给定两个整数“ A”和“ B”以及整数“ k”的序列。任务是通过以下方式修改序列A中的任何一个元素来检查是否可以使两个序列相等：
我们可以将[-k，k]范围内的任何数字添加到A的任何元素。此操作只能执行一次。如果可以，请打印是，否则，请打印否。
例子：
Input: K = 2, A[] = {1, 2, 3}, B[] = {3, 2, 1}
Output: Yes
0 can be added to any element and both the sequences will be equal.
Input: K = 4, A[] = {1, 5}, B[] = {1, 1}
Output: Yes
-4 can be added to 5 then the sequence A becomes {1, 1} which is equal to the sequence B.
方法：请注意，要使两个序列仅移动一遍就必须使两个序列中只有一个不匹配元素，并且它们之间的绝对差必须小于或等于’k’。
- 对两个数组进行排序，然后查找不匹配的元素。
- 如果存在多个不匹配元素，则打印“否”
- 否则，找到元素之间的绝对差异。
- 如果差异<= k，则打印“是”，否则打印“否”。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to check if both
// sequences can be made equal
static bool check(int n, int k,
int *a, int *b)
{
// Sorting both the arrays
sort(a,a+n);
sort(b,b+n);
// Flag to tell if there are
// more than one mismatch
bool fl = false;
// To stores the index
// of mismatched element
int ind = -1;
for (int i = 0; i < n; i++)
{
if (a[i] != b[i])
{
// If there is more than one
// mismatch then return False
if (fl == true)
{
return false;
}
fl = true;
ind = i;
}
}
// If there is no mismatch or the
// difference between the
// mismatching elements is <= k
// then return true
if (ind == -1 | abs(a[ind] - b[ind]) <= k)
{
return true;
}
return false;
}
// Driver code
int main()
{
int n = 2, k = 4;
int a[] = {1, 5};
int b[] = {1, 1};
if (check(n, k, a, b))
{
printf("Yes");
}
else
{
printf("No");
}
return 0;
}
// This code is contributed by mits Java
// Java implementation of the above approach
import java.util.Arrays;
class GFG
{
// Function to check if both
// sequences can be made equal
static boolean check(int n, int k,
int[] a, int[] b)
{
// Sorting both the arrays
Arrays.sort(a);
Arrays.sort(b);
// Flag to tell if there are
// more than one mismatch
boolean fl = false;
// To stores the index
// of mismatched element
int ind = -1;
for (int i = 0; i < n; i++)
{
if (a[i] != b[i])
{
// If there is more than one
// mismatch then return False
if (fl == true)
{
return false;
}
fl = true;
ind = i;
}
}
// If there is no mismatch or the
// difference between the
// mismatching elements is <= k
// then return true
if (ind == -1 | Math.abs(a[ind] - b[ind]) <= k)
{
return true;
}
return false;
}
// Driver code
public static void main(String[] args)
{
int n = 2, k = 4;
int[] a = {1, 5};
int b[] = {1, 1};
if (check(n, k, a, b))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by 29AjayKumarPython 3
# Python implementation of the above approach
# Function to check if both
# sequences can be made equal
def check(n, k, a, b):
# Sorting both the arrays
a.sort()
b.sort()
# Flag to tell if there are
# more than one mismatch
fl = False
# To stores the index
# of mismatched element
ind = -1
for i in range(n):
if(a[i] != b[i]):
# If there is more than one
# mismatch then return False
if(fl == True):
return False
fl = True
ind = i
# If there is no mismatch or the
# difference between the
# mismatching elements is <= k
# then return true
if(ind == -1 or abs(a[ind]-b[ind]) <= k):
return True
return False
n, k = 2, 4
a =[1, 5]
b =[1, 1]
if(check(n, k, a, b)):
print("Yes")
else:
print("No")C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to check if both
// sequences can be made equal
static bool check(int n, int k,
int[] a, int[] b)
{
// Sorting both the arrays
Array.Sort(a);
Array.Sort(b);
// Flag to tell if there are
// more than one mismatch
bool fl = false;
// To stores the index
// of mismatched element
int ind = -1;
for (int i = 0; i < n; i++)
{
if (a[i] != b[i])
{
// If there is more than one
// mismatch then return False
if (fl == true)
{
return false;
}
fl = true;
ind = i;
}
}
// If there is no mismatch or the
// difference between the
// mismatching elements is <= k
// then return true
if (ind == -1 | Math.Abs(a[ind] - b[ind]) <= k)
{
return true;
}
return false;
}
// Driver code
public static void Main()
{
int n = 2, k = 4;
int[] a = {1, 5};
int[] b = {1, 1};
if (check(n, k, a, b))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by Rajput-JiPHP
输出：
Yes