给定分别为N和M的两个数组A []和B [] ，其中每个元素的范围为[0，9] ，任务是使数组A []的每个元素严格大于或小于每个通过将任意元素从任一数组更改为[0，9]范围内的任何数字(最小次数)来更改数组B []中的元素。
例子：
Input: A[] = [0, 1, 0], B[] = [2, 0, 0]
Output: 2
Explanation:
Modifying the array B[] to [2, 2, 2] makes every element in the array A[] strictly less than every element in the array B[].
Hence, the minimum number of changes required = 2.
Input: A[] = [0, 0, 1, 3, 3], B[] = [0, 2, 3]
Output: 3
Explanation:
Modifying the array B[] to [4, 4, 4] makes every element in the array A[] strictly less than every element in the array B[].
Hence, the minimum number of changes required = 3.
方法：解决给定问题的想法是使用大小为10的两个辅助数组prefix_a []和prefix_b [] ，其中prefix_a [i]和prefix_b [i]存储数组A [] ≤i的元素数，并且数组B []≤i中元素的数量。请按照以下步骤解决问题：
- 使用{0}初始化大小为10的两个数组prefix_a []和prefix_b [] 。
- 将每个元素的频率分别存储在数组prefix_a和prefix_b中的数组A []和B []中。
- 通过使用变量i在范围[1，9]中进行迭代，对数组prefix_a执行前缀求和，并将prefix_a [i]更新为(prefix [i] + prefix_a [i – 1]) 。
- 对数组prefix_b []重复上述步骤。
- 在范围内迭代 [0，9]使用变量i
- 存储操作的数目使每个元件阵列A []严格小于数字我更大，使每个元素在数组B []小于位数i的一个变量，表示X。
- 使用prefix_a [i] + M – prefix_b [i]对其进行初始化。
- 类似地，存储操作的数目使每个元件在数组B []严格大于数字我更大，并且使每一个元件在数组A []小于位数i的一个变量，表示Y.
- 用prefix_b [i] + N – prefix_a [i]初始化它。
- 将总的最小操作总数更新为X和Y的最小值。将获得的最小值存储在变量中，例如ans 。
- 完成上述步骤后，输出ans的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimize
// replacements to make every element
// in the array A[] strictly greater
// than every element in B[] or vice-versa
void MinTime(int* a, int* b, int n, int m)
{
// Store the final result
int ans = INT_MAX;
// Create two arrays and
// initialize with 0s
int prefix_a[10] = { 0 };
int prefix_b[10] = { 0 };
// Traverse the array a[]
for (int i = 0; i < n; i++) {
// Increment prefix_a[a[i]] by 1
prefix_a[a[i]]++;
}
// Traverse the array b[]
for (int i = 0; i < m; i++) {
// Increment prefix_b[b[i]] by 1
prefix_b[b[i]]++;
}
// Calculate prefix sum
// of the array a[]
for (int i = 1; i <= 9; i++) {
prefix_a[i] += prefix_a[i - 1];
}
// Calculate prefix sum
// of the array b[]
for (int i = 1; i <= 9; i++) {
prefix_b[i] += prefix_b[i - 1];
}
// Iterate over the range [0, 9]
for (int i = 0; i <= 9; i++) {
// Make every element in array
// a[] strictly greater than digit
// i and make every element in the
// array b[] less than digit i
ans = min(ans, prefix_a[i] + m
- prefix_b[i]);
// Make every element in array
// b[] strictly greater than digit
// i and make every element in
// array a[] less than digit i
ans = min(ans, n - prefix_a[i]
+ prefix_b[i]);
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
int A[] = { 0, 0, 1, 3, 3 };
int B[] = { 2, 0, 3 };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
MinTime(A, B, N, M);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the minimize
// replacements to make every element
// in the array A[] strictly greater
// than every element in B[] or vice-versa
static void MinTime(int []a, int []b, int n, int m)
{
// Store the final result
int ans = 2147483647;
// Create two arrays and
// initialize with 0s
int []prefix_a = new int[10];
int []prefix_b = new int[10];
// Traverse the array a[]
for (int i = 0; i < n; i++) {
// Increment prefix_a[a[i]] by 1
prefix_a[a[i]]++;
}
// Traverse the array b[]
for (int i = 0; i < m; i++) {
// Increment prefix_b[b[i]] by 1
prefix_b[b[i]]++;
}
// Calculate prefix sum
// of the array a[]
for (int i = 1; i <= 9; i++) {
prefix_a[i] += prefix_a[i - 1];
}
// Calculate prefix sum
// of the array b[]
for (int i = 1; i <= 9; i++) {
prefix_b[i] += prefix_b[i - 1];
}
// Iterate over the range [0, 9]
for (int i = 0; i <= 9; i++) {
