对于给定数组中的每个元素，最大化两个最接近的其他数组的乘积
给定大小为M的数组arr1[]和大小为N且长度至少为 2 的数组arr2[] ，任务是针对arr1[] 中的每个元素，最大化arr2[]中最接近arr1中元素的两个元素的乘积[]。最接近的元素必须出现在不同的索引上。
例子：
Input: arr1 = [5, 10, 17, 22, -1], arr2 = [-1, 26, 5, 20, 14, 17, -7]
Output: -5 70 340 520 7
Explanation:
The closest elements to 5 are 5 and -1, therefore the maximum product is -5
The closest elements to 10 are 5 and 14, therefore the maximum product is 70
The closest elements to 17 are 20, 17, and 14, therefore the maximum product of 20 and 17 is 340
The closest elements to 22 are 20 and 26, therefore the maximum product is 520
The closest elements to -1 are -1, 5, and -7, therefore the maximum product of -1 and -7 is 7
Input: arr1 = [3, 9, 4, -1, 22], arr2 = [-1, 1, 21, 8, -3, 20, 25]
Output: -1 8 8 3 420
方法：给定的问题可以使用贪心方法来解决。这个想法是按升序对数组arr2进行排序，然后对于arr1中的每个元素，使用二进制搜索在 arr2中找到最接近它的元素。可以按照以下步骤解决问题：
- 对数组arr2进行升序排序
- 迭代数组arr1并在每次迭代时对arr2应用二进制搜索以找到最接近arr1[i] 的元素的索引，例如x ：
- 如果不存在先前的索引x-1，则返回arr2[x] * arr2[x+1]
- Else 如果不存在下一个索引x+1，则返回arr2[x] * arr2[x-1]
- 否则，检查arr2[x-1]和arr2[x+1]之间的哪个元素更接近arr1[i]：
- 如果arr2[x-1]更接近，则返回arr2[x] * arr2[x-1]
- 否则，如果arr2[x+1]更接近，则返回arr2[x] * arr2[x+1]
- 否则，如果arr2[x-1]和arr2[x+1]都与arr1[i]等距，则返回arr2[x] * arr2[x+1]和arr2[x] * arr2[x+ 1]
下面是上述方法的实现。
C++
// C++ code for the above approach
#include
using namespace std;
// Binary search function to
// find element closest to arr1[i]
int binarySearch(int num,
vector arr)
{
// Initialize left right and mid
int mid = -1, left = 0,
right = arr.size() - 1;
// Initialize closest index and
// smallest difference
int closestInd = -1;
int smallestDiff = INT_MAX;
while (left <= right)
{
mid = (left + right) >> 1;
if (abs(arr[mid] - num) < smallestDiff)
{
// Update smallest difference
smallestDiff = abs(arr[mid] - num);
// Update closest index
closestInd = mid;
}
else if (abs(arr[mid] - num) == smallestDiff)
{
if (arr[mid] > arr[closestInd])
closestInd = mid;
}
if (arr[mid] == num)
{
// This is the closest
// element index
return mid;
}
else if (arr[mid] < num)
{
// Closer element lies
// to the right
left = mid + 1;
}
else
{
// Closer element lies
// to the left
right = mid - 1;
}
}
return closestInd;
}
// Function to find the maximum product of
// Closest two elements in second array
// for every element in the first array
vector maxProdClosest(vector arr1,
vector arr2)
{
// Find the length of both arrays
int M = arr1.size(), N = arr2.size();
// Initialize an array to store
// the result for every element
vector ans(M);
// Sort the second array arr2
sort(arr2.begin(), arr2.end());
// Iterate the array arr1
for (int i = 0; i < M; i++)
{
// Apply binary search and
// find the index of closest
// element to arr1[i] in arr2
int ind = binarySearch(arr1[i],
arr2);
// No element at previous index
if (ind == 0)
{
ans[i] = arr2[ind] * arr2[ind + 1];
}
// No element at the next index
else if (ind == N - 1)
{
ans[i] = arr2[ind] * arr2[ind - 1];
}
// Elements at the next and
// previous indices are present
else
{
// arr2[ind - 1] is closer
// to arr1[i]
if (abs(arr2[ind - 1] - arr1[i]) < abs(arr2[ind + 1] - arr1[i]))
{
ans[i] = arr2[ind] * arr2[ind - 1];
}
else if (
// arr2[ind + 1] is
// closer to arr1[i]
abs(arr2[ind - 1] - arr1[i]) > abs(arr2[ind + 1] - arr1[i]))
{
ans[i] = arr2[ind] * arr2[ind + 1];
}
// If both arr2[ind - 1] and
// arr2[ind + 1] are
// equidistant from arr1[i]
else
{
ans[i] = max(
arr2[ind] * arr2[ind - 1],
arr2[ind] * arr2[ind + 1]);
}
}
}
// Return the resulting array
return ans;
}
// Driver function
int main()
{
// Initialize the arrays
vector arr1 = {5, 10, 17, 22, -1};
vector arr2 = {-1, 26, 5, 20,
14, 17, -7};
// Call the function
vector res = maxProdClosest(arr1,
arr2);
// Iterate the array and
// print the result
for (int i = 0; i < res.size(); i++)
{
cout << res[i] << " ";
}
}
// This code is contributed by Potta Lokesh Java
