循环字符串中从 i 到 j 的最小移动
给定一个循环字符串str和两个整数i和j ，任务是计算从str[i]移动到str[j]所需的最小步数。移动是到达字符串中的任何相邻字符，并且仅当str[start] != start[end]时才计算移动，其中start是移动的起始索引， end是结束（在左侧或左侧相邻）右）索引。由于给定的字符串是圆形的，因此str[0]和str[n – 1]彼此相邻。
例子：
Input: str = “SSNSS”, i = 0, j = 3
Output: 0
From left to right : S -> S -> N -> S
From right to left : S -> S -> S
Input: str = “geeksforgeeks”, i = 0, j = 3
Output: 2
方法：
- 从索引i开始朝正确的方向移动直到索引j并且对于访问的每个字符，如果当前字符不等于前一个字符，则递增steps1 = steps1 + 1 。
- 类似地，从i开始向左移动直到索引0并且对于访问的每个字符，如果当前字符不等于前一个字符，则递增steps2 = steps2 + 1 。访问索引0后，开始从索引n – 1遍历到j ，如果str[0] != str[n – 1]则递增step2 。
- 最后打印min(step1, step2) 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count of steps
// required to move from i to j
int getSteps(string str, int i, int j, int n)
{
// Starting from i + 1
int k = i + 1;
// Count of steps
int steps = 0;
// Current character
char ch = str[i];
while (k <= j) {
// If current character is different from previous
if (str[k] != ch) {
// Increment steps
steps++;
// Update current character
ch = str[k];
}
k++;
}
// Return total steps
return steps;
}
// Function to return the minimum number of steps
// required to reach j from i
int getMinSteps(string str, int i, int j, int n)
{
// Swap the values so that i <= j
if (j < i) {
int temp = i;
i = j;
j = temp;
}
// Steps to go from i to j (left to right)
int stepsToRight = getSteps(str, i, j, n);
// While going from i to j (right to left)
// First go from i to 0
// then from (n - 1) to j
int stepsToLeft = getSteps(str, 0, i, n)
+ getSteps(str, j, n - 1, n);
// If first and last character is different
// then it'll add a step to stepsToLeft
if (str[0] != str[n - 1])
stepsToLeft++;
// Return the minimum of two paths
return min(stepsToLeft, stepsToRight);
}
// Driver code
int main()
{
string str = "SSNSS";
int n = str.length();
int i = 0, j = 3;
cout << getMinSteps(str, i, j, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return the count of steps
// required to move from i to j
static int getSteps(String str, int i, int j, int n)
{
// Starting from i + 1
int k = i + 1;
// Count of steps
int steps = 0;
// Current character
char ch = str.charAt(i);
while (k <= j)
{
// If current character is different from previous
if (str.charAt(k) != ch)
{
// Increment steps
steps++;
// Update current character
ch = str.charAt(k);
}
k++;
}
// Return total steps
return steps;
}
// Function to return the minimum number of steps
// required to reach j from i
static int getMinSteps(String str, int i, int j, int n)
{
// Swap the values so that i <= j
if (j < i)
{
int temp = i;
i = j;
j = temp;
}
// Steps to go from i to j (left to right)
int stepsToRight = getSteps(str, i, j, n);
// While going from i to j (right to left)
// First go from i to 0
// then from (n - 1) to j
int stepsToLeft = getSteps(str, 0, i, n)
+ getSteps(str, j, n - 1, n);
// If first and last character is different
// then it'll add a step to stepsToLeft
if (str.charAt(0) != str.charAt(n - 1))
stepsToLeft++;
// Return the minimum of two paths
return Math.min(stepsToLeft, stepsToRight);
}
// Driver code
public static void main(String []args)
{
String str = "SSNSS";
int n = str.length();
int i = 0, j = 3;
System.out.println(getMinSteps(str, i, j, n));
}
}
// This code is contributed by ihritikPython3
# Python3 implementation of the approach
# Function to return the count of steps
# required to move from i to j
def getSteps( str, i, j, n) :
# Starting from i + 1
k = i + 1
# Count of steps
steps = 0
# Current character
ch = str[i]
while (k <= j):
# If current character is different from previous
if (str[k] != ch):
# Increment steps
steps = steps + 1
# Update current character
ch = str[k]
k = k + 1
# Return total steps
return steps
# Function to return the minimum number of steps
# required to reach j from i
def getMinSteps( str, i, j, n):
# Swap the values so that i <= j
if (j < i):
temp = i
i = j
j = temp
# Steps to go from i to j (left to right)
stepsToRight = getSteps(str, i, j, n)
# While going from i to j (right to left)
# First go from i to 0
# then from (n - 1) to j
stepsToLeft = getSteps(str, 0, i, n) + getSteps(str, j, n - 1, n)
# If first and last character is different
# then it'll add a step to stepsToLeft
if (str[0] != str[n - 1]):
stepsToLeft = stepsToLeft + 1
# Return the minimum of two paths
return min(stepsToLeft, stepsToRight)
# Driver code
str = "SSNSS"
n = len(str)
i = 0
j = 3
print(getMinSteps(str, i, j, n))
# This code is contributed by ihritikC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count of steps
// required to move from i to j
static int getSteps(string str, int i, int j, int n)
{
// Starting from i + 1
int k = i + 1;
// Count of steps
int steps = 0;
// Current character
char ch = str[i];
while (k <= j)
{
// If current character is different from previous
if (str[k] != ch)
{
// Increment steps
steps++;
// Update current character
ch = str[k];
}
k++;
}
// Return total steps
return steps;
}
// Function to return the minimum number of steps
// required to reach j from i
static int getMinSteps(string str, int i, int j, int n)
{
// Swap the values so that i <= j
if (j < i)
{
int temp = i;
i = j;
j = temp;
}
// Steps to go from i to j (left to right)
int stepsToRight = getSteps(str, i, j, n);
// While going from i to j (right to left)
// First go from i to 0
// then from (n - 1) to j
int stepsToLeft = getSteps(str, 0, i, n)
+ getSteps(str, j, n - 1, n);
// If first and last character is different
// then it'll add a step to stepsToLeft
if (str[0] != str[n - 1])
stepsToLeft++;
// Return the minimum of two paths
return Math.Min(stepsToLeft, stepsToRight);
}
// Driver code
public static void Main()
{
string str = "SSNSS";
int n = str.Length;
int i = 0, j = 3;
Console.WriteLine(getMinSteps(str, i, j, n));
}
}
// This code is contributed by ihritikPHP
输出：
0