给定大小为N的整数数组A [] ,任务是找到大小为B的子序列,以使它们中的任意两个之间的最小差异最大,并打印出该最大最小差异。
例子:
Input: A[ ] = {1, 2, 3, 5}, B = 3
Output: 2
Explanation:
Possible subsequences of size 3 are {1, 2, 3}, {1, 2, 5}, {1, 3, 5} and {2, 3, 5}.
For {1, 3, 5}, possible differences are (|1 – 3| = 2), (|3 – 5| = 2) and (|1 – 5| = 4), Minimum(2, 2, 4) = 2
For the remaining subsequences, the minimum difference comes out to be 1.
Hence, the maximum of all minimum differences is 2.
Input: A[ ] = {5, 17, 11}, B = 2
Output: 12
Explanation:
Possible subsequences of size 2 are {5, 17}, {17, 11} and {5, 11}.
For {5, 17}, possible difference is (|5 – 17| = 12), Minimum = 12
For {17, 11}, possible difference is (|17 – 11| = 6), Minimum = 6
For {5, 11}, possible difference is (|5 – 11| = 6), Minimum = 6
Maximum(12, 6, 6) = 12
Hence, the maximum of all minimum differences is 12.
天真的方法:
解决此问题的最简单方法是生成大小为B的所有可能子序列,并在所有可能的对之间找到最小差异。最后,在所有最小差异中找到最大值。
时间复杂度: O(2 N * N 2 )
辅助空间: O(N)
高效方法:
请按照以下步骤使用Binary Search优化上述方法:
- 将搜索空间从0设置为array( maxm )中的最大元素
- 对于每一个计算出的中间,检查是否有可能获得大小B的子序列与任何一对之间的最小差值等于中旬。
- 如果可能的话,然后存储在中期的变量,并找到在右半更好的答案,并丢弃中间的左半
- 否则,遍历中部的左半部分,以检查是否存在具有最小成对最小差异的子序列。
- 最后,在二分查找终止后,打印出最高的中位数,在该最高中位数中发现任何具有等于中位数对的最小差的子序列。
Illustration:
A[ ] = {1, 2, 3, 4, 5}, B = 3
Search space: {0, 1, 2, 3, 4, 5}
Steps involved in binary search are as follows:
- start = 0, end = 5, mid = (0 + 5) / 2 = 2
Subsequence of size B with minimum difference of mid(= 2) is {1, 3, 5}.
Therefore, ans = 2 - Now, traverse the right half.
start = mid +1 = 3, end = 5, mid = (3 + 5) / 2 = 4
Subsequence of size B with minimum difference of mid(= 4) is not possible.
Therefore, ans is still 2. - Now, traverse the left half
start = 3, end = mid – 1 = 3, mid = (3 + 3) / 2 = 3
Subsequence of size B with minimum difference of mid(= 3) is not possible.
Therefore, ans is still 2. - Again, traverse left half.
start = 3, end = mid – 1 = 2.
Since start exceeds end, binary search teminates. - Finally, the largest possible minimum difference is 2.
