计算数组中的对(I，j)的数量，使得 arr[i] * j = arr[j] * i
原文:https://www . geesforgeks . org/count-对数-I-j-from-a-arri-j-arrj-I/
给定大小为 N 的数组 arr[] ，任务是计算数组中可能的对 (i，j) 的数量，使得 arr[j] * i = arr[i] * j ，其中 1 ≤ i < j ≤ N.
示例:
输入: arr[] = {1，3，5，6，5} 输出: 2 说明: 对(1，5)满足条件，因为 arr[1] * 5 = arr[5] * 1。 对(2，4)满足条件，因为 arr[2] * 4 = arr[4] * 2。 因此，满足给定条件的对的总数是 2。
输入: arr[] = {2，1，3 } T3】输出: 0
天真方法:解决问题最简单的方法是从数组中生成所有可能的对，并检查每个对，给定的条件是否满足。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(N2) 辅助空间: O(1)
有效方法:为了优化上述方法，该思想基于给定等式 arr[i] * j = arr[j] * i 到 arr[i] / i = arr[j] / j 的重排。 按照以下步骤解决问题:
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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<double, int> 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 语言(一种计算机语言，尤用于创建网站)
// 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<Double, Integer> mp
= new HashMap<Double, Integer>();
// 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.
Python 3
# 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<double,
int> mp = new Dictionary<double,
int>();
// 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
java 描述语言
<script>
// Javascript program for the above approach
// Function to count pairs from an
// array satisfying given conditions
function countPairs(arr, N)
{
// Stores the total
// count of pairs
var count = 0;
// Stores count of a[i] / i
var mp = new Map();
// Traverse the array
for (var i = 0; i < N; i++) {
var val = 1.0 * arr[i];
var idx = 1.0 * (i + 1);
// Updating count
count += mp.has(val/idx)?mp.get(val/idx):0
// Update frequency
// in the Map
if(mp.has(val/idx))
mp.set(val/idx, mp.get(val/idx)+1)
else
mp.set(val/idx, 1)
}
// Print count of pairs
document.write( count);
}
// Driver Code
// Given array
var arr = [1, 3, 5, 6, 5];
// Size of the array
var N = arr.length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
// This code is contributed by itsok.
</script>
Output:
2
时间复杂度:O(N) T5辅助空间:** O(N)
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