给定两个数组A []和B []以及一个整数K ，任务是通过添加或减去范围[0 ]中的任何整数来最大化可以与数组A []相等的数组B []的整数数量，K] 。
例子：
Input: K=5, A[] = [100, 65, 35, 85, 55], B[] = [30, 60, 75, 95]
Output: 3
Explanation:
30 + 5, 60 + 5, 95 + 5 gives the values which are equal with few elements of array A[].
Input: K = 5, A[] = [10, 20, 30, 40, 50], B[] = [1, 20, 3]
Output: 1
Explanation:
Only the 2nd value can be made equal, Since its value [20] can be changed to [20] by adding/subtracting 0 from it.
方法：想法是检查数组B []的元素与数组A []的任何元素之间的绝对差是否小于或等于K。如果是，则将其包括在计数中。在检查完数组B []中所有元素的上述条件后，打印所有这些元素的计数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
void countElement(int A[], int N,
int B[], int M, int K)
{
// To store the count of element
int cnt = 0;
// Traverse the array B[]
for (int i = 0; i < M; i++) {
int currentElement = B[i];
// Traverse the array A[]
for (int j = 0; j < N; j++) {
// Find the difference
int diff
= abs(currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K) {
cnt++;
break;
}
}
}
// Print the count
cout << cnt;
}
// Driver Code
int main()
{
// Given array A[] and B[]
int A[] = { 100, 65, 35, 85, 55 };
int B[] = { 30, 60, 75, 95 };
// Given K
int K = 5;
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
// Function Call
countElement(A, N, B, M, K);
return 0;
} Java
// Java program for the above approach
class GFG{
// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
static void countElement(int A[], int N,
int B[], int M, int K)
{
// To store the count of element
int cnt = 0;
// Traverse the array B[]
for(int i = 0; i < M; i++)
{
int currentElement = B[i];
// Traverse the array A[]
for(int j = 0; j < N; j++)
{
// Find the difference
int diff = Math.abs(
currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K)
{
cnt++;
break;
}
}
}
// Print the count
System.out.print(cnt);
}
// Driver Code
public static void main(String[] args)
{
// Given array A[] and B[]
int A[] = { 100, 65, 35, 85, 55 };
int B[] = { 30, 60, 75, 95 };
// Given K
int K = 5;
int N = A.length;
int M = B.length;
// Function call
countElement(A, N, B, M, K);
}
}
// This code is contributed by Rajput-JiPython3
# Python3 program to implement
# the above approach
# Function that count the number of
# integers from array B such that
# subtracting element in the range
# [0, K] given any element in A
def countElement(A, N, B, M, K):
# To store the count of element
cnt = 0
# Traverse the array B
for i in range(M):
currentElement = B[i]
# Traverse the array A
for j in range(N):
# Find the difference
diff = abs(currentElement - A[j])
# If difference is atmost
# K then increment the cnt
if(diff <= K):
cnt += 1
break
# Print the count
print(cnt)
# Driver Code
if __name__ == '__main__':
# Given array A and B
A = [ 100, 65, 35, 85, 55 ]
B = [ 30, 60, 75, 95 ]
N = len(A)
M = len(B)
# Given K
K = 5
# Function call
countElement(A, N, B, M, K)
# This code is contributed by Shivam SinghC#
// C# program for the above approach
using System;
class GFG{
// Function that count the number of
// integers from array []B such that
// subtracting element in the range
// [0, K] given any element in []A
static void countElement(int []A, int N,
int []B, int M, int K)
{
// To store the count of element
int cnt = 0;
// Traverse the array []B
for(int i = 0; i < M; i++)
{
int currentElement = B[i];
// Traverse the array []A
for(int j = 0; j < N; j++)
{
// Find the difference
int diff = Math.Abs(
currentElement - A[j]);
// If difference is atmost
// K then increment the cnt
if (diff <= K)
{
cnt++;
break;
}
}
}
// Print the count
Console.Write(cnt);
}
// Driver Code
public static void Main(String[] args)
{
// Given array []A and []B
int []A = { 100, 65, 35, 85, 55 };
int []B = { 30, 60, 75, 95 };
// Given K
int K = 5;
int N = A.Length;
int M = B.Length;
// Function call
countElement(A, N, B, M, K);
}
}
// This code is contributed by Rohit_ranjan输出：
3
时间复杂度： O(N * M)，其中N和M是数组A []和B []的长度。
辅助空间： O(1)