给定2D平面上的坐标(x，y) 。我们必须从原点即(0，0)的当前位置到达(x，y)。在每个步骤中，我们都可以在平面上垂直或水平移动。在每个步骤中水平移动时，我们都写为“ H”，在每个步骤中垂直移动时，我们都写为“ V”。因此，可能有许多包含’H’和’V’的字符串，它们表示从(0，0)到(x，y)的路径。任务是在所有可能的字符串找到按字典顺序排列的第K个最小字符串。
例子：
Input: x = 2, y = 2, k = 2
Output: HVVH
Explanation: There are 6 ways to reach (2, 2) from (0, 0). The possible list of strings in lexicographically sorted order: [“HHVV”, “HVHV”, “HVVH”, “VHHV”, “VHVH”, “VVHH”]. Hence, the lexicographically 2nd smallest string is HVHV.
Input : x = 2, y = 2, k = 3
Output : VHHV
先决条件：从起点到达终点的方法
方法：想法是使用递归来解决问题。从原点到达(x，y)的方式为x + y C x 。
现在观察，从(1，0)到达(x，y)的方式将是(x + y – 1，x – 1)，因为我们已经在水平方向上进行了一步，因此从中减去1 X。同样，从(0，1)到达(x，y)的方式将是(x + y – 1，y – 1)，因为我们已经在垂直方向上迈出了一步，所以从y减去1 。由于“ H”在字典上小于“ V”，因此在所有字符串中，起始字符串的开头将包含“ H”，即，初始运动将为“水平”。
因此，如果K <= x + y – 1 C x – 1 ，我们将“ H”作为第一步，否则将“ V”作为第一步，并求解从(1 ，0)将为K = K – x + y – 1 C x – 1 。
以下是此方法的实现：
C++
// CPP Program to find Lexicographically Kth
// smallest way to reach given coordinate from origin
#include
using namespace std;
// Return (a+b)!/a!b!
int factorial(int a, int b)
{
int res = 1;
// finding (a+b)!
for (int i = 1; i <= (a + b); i++)
res = res * i;
// finding (a+b)!/a!
for (int i = 1; i <= a; i++)
res = res / i;
// finding (a+b)!/b!
for (int i = 1; i <= b; i++)
res = res / i;
return res;
}
// Return the Kth smallest way to reach given coordinate from origin
void Ksmallest(int x, int y, int k)
{
// if at origin
if (x == 0 && y == 0)
return;
// if on y-axis
else if (x == 0) {
// decrement y.
y--;
// Move vertical
cout << "V";
// recursive call to take next step.
Ksmallest(x, y, k);
}
// If on x-axis
else if (y == 0) {
// decrement x.
x--;
// Move horizontal.
cout << "H";
// recursive call to take next step.
Ksmallest(x, y, k);
}
else {
// If x + y C x is greater than K
if (factorial(x - 1, y) > k) {
// Move Horizontal
cout << "H";
// recursive call to take next step.
Ksmallest(x - 1, y, k);
}
else {
// Move vertical
cout << "V";
// recursive call to take next step.
Ksmallest(x, y - 1, k - factorial(x - 1, y));
}
}
}
// Driven Program
int main()
{
int x = 2, y = 2, k = 2;
Ksmallest(x, y, k);
return 0;
} Java
// Java Program to find
// Lexicographically Kth
// smallest way to reach
// given coordinate from origin
import java.io.*;
class GFG
{
// Return (a+b)!/a!b!
static int factorial(int a,
int b)
{
int res = 1;
// finding (a+b)!
for (int i = 1;
i <= (a + b); i++)
res = res * i;
// finding (a+b)!/a!
for (int i = 1; i <= a; i++)
res = res / i;
// finding (a+b)!/b!
for (int i = 1; i <= b; i++)
res = res / i;
return res;
}
// Return the Kth smallest
// way to reach given
// coordinate from origin
static void Ksmallest(int x,
int y, int k)
{
// if at origin
if (x == 0 && y == 0)
return;
// if on y-axis
else if (x == 0)
{
// decrement y.
y--;
// Move vertical
System.out.print("V");
// recursive call to
// take next step.
Ksmallest(x, y, k);
}
// If on x-axis
else if (y == 0)
{
// decrement x.
x--;
// Move horizontal.
