Dada una array de tamaño NXN de enteros. La tarea es encontrar el número de caminos decrecientes en la array. Puede comenzar desde cualquier celda y desde la celda (i, j), puede moverse a (i + 1, j), (i – 1, j), (i, j + 1) y (i , j – 1) celda.
Ejemplos:
Input : m[][] = { { 1, 2 },
{ 1, 3 } }
Output : 8
Explanation : Decreasing paths are { 1 }, { 1 }, { 2 }, { 3 },
{ 2, 1 }, { 3, 1 }, { 3, 2 }, { 3, 2, 1 }
Input : m[][] = { { 1, 2, 3 },
{ 1, 3, 4 },
{ 1, 5, 6 } }
Output : 41
La idea para solucionar este problema es utilizar Programación Dinámica. Declare una array dp[][] , donde dp[i][j] almacena el número de rutas decrecientes que se pueden formar a partir de la celda (i, j) . Entonces, definiremos una función recursiva para evaluar el número de caminos decrecientes con parámetros, digamos i, j, el número de fila y el número de columna de la celda actual. Realice todos los movimientos posibles desde la celda (i,j) y lleve la cuenta del número total de caminos. Primero, verificaremos en la función que el número de caminos decrecientes para la posición de entrada (i, j) ya está calculado o no. En caso afirmativo, devuelva el valor dp[i][j]de lo contrario, encuentre el número de secuencias decrecientes permitidas en cuatro direcciones y devuelva el valor. Mientras tanto, también almacenaremos el número de decrecientes para las celdas intermedias. Dado que DP[i][j] almacena el número de caminos decrecientes para cada celda, la suma de todas las celdas de DP[][] responderá al conteo de caminos decrecientes en la array completa.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to count number
// of decreasing path in a matrix
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
// Function that returns the number of
// decreasing paths from a cell(i, j)
int CountDecreasingPathsCell(int mat[MAX][MAX], int dp[MAX][MAX],
int n, int x, int y)
{
// checking if already calculated
if (dp[x][y] != -1)
return dp[x][y];
// all possible paths
int delta[4][2] = { { 0, 1 }, { 1, 0 }, { -1, 0 }, { 0, -1 } };
int newx, newy;
// counts the total number of paths
int ans = 1;
// In all four allowed direction.
for (int i = 0; i < 4; i++) {
// new co-ordinates
newx = x + delta[i][0];
newy = y + delta[i][1];
// Checking if not going out of matrix and next
// cell value is less than current cell value.
if (newx >= 0 && newx < n && newy >= 0
&& newy < n && mat[newx][newy] < mat[x][y]) {
ans += CountDecreasingPathsCell(mat, dp, n, newx, newy);
}
}
// function that returns the answer
return dp[x][y] = ans;
}
// Function that counts the total
// decreasing path in the matrix
int countDecreasingPathsMatrix(int n,
int mat[MAX][MAX])
{
int dp[MAX][MAX];
// Initialising dp[][] to -1.
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
dp[i][j] = -1;
int sum = 0;
// Calculating number of decreasing path from each cell.
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
sum += CountDecreasingPathsCell(mat, dp, n, i, j);
return sum;
}
// Driver Code
int main()
{
int n = 2;
int mat[MAX][MAX] = { { 1, 2 }, { 1, 3 } };
// function call that returns the
// count of decreasing paths in a matrix
cout << countDecreasingPathsMatrix(n, mat)
<< endl;
return 0;
}
Python3
# Python3 program to count number # of decreasing path in a matrix MAX = 100 # Function that returns the number of # decreasing paths from a cell(i, j) def CountDecreasingPathsCell(mat, dp, n, x, y): # checking if already calculated if (dp[x][y] != -1): return dp[x][y] # all possible paths delta = [[0, 1], [1, 0], [-1, 0], [0, -1]] newx, newy = 0, 0 # counts the total number of paths ans = 1 # In all four allowed direction. for i in range(4): # new co-ordinates newx = x + delta[i][0] newy = y + delta[i][1] # Checking if not going out of matrix and next # cell value is less than current cell value. if (newx >= 0 and newx < n and newy >= 0 and newy < n and mat[newx][newy] < mat[x][y]): ans += CountDecreasingPathsCell(mat, dp, n, newx, newy) # function that returns the answer dp[x][y] = ans return dp[x][y] # Function that counts the total # decreasing path in the matrix def countDecreasingPathsMatrix(n,mat): dp = [] # Initialising dp[][] to -1. for i in range(n): l = [] for j in range(n): l.append(-1) dp.append(l) sum = 0 # Calculating number of decreasing # path from each cell. for i in range(n): for j in range(n): sum += CountDecreasingPathsCell(mat, dp, n, i, j) return sum # Driver Code n = 2 mat = [[1, 2], [1, 3]] # function call that returns the # count of decreasing paths in a matrix print(countDecreasingPathsMatrix(n, mat)) # This code is contributed by SHUBHAMSINGH10
C#
// C# program to count number
// of decreasing path in a matrix
using System;
class GFG
{
// Function that returns
// the number of decreasing
// paths from a cell(i, j)
public static int CountDecreasingPathsCell(int[,] mat, int[,] dp,
int n, int x, int y)
{
// checking if already
// calculated
if (dp[x, y] != -1)
return dp[x, y];
// all possible paths
int[,] delta = {{0, 1}, {1, 0},
{-1, 0},{0, -1}};
int newx, newy;
// counts the total
// number of paths
int ans = 1;
// In all four
// allowed direction.
