给定大小为N的数组arr [] ,任务是计算该数组中可能存在的对数(i,j) ,以便arr [j] * i = arr [i] * j ,其中1≤i
例子:
Input: arr[] = {1, 3, 5, 6, 5}
Output: 2
Explanation:
Pair (1, 5) satisfies the condition, since arr[1] * 5 = arr[5] * 1.
Pair (2, 4) satisfies the condition, since arr[2] * 4 = arr[4] * 2.
Therefore, total number of pairs satisfying the given condition is 2.
Input: arr[] = {2, 1, 3}
Output: 0
天真的方法:解决问题的最简单方法是从数组中生成所有可能的对,并检查每个对,是否满足给定条件。增加满足条件的对的计数。最后,打印所有这些对的计数。
时间复杂度: O(N 2 )
辅助空间: O(1)
高效方法:为了优化上述方法,该思想基于将给定方程arr [i] * j = arr [j] * i重排为arr [i] / i = arr [j] / j 。
请按照以下步骤解决问题:
- 初始化一个变量,例如count ,以存储满足给定条件的对的总数。
- 初始化一个无序映射,例如mp,以计算值arr [i] / i的频率。
- 遍历数组arr []并更新Map中arr [i] / i的频率。
- 打印计数作为答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count pairs from an
// array satisfying given conditions
void countPairs(int arr[], int N)
{
// Stores the total
// count of pairs
int count = 0;
// Stores count of a[i] / i
unordered_map mp;
// Traverse the array
for (int i = 0; i < N; i++) {
double val = 1.0 * arr[i];
double idx = 1.0 * (i + 1);
// Updating count
count += mp[val / idx];
// Update frequency
// in the Map
mp[val / idx]++;
}
// Print count of pairs
cout << count;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 3, 5, 6, 5 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count pairs from an
// array satisfying given conditions
static void countPairs(int []arr, int N)
{
// Stores the total
// count of pairs
int count = 0;
// Stores count of a[i]/i
Map mp
= new HashMap();
// Traverse the array
for(int i = 0; i < N; i++)
{
Double val = 1.0 * arr[i];
Double idx = 1.0 * (i + 1);
// Updating count
if (mp.containsKey(val / idx))
count += mp.get(val/idx);
// Update frequency
// in the Map
if (mp.containsKey(val / idx))
mp.put(val / idx, mp.getOrDefault(val / idx, 0) + 1);
else
mp.put(val/idx, 1);
}
// Print count of pairs
System.out.print(count);
}
// Driver Code
public static void main(String args[])
{
// Given array
int []arr = { 1, 3, 5, 6, 5 };
// Size of the array
int N = arr.length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
}
}
// This code is contributed by ipg2016107.
Python3
# Python3 program for the above approach
from collections import defaultdict
# Function to count pairs from an
# array satisfying given conditions
def countPairs(arr, N):
# Stores the total
# count of pairs
count = 0
# Stores count of a[i] / i
mp = defaultdict(int)
# Traverse the array
for i in range(N):
val = 1.0 * arr[i]
idx = 1.0 * (i + 1)
# Updating count
count += mp[val / idx]
# Update frequency
# in the Map
mp[val / idx] += 1
# Print count of pairs
print(count)
# Driver Code
if __name__ == "__main__":
# Given array
arr = [1, 3, 5, 6, 5]
# Size of the array
N = len(arr)
# Function call to count pairs
# satisfying given conditions
countPairs(arr, N)
# This code is contributed by ukasp
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count pairs from an
// array satisfying given conditions
static void countPairs(int []arr, int N)
{
// Stores the total
// count of pairs
int count = 0;
// Stores count of a[i]/i
Dictionary mp = new Dictionary();
// Traverse the array
for(int i = 0; i < N; i++)
{
double val = 1.0 * arr[i];
double idx = 1.0 * (i + 1);
// Updating count
if (mp.ContainsKey(val / idx))
count += mp[val/idx];
// Update frequency
// in the Map
if (mp.ContainsKey(val / idx))
mp[val / idx]++;
else
mp[val/idx] = 1;
}
// Print count of pairs
Console.WriteLine(count);
}
// Driver Code
public static void Main()
{
// Given array
int []arr = { 1, 3, 5, 6, 5 };
// Size of the array
int N = arr.Length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
}
}
// This code is contributed by SURENDRA_GANGWAR
输出:
2
时间复杂度: O(N)
辅助空间: O(N)