System.out.print("H");
// recursive call to
// take next step.
Ksmallest(x, y, k);
}
else
{
// If x + y C x is
// greater than K
if (factorial(x - 1, y) > k)
{
// Move Horizontal
System.out.print( "H");
// recursive call to
// take next step.
Ksmallest(x - 1, y, k);
}
else
{
// Move vertical
System.out.print("V");
// recursive call to
// take next step.
Ksmallest(x, y - 1, k -
factorial(x - 1, y));
}
}
}
// Driver Code
public static void main (String[] args)
{
int x = 2, y = 2, k = 2;
Ksmallest(x, y, k);
}
}
// This code is contributed
// by anuj_67.Python3
# Python3 Program to find Lexicographically Kth
# smallest way to reach given coordinate from origin
# Return (a+b)!/a!b!
def factorial(a, b):
res = 1
# finding (a+b)!
for i in range(1, a + b + 1):
res = res * i
# finding (a+b)!/a!
for i in range(1, a + 1):
res = res // i
# finding (a+b)!/b!
for i in range(1, b + 1):
res = res // i
return res
# Return the Kth smallest way to reach
# given coordinate from origin
def Ksmallest(x, y, k):
# if at origin
if x == 0 and y == 0:
return
# if on y-axis
elif x == 0:
# decrement y.
y -= 1
# Move vertical
print("V", end = "")
# recursive call to take next step.
Ksmallest(x, y, k)
# If on x-axis
elif y == 0:
# decrement x.
x -= 1
# Move horizontal.
print("H", end = "")
# recursive call to take next step.
Ksmallest(x, y, k)
else:
# If x + y C x is greater than K
if factorial(x - 1, y) > k:
# Move Horizontal
print("H", end = "")
# recursive call to take next step.
Ksmallest(x - 1, y, k)
else:
# Move vertical
print("V", end = "")
# recursive call to take next step.
Ksmallest(x, y - 1, k - factorial(x - 1, y))
# Driver Code
if __name__ == "__main__":
x, y, k = 2, 2, 2
Ksmallest(x, y, k)
# This code is contributed by Rituraj JainC#
// C# Program to find
// Lexicographically Kth
// smallest way to reach
// given coordinate from origin
using System;
class GFG
{
// Return (a+b)!/a!b!
static int factorial(int a,
int b)
{
int res = 1;
// finding (a+b)!
for (int i = 1;
i <= (a + b); i++)
res = res * i;
// finding (a+b)!/a!
for (int i = 1; i <= a; i++)
res = res / i;
// finding (a+b)!/b!
for (int i = 1; i <= b; i++)
res = res / i;
return res;
}
// Return the Kth smallest
// way to reach given
// coordinate from origin
static void Ksmallest(int x,
int y, int k)
{
// if at origin
if (x == 0 && y == 0)
return;
// if on y-axis
else if (x == 0)
{
// decrement y.
y--;
// Move vertical
Console.Write("V");
// recursive call to
// take next step.
Ksmallest(x, y, k);
}
// If on x-axis
else if (y == 0)
{
// decrement x.
x--;
// Move horizontal.
Console.Write("H");
// recursive call to
// take next step.
Ksmallest(x, y, k);
}
else
{
// If x + y C x is
// greater than K
if (factorial(x - 1, y) > k)
{
// Move Horizontal
Console.Write( "H");
// recursive call to
// take next step.
Ksmallest(x - 1, y, k);
}
else
{
// Move vertical
Console.Write("V");
// recursive call to
// take next step.
Ksmallest(x, y - 1, k -
factorial(x - 1, y));
}
}
}
// Driver Code
public static void Main (String[] args)
{
int x = 2, y = 2, k = 2;
Ksmallest(x, y, k);
}
}
// This code is contributed by 29AjayKumarPHP
$k)
{
// Move Horizontal
echo("H");
// recursive call to
// take next step.
Ksmallest($x - 1, $y, $k);
}
else
{
// Move vertical
echo("V");
// recursive call to
// take next step.
Ksmallest($x, $y - 1, $k -
factorial($x - 1, $y));
}
}
}
// Driver Code
$x = 2; $y = 2;$k = 2;
Ksmallest($x, $y, $k);
// This code is contributed
// by Code_Mech.
?>输出
HVVH