总和为完全平方的子数组的计数

给定一个包含正元素和负元素的数组arr[] ，任务是计算总和为完全平方数的所有子数组。
例子：

Input: arr[] = {2, 3, -5, 6, -7, 4}; 
Output: 5 
Explanation: 
Subarrays {2, 3, -5}, {-5, 6}, {3, -5, 6}, {3, -5, 6, -7, 4} and {4} with sum is 0, 1, 4, 1 and 4 respectively have perfect square sum.
Input: arr[] = {3, -6, 4, -2, 7}; 
Output: 3 
Explanation: {3, -6, 4}, {4}, {4, -2, 7} are the subarrays with perfect square sum. 

天真的方法：
一个简单的解决方案是生成所有可能的子数组。在遍历时，跟踪子数组和。保持总和为完美平方的所有子数组的计数。
有效的解决方案：想法是使用前缀和数组来解决给定的问题。

• 创建一个prefixSum 数组并存储它的前缀和。
• 遍历prefixSum 数组并确定它的最小值，即（ prefixMin ）。
• 现在，创建一个无序映射，可用于存储当前 prefixSum 的频率，同时遍历prefixSum 数组。
• 用值 1 初始化映射的第 0 个键索引，因为 0 是一个完美的正方形。
• 使用嵌套循环遍历prefixSum 数组。
• 对于每个 prefixSum 元素，嵌套循环将查找mapKey = (prefixSum[i] – j*j) （如果在地图索引中可用） 。
• 如果(prefixSum[i] – j*j)在地图中已经可用，我们用(prefixSum[i] – j*j)的索引值更新我们的计数器。
• 这个想法是用所有的平方 (j*j) 检查当前的 prefixSum 值，直到差异达到prefixMin 。
• 现在，在外循环的每次迭代中，将带有当前 prefixSum索引的映射增加 1。
• 基本概念是我们继续从(prefixSum[i] – j*j ) 搜索，因为如果数组的一部分是(prefixSum[i] – j*j ) ，那么数组的另一部分将是(j *j)即一个完美的平方和。
• 您可以在上图中看到 totalSum 实际上是用于此目的的 prefixSum。

下面是上述方法的实现：

C++

// C++ code for the above approach.
#include 
using namespace std;
 
#define lli long long int
 
// Function to find count of subarrays
// whose sum is a perfect square.
lli countSubarrays(int arr[],
                   int n)
{
    // to search for index with
    // (current prefix sum - j*j)
    unordered_map mp;
 
    // storing the prefix sum
    int prefixSum[n];
 
    // used to track the minimum
    // value in prefixSum
    int prefixMin = 0;
 
    prefixSum[0] = arr[0];
    prefixMin = min(prefixMin,
                    prefixSum[0]);
 
    // Calculating the prefixSum
    // and tracking the prefixMin
    for (int i = 1; i < n; i++) {
 
        prefixSum[i] = prefixSum[i - 1]
                       + arr[i];
 
        // below statement is used if
        // array contains
        // negative numbers
        prefixMin = min(prefixMin,
                        prefixSum[i]);
    }
 
    // counts the no of subarrays
    // with perfect square sum
    lli countSubs = 0;
 
    // as 0 is a perfect square,
    // so we initialize 0th
    // index-key with value 1
    mp[0] = 1;
 
    // Here we count the perfect
    // square subarray sum by
    // searching if there is a
    // prefix with
    // sum = (current prefixSum - (sq*sq))
    for (int i = 0; i < n; i++) {
        for (int j = 0;
             prefixSum[i] - j * j >= prefixMin;
             j++) {
 
            if (mp.find(prefixSum[i] - j * j)
                != mp.end())
 
                // increasing our subarray count
                countSubs += mp[prefixSum[i]
                                - j * j];
        }
 
        // increasing the current prefixSum
        // index value in map by 1 to count
        // the other perfect squares while
        // traversing further
        mp[prefixSum[i]]++;
    }
 
    return countSubs;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, -5,
                  6, -7, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    lli ans = countSubarrays(arr, n);
 
    // printing the result
    cout << ans;
 
    return 0;
}

Java

// Java code for
// the above approach.
import java.util.*;
class GFG{
   
// Function to find count of
// subarrays whose sum is
// a perfect square.
static long countSubarrays(int arr[],
                           int n)
{
  // To search for index with
  // (current prefix sum - j*j)
  HashMap mp = new HashMap();
 
  // Storing the prefix sum
  int []prefixSum = new int[n];
 
  // Used to track the minimum
  // value in prefixSum
  int prefixMin = 0;
 
  prefixSum[0] = arr[0];
  prefixMin = Math.min(prefixMin,
                       prefixSum[0]);
 
  // Calculating the prefixSum
  // and tracking the prefixMin
  for (int i = 1; i < n; i++)
  {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
 
    // Below statement is used if
    // array contains
    // negative numbers
    prefixMin = Math.min(prefixMin,
                         prefixSum[i]);
  }
 