for (int i = 0; i < 4; i++)
{
// new co-ordinates
newx = x + delta[i,0];
newy = y + delta[i,1];
// Checking if not going out
// of matrix and next cell
// value is less than current
// cell value.
if (newx >= 0 && newx < n &&
newy >= 0 && newy < n &&
mat[newx,newy] < mat[x,y])
{
ans += CountDecreasingPathsCell(mat, dp, n,
newx, newy);
}
}
// function that
// returns the answer
return dp[x,y] = ans;
}
// Function that counts the total
// decreasing path in the matrix
public static int countDecreasingPathsMatrix(int n,
int[,] mat)
{
int[,] dp = new int[n, n];
// Initialising dp[][] to -1.
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
dp[i, j] = -1;
int sum = 0;
// Calculating number of
// decreasing path from each cell.
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
sum += CountDecreasingPathsCell(mat, dp,
n, i, j);
return sum;
}
// Driver code
static public void Main ()
{
int n = 2;
int[,] mat= {{1, 2},
{1, 3}};
// function call that returns the
// count of decreasing paths in a matrix
Console.WriteLine(countDecreasingPathsMatrix(n, mat));
}
}
// This code is contributed by vij.
Javascript
<script>
// Javascript program to count number
// of decreasing path in a matrix
var MAX = 100
// Function that returns the number of
// decreasing paths from a cell(i, j)
function CountDecreasingPathsCell(mat, dp, n, x, y)
{
// checking if already calculated
if (dp[x][y] != -1)
return dp[x][y];
// all possible paths
var delta = [ [ 0, 1 ], [ 1, 0 ], [ -1, 0 ],
[ 0, -1 ] ];
var newx, newy;
// counts the total number of paths
var ans = 1;
// In all four allowed direction.
for (var i = 0; i < 4; i++) {
// new co-ordinates
newx = x + delta[i][0];
newy = y + delta[i][1];
// Checking if not going out of matrix and next
// cell value is less than current cell value.
if (newx >= 0 && newx < n && newy >= 0
&& newy < n && mat[newx][newy] < mat[x][y])
{
ans += CountDecreasingPathsCell
(mat, dp, n, newx, newy);
}
}
// function that returns the answer
dp[x][y] = ans;
return ans;
}
// Function that counts the total
// decreasing path in the matrix
function countDecreasingPathsMatrix(n, mat)
{
var dp = Array.from(Array(MAX),
()=> Array(MAX));
// Initialising dp[][] to -1.
for (var i = 0; i < n; i++)
for (var j = 0; j < n; j++)
dp[i][j] = -1;
var sum = 0;
// Calculating number of decreasing
// path from each cell.
for (var i = 0; i < n; i++)
for (var j = 0; j < n; j++)
sum += CountDecreasingPathsCell
(mat, dp, n, i, j);
return sum;
}
// Driver Code
var n = 2;
var mat = [ [ 1, 2 ], [ 1, 3 ] ];
// function call that returns the
// count of decreasing paths in a matrix
document.write( countDecreasingPathsMatrix(n, mat));
</script>
Producción:
8
Complejidad de Tiempo : O(N 2 )
Espacio Auxiliar : O(N 2 )