// Java implementation for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the maximum product of
// Closest two elements in second array
// for every element in the first array
public static int[] maxProdClosest(int[] arr1,
int[] arr2)
{
// Find the length of both arrays
int M = arr1.length, N = arr2.length;
// Initialize an array to store
// the result for every element
int[] ans = new int[M];
// Sort the second array arr2
Arrays.sort(arr2);
// Iterate the array arr1
for (int i = 0; i < M; i++) {
// Apply binary search and
// find the index of closest
// element to arr1[i] in arr2
int ind = binarySearch(arr1[i],
arr2);
// No element at previous index
if (ind == 0) {
ans[i] = arr2[ind] * arr2[ind + 1];
}
// No element at the next index
else if (ind == N - 1) {
ans[i] = arr2[ind] * arr2[ind - 1];
}
// Elements at the next and
// previous indices are present
else {
// arr2[ind - 1] is closer
// to arr1[i]
if (Math.abs(arr2[ind - 1]
- arr1[i])
< Math.abs(arr2[ind + 1]
- arr1[i])) {
ans[i] = arr2[ind] * arr2[ind - 1];
}
else if (
// arr2[ind + 1] is
// closer to arr1[i]
Math.abs(arr2[ind - 1]
- arr1[i])
> Math.abs(arr2[ind + 1]
- arr1[i])) {
ans[i] = arr2[ind] * arr2[ind + 1];
}
// If both arr2[ind - 1] and
// arr2[ind + 1] are
// equidistant from arr1[i]
else {
ans[i] = Math.max(
arr2[ind] * arr2[ind - 1],
arr2[ind] * arr2[ind + 1]);
}
}
}
// Return the resulting array
return ans;
}
// Binary search function to
// find element closest to arr1[i]
public static int binarySearch(int num,
int[] arr)
{
// Initialize left right and mid
int mid = -1, left = 0,
right = arr.length - 1;
// Initialize closest index and
// smallest difference
int closestInd = -1;
int smallestDiff = Integer.MAX_VALUE;
while (left <= right) {
mid = (left + right) >> 1;
if (Math.abs(arr[mid] - num)
< smallestDiff) {
// Update smallest difference
smallestDiff = Math.abs(arr[mid] - num);
// Update closest index
closestInd = mid;
}
else if (Math.abs(arr[mid] - num)
== smallestDiff) {
if (arr[mid] > arr[closestInd])
closestInd = mid;
}
if (arr[mid] == num) {
// This is the closest
// element index
return mid;
}
else if (arr[mid] < num) {
// Closer element lies
// to the right
left = mid + 1;
}
else {
// Closer element lies
// to the left
right = mid - 1;
}
}
return closestInd;
}
// Driver function
public static void main(String[] args)
{
// Initialize the arrays
int[] arr1 = { 5, 10, 17, 22, -1 };
int[] arr2 = { -1, 26, 5, 20,
14, 17, -7 };
// Call the function
int[] res = maxProdClosest(arr1,
arr2);
// Iterate the array and
// print the result
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
}Python3
# python3 code for the above approach
INT_MAX = 2147483647
# Binary search function to
# find element closest to arr1[i]
def binarySearch(num, arr):
# Initialize left right and mid
mid, left, right = -1, 0, len(arr) - 1
# Initialize closest index and
# smallest difference
closestInd = -1
smallestDiff = INT_MAX
while (left <= right):
mid = (left + right) >> 1
if (abs(arr[mid] - num) < smallestDiff):
# Update smallest difference
smallestDiff = abs(arr[mid] - num)
# Update closest index
closestInd = mid
elif (abs(arr[mid] - num) == smallestDiff):
if (arr[mid] > arr[closestInd]):
closestInd = mid
if (arr[mid] == num):
# This is the closest
# element index
return mid
elif (arr[mid] < num):
# Closer element lies
# to the right
left = mid + 1
else:
# Closer element lies
# to the left
right = mid - 1
return closestInd
# Function to find the maximum product of
# Closest two elements in second array
# for every element in the first array
def maxProdClosest(arr1, arr2):
# Find the length of both arrays
M, N = len(arr1), len(arr2)
# Initialize an array to store
# the result for every element
ans = [0 for _ in range(M)]
# Sort the second array arr2
arr2.sort()
# Iterate the array arr1
for i in range(0, M):
