
{
// Stores the prefix sum of each
// row and column
let prefixCnt = new Array(M);
// Loop to create 2D array using 1D array
for (var i = 0; i < prefixCnt.length; i++) {
prefixCnt[i] = new Array(2);
}
// Compute matrix prefixCnt[][]
preCompute(mat, prefixCnt);
for(let i = 0; i < sizeQ; i++)
{
// Function Call for each query
document.write(countQuery(prefixCnt, Q[i][0],
Q[i][1],
Q[i][2],
Q[i][3]) + " ");
}
}
// Driver Code
// Given matrix
let mat = [[ 0, 1, 0, 1, 1, 1, 0 ],
[ 1, 0, 1, 1, 1, 0, 1 ],
[ 1, 1, 0, 0, 1, 1, 0 ],
[ 1, 1, 1, 1, 1, 0, 1 ],
[ 0, 0, 1, 0, 1, 1, 1 ],
[ 1, 1, 0, 1, 1, 0, 1 ]];
let Q = [[ 0, 1, 3, 2 ],
[ 2, 2, 4, 5 ],
[ 4, 3, 5, 6 ]];
let sizeQ = Q.length;
// Function Call
count0s(mat, Q, sizeQ);
// This code is contributed by target_2.
</script>
Producción:
3 4 2
Complejidad temporal: O(M*N + Q)
Espacio auxiliar: O(M*N)
Publicación traducida automáticamente
Artículo escrito por sunidhichandra27 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA