给定一个大小为N的数组arr[]和一个整数K ，任务是从乘积数组中找到第K个数字。
注意：数组的乘积数组prod[]是大小为(N*(N-1))/2的排序数组，其中每个元素形成为prod[k] = arr[i] * arr[j] ，其中0 ≤ i < j < N 。
例子：
Input: arr[] = {-4, -2, 3, 3}, K = 3
Output: -6
Final prod[] array = {-12, -12, -6, -6, 8, 9}
where prod[K] = -6
Input: arr[] = {5, 4, 3, 2, -1, 0, 0}, K = 20
Output: 15
朴素的方法：通过迭代给定数组两次来生成 prod[] 数组，然后对 prod[] 数组进行排序并从数组中找到第 K个元素。
时间复杂度： O(N 2 * log(N))
有效的方法：可以轻松确定负数、零和正数对的数量，因此您可以判断答案是负数、零还是正数。如果答案是否定的，则可以通过一一选择一个负数和一个正数来衡量大于或等于K的对数，因此使用二分搜索得到答案。当答案为正时，答案完全相同，但考虑选择相同的元素两次，减去它会精确计算每对两次。
下面是上述方法的实现：
C++
// C++ implementation to find the
// Kth number in the list formed
// from product of any two numbers
// in the array and sorting them
#include
using namespace std;
// Function to find number of pairs
bool check(long long x,
vector& pos,
vector& neg, int k)
{
long long pairs = 0;
int p = neg.size() - 1;
int nn = neg.size() - 1;
int pp = pos.size() - 1;
// Negative and Negative
for (int i = 0; i < neg.size(); i++) {
while (p >= 0 and neg[i] * neg[p] <= x)
p--;
// Add Possible Pairs
pairs += min(nn - p, nn - i);
}
// Positive and Positive
p = 0;
for (int i = pos.size() - 1; i >= 0; i--) {
while (p < pos.size() and pos[i] * pos[p] <= x)
p++;
// Add Possible pairs
pairs += min(p, i);
}
// Negative and Positive
p = pos.size() - 1;
for (int i = neg.size() - 1;
i >= 0;
i--) {
while (p >= 0 and neg[i] * pos[p] <= x)
p--;
// Add Possible pairs
pairs += pp - p;
}
return (pairs >= k);
}
// Function to find the kth
// element in the list
long long kth_element(int a[],
int n, int k)
{
vector pos, neg;
// Separate Positive and
// Negative elements
for (int i = 0; i < n; i++) {
if (a[i] >= 0)
pos.push_back(a[i]);
else
neg.push_back(a[i]);
}
// Sort the Elements
sort(pos.begin(), pos.end());
sort(neg.begin(), neg.end());
long long l = -1e18,
ans = 0, r = 1e18;
// Binary search
while (l <= r) {
long long mid = (l + r) >> 1;
if (check(mid, pos, neg, k)) {
ans = mid;
r = mid - 1;
}
else
l = mid + 1;
}
// Return the required answer
return ans;
}
// Driver code
int main()
{
int a[] = { -4, -2, 3, 3 }, k = 3;
int n = sizeof(a) / sizeof(a[0]);
// Function call
cout << kth_element(a, n, k);
return 0;
} Java
// Java implementation to find the
// Kth number in the list formed
// from product of any two numbers
// in the array and sorting them
import java.util.*;
class GFG
{
// Function to find number of pairs
static boolean check(int x, Vector pos, Vector neg, int k)
{
int pairs = 0;
int p = neg.size() - 1;
int nn = neg.size() - 1;
int pp = pos.size() - 1;
// Negative and Negative
for (int i = 0; i < neg.size(); i++)
{
while ((p >= 0) && ((int)neg.get(i) *
(int)neg.get(p) <= x))
p--;
// Add Possible Pairs
pairs += Math.min(nn - p, nn - i);
}
// Positive and Positive
p = 0;
for (int i = pos.size() - 1; i >= 0; i--)
{
while ((p < pos.size()) && ((int)pos.get(i) *
(int)pos.get(p) <= x))
p++;
// Add Possible pairs
pairs += Math.min(p, i);
}
// Negative and Positive
p = pos.size() - 1;
for (int i = neg.size() - 1;
i >= 0; i--) {
while ((p >= 0) && ((int)neg.get(i) *
(int)pos.get(p) <= x))
p--;
// Add Possible pairs
pairs += pp - p;
}
return (pairs >= k);
}
// Function to find the kth
// element in the list
static int kth_element(int a[], int n, int k)
{
Vector pos = new Vector();
Vector neg = new Vector();;
// Separate Positive and
// Negative elements
for (int i = 0; i < n; i++)
{
if (a[i] >= 0)
pos.add(a[i]);
else
neg.add(a[i]);
}
// Sort the Elements
//sort(pos.begin(), pos.end());
//sort(neg.begin(), neg.end());
Collections.sort(pos);
Collections.sort(neg);
int l = (int)-1e8, ans = 0, r = (int)1e8;
// Binary search
while (l <= r)
{
int mid = (l + r) >> 1;
if (check(mid, pos, neg, k))
{
ans = mid;
r = mid - 1;
}
else
l = mid + 1;
}
// Return the required answer
return ans;
}
// Driver code
public static void main (String[] args)
{
int a[] = { -4, -2, 3, 3 }, k = 3;
int n = a.length;
// Function call
System.out.println(kth_element(a, n, k));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation to find the
# Kth number in the list formed
# from product of any two numbers
# in the array and sorting them
# Function to find number of pairs
def check(x, pos, neg, k):
pairs = 0
p = len(neg) - 1
nn = len(neg) - 1
pp = len(pos) - 1
# Negative and Negative
for i in range(len(neg)):
while (p >= 0 and neg[i] * neg[p] <= x):
p -= 1
# Add Possible Pairs
pairs += min(nn - p, nn - i)
# Positive and Positive
p = 0
for i in range(len(pos) - 1, -1, -1):
while (p < len(pos) and pos[i] * pos[p] <= x):
p += 1
# Add Possible pairs
pairs += min(p, i)
# Negative and Positive
p = len(pos) - 1
for i in range(len(neg) - 1, -1, -1):
while (p >= 0 and neg[i] * pos[p] <= x):
p -= 1
# Add Possible pairs
pairs += pp - p
return (pairs >= k)
# Function to find the kth
# element in the list
def kth_element(a, n, k):
pos, neg = [],[]
# Separate Positive and
# Negative elements
for i in range(n):
if (a[i] >= 0):
pos.append(a[i])
else:
neg.append(a[i])
# Sort the Elements
pos = sorted(pos)
neg = sorted(neg)
l = -10**18
ans = 0
r = 10**18
# Binary search
while (l <= r):
mid = (l + r) >> 1
if (check(mid, pos, neg, k)):
ans = mid
r = mid - 1
else:
l = mid + 1
# Return the required answer
return ans
# Driver code
a = [-4, -2, 3, 3]
k = 3
n = len(a)
# Function call
print(kth_element(a, n, k))
# This code is contributed by mohit kumar 29C#
// C# implementation to find the
// Kth number in the list formed
// from product of any two numbers
// in the array and sorting them
using System;
using System.Collections.Generic;
class GFG
{
// Function to find number of pairs
static bool check(int x, List pos, List neg, int k)
{
int pairs = 0;
int p = neg.Count - 1;
int nn = neg.Count - 1;
int pp = pos.Count - 1;
// Negative and Negative
for (int i = 0; i < neg.Count; i++)
{
while ((p >= 0) && ((int)neg[i] *
(int)neg[p] <= x))
p--;
// Add Possible Pairs
pairs += Math.Min(nn - p, nn - i);
}
// Positive and Positive
p = 0;
for (int i = pos.Count - 1; i >= 0; i--)
{
while ((p < pos.Count) && ((int)pos[i] *
(int)pos[p] <= x))
p++;
// Add Possible pairs
pairs += Math.Min(p, i);
}
// Negative and Positive
p = pos.Count - 1;
for (int i = neg.Count - 1; i >= 0; i--)
{
while ((p >= 0) && ((int)neg[i] *
(int)pos[p] <= x))
p--;
// Add Possible pairs
pairs += pp - p;
}
return (pairs >= k);
}
// Function to find the kth
// element in the list
static int kth_element(int []a, int n, int k)
{
List pos = new List();
List neg = new List();;
// Separate Positive and
// Negative elements
for (int i = 0; i < n; i++)
{
if (a[i] >= 0)
pos.Add(a[i]);
else
neg.Add(a[i]);
}
// Sort the Elements
//sort(pos.begin(), pos.end());
//sort(neg.begin(), neg.end());
pos.Sort();
neg.Sort();
int l = (int)-1e8, ans = 0, r = (int)1e8;
// Binary search
while (l <= r)
{
int mid = (l + r) >> 1;
if (check(mid, pos, neg, k))
{
ans = mid;
r = mid - 1;
}
else
l = mid + 1;
}
// Return the required answer
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []a = { -4, -2, 3, 3 };
int k = 3;
int n = a.Length;
// Function call
Console.WriteLine(kth_element(a, n, k));
}
}
// This code is contributed by 29AjayKumar 输出：
-6
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live