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to check a subsequence can
// be formed with min difference mid
bool can_place(int A[], int n,
int B, int mid)
{
int count = 1;
int last_position = A[0];
// If a subsequence of size B
// with min diff = mid is possible
// return true else false
for (int i = 1; i < n; i++) {
if (A[i] - last_position
>= mid) {
last_position = A[i];
count++;
if (count == B) {
return true;
}
}
}
return false;
}
// Function to find the maximum of
// all minimum difference of pairs
// possible among the subsequence
int find_min_difference(int A[],
int n, int B)
{
// Sort the Array
sort(A, A + n);
// Stores the boundaries
// of the search space
int s = 0;
int e = A[n - 1] - A[0];
// Store the answer
int ans = 0;
// Binary Search
while (s <= e) {
long long int mid = (s + e) / 2;
// If subsequence can be formed
// with min diff mid and size B
if (can_place(A, n, B, mid)) {
ans = mid;
// Right half
s = mid + 1;
}
else {
// Left half
e = mid - 1;
}
}
return ans;
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3, 5 };
int n = sizeof(A) / sizeof(A[0]);
int B = 3;
int min_difference
= find_min_difference(A, n, B);
cout << min_difference;
return 0;
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to check a subsequence can
// be formed with min difference mid
static boolean can_place(int A[], int n,
int B, int mid)
{
int count = 1;
int last_position = A[0];
// If a subsequence of size B
// with min diff = mid is possible
// return true else false
for(int i = 1; i < n; i++)
{
if (A[i] - last_position >= mid)
{
last_position = A[i];
count++;
if (count == B)
{
return true;
}
}
}
return false;
}
// Function to find the maximum of
// all minimum difference of pairs
// possible among the subsequence
static int find_min_difference(int A[],
int n, int B)
{
// Sort the Array
Arrays.sort(A);
// Stores the boundaries
// of the search space
int s = 0;
int e = A[n - 1] - A[0];
// Store the answer
int ans = 0;
// Binary Search
while (s <= e)
{
int mid = (s + e) / 2;
// If subsequence can be formed
// with min diff mid and size B
if (can_place(A, n, B, mid))
{
ans = mid;
// Right half
s = mid + 1;
}
else
{
// Left half
e = mid - 1;
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 2, 3, 5 };
int n = A.length;
int B = 3;
int min_difference = find_min_difference(A, n, B);
System.out.print(min_difference);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to implement
# the above approach
# Function to check a subsequence can
# be formed with min difference mid
def can_place(A, n, B, mid):
count = 1
last_position = A[0]
# If a subsequence of size B
# with min diff = mid is possible
# return true else false
for i in range(1, n):
if (A[i] - last_position >= mid):
last_position = A[i]
count = count + 1
if (count == B):
return bool(True)
return bool(False)
# Function to find the maximum of
# all minimum difference of pairs
# possible among the subsequence
def find_min_difference(A, n, B):
# Sort the Array
A.sort()
# Stores the boundaries
# of the search space
s = 0
e = A[n - 1] - A[0]
# Store the answer
ans = 0
# Binary Search
while (s <= e):
mid = (int)((s + e) / 2)
# If subsequence can be formed
# with min diff mid and size B
if (can_place(A, n, B, mid)):
ans = mid
# Right half
s = mid + 1
else:
# Left half
e = mid - 1
return ans
# Driver code
A = [ 1, 2, 3, 5 ]
n = len(A)
B = 3
min_difference = find_min_difference(A, n, B)
print(min_difference)
# This code is contributed by divyeshrabadiya07
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check a subsequence can
// be formed with min difference mid
static bool can_place(int[] A, int n,
int B, int mid)
{
int count = 1;
int last_position = A[0];
// If a subsequence of size B
// with min diff = mid is possible
// return true else false
for(int i = 1; i < n; i++)
{
if (A[i] - last_position >= mid)
{
last_position = A[i];
count++;
if (count == B)
{
return true;
}
}
}
return false;
}
// Function to find the maximum of
// all minimum difference of pairs
// possible among the subsequence
static int find_min_difference(int[] A,
int n, int B)
{
// Sort the Array
Array.Sort(A);
// Stores the boundaries
// of the search space
int s = 0;
int e = A[n - 1] - A[0];
// Store the answer
int ans = 0;
// Binary Search
while (s <= e)
{
int mid = (s + e) / 2;
// If subsequence can be formed
// with min diff mid and size B
if (can_place(A, n, B, mid))
{
ans = mid;
// Right half
s = mid + 1;
}
else
{
// Left half
e = mid - 1;
}
}
return ans;
}
// Driver Code
public static void Main(string[] args)
{
int[] A = { 1, 2, 3, 5 };
int n = A.Length;
int B = 3;
int min_difference = find_min_difference(A, n, B);
Console.Write(min_difference);
}
}
// This code is contributed by rock_cool
Javascript
2
时间复杂度: O(NlogN)
辅助空间: O(1)