arr[i] >= arr[j] 的所有数组对的最大模数
给定一个包含 n 个整数的数组。求 arr[i] mod arr[j] 的最大值，其中 arr[i] >= arr[j] and 1 <= i, j <= n
例子：
Input: arr[] = {3, 4, 7}
Output: 3
Explanation:
There are 3 pairs which satisfies arr[i] >= arr[j] are:-
4, 3 => 4 % 3 = 1
7, 3 => 7 % 3 = 1
7, 4 => 7 % 4 = 3
Hence Maximum value among all is 3.
Input: arr[] = {3, 7, 4, 11}
Output: 4
Input: arr[] = {4, 4, 4}
Output: 0一种朴素的方法是运行两个嵌套的 for 循环，并在取模后选择每个可能对的最大值。这种方法的时间复杂度将是 O(n 2 )，这对于较大的 n 值是不够的。
一种有效的方法（当元素来自小范围时）是使用排序和二进制搜索方法。首先，我们将对数组进行排序，以便我们能够对其应用二进制搜索。由于我们需要最大化 arr[i] mod arr[j] 的值，所以我们遍历从 2*arr[j] 到 M+arr[j] 范围内的每个 x（例如 x 可被 arr[j] 整除），其中 M 是序列的最大值。对于 x 的每个值，我们需要找到 arr[i] 的最大值，使得 arr[i] < x。
通过这样做，我们将确保我们只选择了那些将给出 arr[i] mod arr[j] 最大值的 arr[i] 值。之后，我们只需对 arr[j] 的其他值重复上述过程，并通过值 a[i] mod arr[j] 更新答案。例如：-
If arr[] = {4, 6, 7, 8, 10, 12, 15} then for
first element, i.e., arr[j] = 4 we iterate
through x = {8, 12, 16}.
Therefore for each value of x, a[i] will be:-
x = 8, arr[i] = 7 (7 < 8)
ans = 7 mod 4 = 3
x = 12, arr[i] = 10 (10 < 12)
ans = 10 mod 4 = 2 (Since 2 < 3,
No update)
x = 16, arr[i] = 15 (15 < 16)
ans = 15 mod 4 = 3 (Since 3 == 3,
No need to update)C++
// C++ program to find Maximum modulo value
#include
using namespace std;
int maxModValue(int arr[], int n)
{
int ans = 0;
// Sort the array[] by using inbuilt sort function
sort(arr, arr + n);
for (int j = n - 2; j >= 0; --j) {
// Break loop if answer is greater or equals to
// the arr[j] as any number modulo with arr[j]
// can only give maximum value up-to arr[j]-1
if (ans >= arr[j])
break;
// If both elements are same then skip the next
// loop as it would be worthless to repeat the
// rest process for same value
if (arr[j] == arr[j + 1])
continue;
for (int i = 2 * arr[j]; i <= arr[n - 1] + arr[j]; i += arr[j]) {
// Fetch the index which is greater than or
// equals to arr[i] by using binary search
// inbuilt lower_bound() function of C++
int ind = lower_bound(arr, arr + n, i) - arr;
// Update the answer
ans = max(ans, arr[ind - 1] % arr[j]);
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 3, 4, 5, 9, 11 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxModValue(arr, n);
} Java
// Java program to find Maximum modulo value
import java.util.Arrays;
class Test {
static int maxModValue(int arr[], int n)
{
int ans = 0;
// Sort the array[] by using inbuilt sort function
Arrays.sort(arr);
for (int j = n - 2; j >= 0; --j) {
// Break loop if answer is greater or equals to
// the arr[j] as any number modulo with arr[j]
// can only give maximum value up-to arr[j]-1
if (ans >= arr[j])
break;
// If both elements are same then skip the next
// loop as it would be worthless to repeat the
// rest process for same value
if (arr[j] == arr[j + 1])
continue;
for (int i = 2 * arr[j]; i <= arr[n - 1] + arr[j]; i += arr[j]) {
// Fetch the index which is greater than or
// equals to arr[i] by using binary search
int ind = Arrays.binarySearch(arr, i);
if (ind < 0)
ind = Math.abs(ind + 1);
else {
while (arr[ind] == i) {
ind--;
if (ind == 0) {
ind = -1;
break;
}
}
ind++;
}
// Update the answer
ans = Math.max(ans, arr[ind - 1] % arr[j]);
}
}
return ans;
}
// Driver method
public static void main(String args[])
{
int arr[] = { 3, 4, 5, 9, 11 };
System.out.println(maxModValue(arr, arr.length));
}
}Python3
# Python3 program to find Maximum modulo value
def maxModValue(arr, n):
ans = 0
# Sort the array[] by using inbuilt
# sort function
arr = sorted(arr)
for j in range(n - 2, -1, -1):
# Break loop if answer is greater or equals to
# the arr[j] as any number modulo with arr[j]
# can only give maximum value up-to arr[j]-1
if (ans >= arr[j]):
break
# If both elements are same then skip the next
# loop as it would be worthless to repeat the
# rest process for same value
if (arr[j] == arr[j + 1]) :
continue
i = 2 * arr[j]
while(i <= arr[n - 1] + arr[j]):
# Fetch the index which is greater than or
# equals to arr[i] by using binary search
# inbuilt lower_bound() function of C++
ind = 0
for k in arr:
if k >= i:
ind = arr.index(k)
# Update the answer
ans = max(ans, arr[ind - 1] % arr[j])
i += arr[j]
return ans
# Driver Code
arr = [3, 4, 5, 9, 11 ]
n = 5
print(maxModValue(arr, n))
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)C#
// C# program to find Maximum modulo value
using System;
public class GFG {
static int maxModValue(int[] arr, int n)
{
int ans = 0;
// Sort the array[] by using inbuilt
// sort function
Array.Sort(arr);
for (int j = n - 2; j >= 0; --j)
{
// Break loop if answer is greater
// or equals to the arr[j] as any
// number modulo with arr[j] can
// only give maximum value up-to
// arr[j]-1
if (ans >= arr[j])
break;
// If both elements are same then
// skip the next loop as it would
// be worthless to repeat the
// rest process for same value
if (arr[j] == arr[j + 1])
continue;
for (int i = 2 * arr[j];
i <= arr[n - 1] + arr[j];
i += arr[j])
{
// Fetch the index which is
// greater than or equals to
// arr[i] by using binary search
int ind = Array.BinarySearch(arr, i);
if (ind < 0)
ind = Math.Abs(ind + 1);
else {
while (arr[ind] == i) {
ind--;
if (ind == 0) {
ind = -1;
break;
}
}
ind++;
}
// Update the answer
ans = Math.Max(ans, arr[ind - 1]
% arr[j]);
}
}
return ans;
}
// Driver method
public static void Main()
{
int[] arr = { 3, 4, 5, 9, 11 };
Console.WriteLine(
maxModValue(arr, arr.Length));
}
}
// This code is contributed by Sam007.Javascript
输出：
4时间复杂度： O(nlog(n) + Mlog(M)) 其中 n 是元素的总数，M 是所有元素的最大值。
辅助空间： O(1)
本博客由 Shubham Bansal 提供。如果您喜欢 GeeksforGeeks 并愿意做出贡献，您还可以使用 write.geeksforgeeks.org 撰写文章或将您的文章邮寄至 review-team@geeksforgeeks.org。在 GeeksforGeeks 主页上查看您的文章并帮助其他 Geeks。