// Make every element in array
// a[] strictly greater than digit
// i and make every element in the
// array b[] less than digit i
ans = Math.min(ans, prefix_a[i] + m
- prefix_b[i]);
// Make every element in array
// b[] strictly greater than digit
// i and make every element in
// array a[] less than digit i
ans = Math.min(ans, n - prefix_a[i]
+ prefix_b[i]);
}
// Print the answer
System.out.println(ans);
}
// Driver Code
public static void main(String [] args)
{
int []A = { 0, 0, 1, 3, 3 };
int []B = { 2, 0, 3 };
int N = A.length;
int M = B.length;
MinTime(A, B, N, M);
}
}
// This code is contributed by chitranayal.Python3
# Python program for the above approach
# Function to find the minimize
# replacements to make every element
# in the array A[] strictly greater
# than every element in B[] or vice-versa
def MinTime(a, b, n, m):
# Store the final result
ans = float('inf')
# Create two arrays and
# initialize with 0s
prefix_a = [ 0 ]*10
prefix_b = [ 0 ]*10
# Traverse the array a[]
for i in range(n):
# Increment prefix_a[a[i]] by 1
prefix_a[a[i]] += 1
# Traverse the array b[]
for i in range(m):
# Increment prefix_b[b[i]] by 1
prefix_b[b[i]] += 1
# Calculate prefix sum
# of the array a[]
for i in range(1, 10):
prefix_a[i] += prefix_a[i - 1]
# Calculate prefix sum
# of the array b[]
for i in range(1, 10):
prefix_b[i] += prefix_b[i - 1]
# Iterate over the range [0, 9]
for i in range(1, 10):
# Make every element in array
# a[] strictly greater than digit
# i and make every element in the
# array b[] less than digit i
ans = min(ans, prefix_a[i] + m- prefix_b[i])
# Make every element in array
# b[] strictly greater than digit
# i and make every element in
# array a[] less than digit i
ans = min(ans, n - prefix_a[i] + prefix_b[i])
# Print the answer
print(ans)
# Driver Code
A = [ 0, 0, 1, 3, 3 ]
B = [ 2, 0, 3 ]
N = len(A)
M = len(B)
MinTime(A, B, N, M)
# This code is contributed by rohitsingh07052.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimize
// replacements to make every element
// in the array A[] strictly greater
// than every element in B[] or vice-versa
static void MinTime(int []a, int []b, int n, int m)
{
// Store the final result
int ans = 2147483647;
// Create two arrays and
// initialize with 0s
int []prefix_a = new int[10];
int []prefix_b = new int[10];
Array.Clear(prefix_a,0,prefix_a.Length);
Array.Clear(prefix_b,0,prefix_b.Length);
// Traverse the array a[]
for (int i = 0; i < n; i++) {
// Increment prefix_a[a[i]] by 1
prefix_a[a[i]]++;
}
// Traverse the array b[]
for (int i = 0; i < m; i++) {
// Increment prefix_b[b[i]] by 1
prefix_b[b[i]]++;
}
// Calculate prefix sum
// of the array a[]
for (int i = 1; i <= 9; i++) {
prefix_a[i] += prefix_a[i - 1];
}
// Calculate prefix sum
// of the array b[]
for (int i = 1; i <= 9; i++) {
prefix_b[i] += prefix_b[i - 1];
}
// Iterate over the range [0, 9]
for (int i = 0; i <= 9; i++) {
// Make every element in array
// a[] strictly greater than digit
// i and make every element in the
// array b[] less than digit i
ans = Math.Min(ans, prefix_a[i] + m
- prefix_b[i]);
// Make every element in array
// b[] strictly greater than digit
// i and make every element in
// array a[] less than digit i
ans = Math.Min(ans, n - prefix_a[i]
+ prefix_b[i]);
}
// Print the answer
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
int []A = { 0, 0, 1, 3, 3 };
int []B = { 2, 0, 3 };
int N = A.Length;
int M = B.Length;
MinTime(A, B, N, M);
}
}
// This code is contributed by bgangwar59.输出：
3时间复杂度： O(N + M)
辅助空间： O(1)