# Apply binary search and
# find the index of closest
# element to arr1[i] in arr2
ind = binarySearch(arr1[i], arr2)
# No element at previous index
if (ind == 0):
ans[i] = arr2[ind] * arr2[ind + 1]
# No element at the next index
elif (ind == N - 1):
ans[i] = arr2[ind] * arr2[ind - 1]
# Elements at the next and
# previous indices are present
else:
# arr2[ind - 1] is closer
# to arr1[i]
if (abs(arr2[ind - 1] - arr1[i]) < abs(arr2[ind + 1] - arr1[i])):
ans[i] = arr2[ind] * arr2[ind - 1]
elif (
# arr2[ind + 1] is
# closer to arr1[i]
abs(arr2[ind - 1] - arr1[i]) > abs(arr2[ind + 1] - arr1[i])):
ans[i] = arr2[ind] * arr2[ind + 1]
# If both arr2[ind - 1] and
# arr2[ind + 1] are
# equidistant from arr1[i]
else:
ans[i] = max(
arr2[ind] * arr2[ind - 1],
arr2[ind] * arr2[ind + 1])
# Return the resulting array
return ans
# Driver function
if __name__ == "__main__":
# Initialize the arrays
arr1 = [5, 10, 17, 22, -1]
arr2 = [-1, 26, 5, 20, 14, 17, -7]
# Call the function
res = maxProdClosest(arr1, arr2)
# Iterate the array and
# print the result
for i in range(0, len(res)):
print(res[i], end=" ")
# This code is contributed by rakeshsahniC#
// C# implementation for the above approach
using System;
class GFG {
// Function to find the maximum product of
// Closest two elements in second array
// for every element in the first array
public static int[] maxProdClosest(int[] arr1,
int[] arr2)
{
// Find the length of both arrays
int M = arr1.Length, N = arr2.Length;
// Initialize an array to store
// the result for every element
int[] ans = new int[M];
// Sort the second array arr2
Array.Sort(arr2);
// Iterate the array arr1
for (int i = 0; i < M; i++) {
// Apply binary search and
// find the index of closest
// element to arr1[i] in arr2
int ind = binarySearch(arr1[i], arr2);
// No element at previous index
if (ind == 0) {
ans[i] = arr2[ind] * arr2[ind + 1];
}
// No element at the next index
else if (ind == N - 1) {
ans[i] = arr2[ind] * arr2[ind - 1];
}
// Elements at the next and
// previous indices are present
else {
// arr2[ind - 1] is closer
// to arr1[i]
if (Math.Abs(arr2[ind - 1] - arr1[i])
< Math.Abs(arr2[ind + 1] - arr1[i])) {
ans[i] = arr2[ind] * arr2[ind - 1];
}
else if (
// arr2[ind + 1] is
// closer to arr1[i]
Math.Abs(arr2[ind - 1] - arr1[i])
> Math.Abs(arr2[ind + 1] - arr1[i])) {
ans[i] = arr2[ind] * arr2[ind + 1];
}
// If both arr2[ind - 1] and
// arr2[ind + 1] are
// equidistant from arr1[i]
else {
ans[i] = Math.Max(
arr2[ind] * arr2[ind - 1],
arr2[ind] * arr2[ind + 1]);
}
}
}
// Return the resulting array
return ans;
}
// Binary search function to
// find element closest to arr1[i]
public static int binarySearch(int num, int[] arr)
{
// Initialize left right and mid
int mid = -1, left = 0, right = arr.Length - 1;
// Initialize closest index and
// smallest difference
int closestInd = -1;
int smallestDiff = Int32.MaxValue;
while (left <= right) {
mid = (left + right) >> 1;
if (Math.Abs(arr[mid] - num) < smallestDiff) {
// Update smallest difference
smallestDiff = Math.Abs(arr[mid] - num);
// Update closest index
closestInd = mid;
}
else if (Math.Abs(arr[mid] - num)
== smallestDiff) {
if (arr[mid] > arr[closestInd])
closestInd = mid;
}
if (arr[mid] == num) {
// This is the closest
// element index
return mid;
}
else if (arr[mid] < num) {
// Closer element lies
// to the right
left = mid + 1;
}
else {
// Closer element lies
// to the left
right = mid - 1;
}
}
return closestInd;
}
// Driver function
public static void Main(string[] args)
{
// Initialize the arrays
int[] arr1 = { 5, 10, 17, 22, -1 };
int[] arr2 = { -1, 26, 5, 20, 14, 17, -7 };
// Call the function
int[] res = maxProdClosest(arr1, arr2);
// Iterate the array and
// print the result
for (int i = 0; i < res.Length; i++) {
Console.Write(res[i] + " ");
}
}
}
// This code is contributed by ukasp.Javascript
输出
-5 70 340 520 7 时间复杂度： O(N * log N + M * log N)
辅助空间： O(1)