  // Counts the no of subarrays
  // with perfect square sum
  long countSubs = 0;
 
  // As 0 is a perfect square,
  // so we initialize 0th
  // index-key with value 1
  mp.put(0, 1);
 
  // Here we count the perfect
  // square subarray sum by
  // searching if there is a
  // prefix with
  // sum = (current prefixSum - (sq*sq))
  for (int i = 0; i < n; i++)
  {
    for (int j = 0;
             prefixSum[i] - j *
             j >= prefixMin; j++)
    {
      if (mp.containsKey(prefixSum[i] - j * j))
 
        // Increasing our subarray count
        countSubs += mp.get(prefixSum[i] -
                            j * j);
    }
 
    // Increasing the current prefixSum
    // index value in map by 1 to count
    // the other perfect squares while
    // traversing further
    if(mp.containsKey(prefixSum[i]))
    {
      mp.put(prefixSum[i],
      mp.get(prefixSum[i]) + 1);
    }
    else
    {
      mp.put(prefixSum[i], 1);
    }
  }
 
  return countSubs;
}
 
// Driver code
public static void main(String[] args)
{
  int arr[] = {2, 3, -5,
               6, -7, 4};
  int n = arr.length;
  long ans = countSubarrays(arr, n);
 
  // Printing the result
  System.out.print(ans);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 code for the above approach.
from collections import defaultdict
 
# Function to find count of subarrays
# whose sum is a perfect square.
def countSubarrays(arr, n):
     
    # To search for index with
    # (current prefix sum - j*j)
    mp = defaultdict(lambda:0)
     
    # Storing the prefix sum
    prefixSum = [0] * n
     
    # Used to track the minimum
    # value in prefixSum
    prefixMin = 0
     
    prefixSum[0] = arr[0]
    prefixMin = min(prefixMin, prefixSum[0])
     
    # Calculating the prefixSum
    # and tracking the prefixMin
    for i in range(1, n):
        prefixSum[i] = prefixSum[i - 1] + arr[i]
         
        # Below statement is used if
        # array contains negative numbers
        prefixMin = min(prefixMin, prefixSum[i])
         
    # Counts the no of subarrays
    # with perfect square sum
    countSubs = 0
     
    # As 0 is a perfect square,
    # so we initialize 0th
    # index-key with value 1
    mp[0] = 1
     
    # Here we count the perfect
    # square subarray sum by
    # searching if there is a
    # prefix with
    # sum = (current prefixSum - (sq*sq))
    for i in range(n):
        j = 0
         
        while prefixSum[i] - j * j >= prefixMin:
            if prefixSum[i] - j * j in mp:
                 
                # Increasing our subarray count
                countSubs += mp[prefixSum[i] - j * j]
            j += 1
             
        # Increasing the current prefixSum
        # index value in map by 1 to count
        # the other perfect squares while
        # traversing further
        mp[prefixSum[i]] += 1
         
    return countSubs
 
# Driver code
arr = [ 2, 3, -5, 6, -7, 4 ]
n = len(arr)
ans = countSubarrays(arr, n)
     
# Printing the result
print(ans)
 
# This code is contributed by Shivam Singh

C#

// C# code for
// the above approach.
using System;
using System.Collections.Generic;
class GFG{
   
// Function to find count of
// subarrays whose sum is
// a perfect square.
static long countSubarrays(int []arr,
                           int n)
{
  // To search for index with
  // (current prefix sum - j*j)
  Dictionary mp =
             new Dictionary();
 
  // Storing the prefix sum
  int []prefixSum = new int[n];
 
  // Used to track the minimum
  // value in prefixSum
  int prefixMin = 0;
 
  prefixSum[0] = arr[0];
  prefixMin = Math.Min(prefixMin,
                       prefixSum[0]);
 
  // Calculating the prefixSum
  // and tracking the prefixMin
  for (int i = 1; i < n; i++)
  {
    prefixSum[i] = prefixSum[i - 1] +
                   arr[i];
 
    // Below statement is used if
    // array contains
    // negative numbers
    prefixMin = Math.Min(prefixMin,
                         prefixSum[i]);
  }
 
  // Counts the no of subarrays
  // with perfect square sum
  long countSubs = 0;
 
  // As 0 is a perfect square,
  // so we initialize 0th
  // index-key with value 1
  mp.Add(0, 1);
 
  // Here we count the perfect
  // square subarray sum by
  // searching if there is a
  // prefix with
  // sum = (current prefixSum -
  // (sq*sq))
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; prefixSum[i] - j *
             j >= prefixMin; j++)
    {
      if (mp.ContainsKey(prefixSum[i] -
                         j * j))
 
        // Increasing our subarray count
        countSubs += mp[prefixSum[i] -
                        j * j];
    }
 
    // Increasing the current prefixSum
    // index value in map by 1 to count
    // the other perfect squares while
    // traversing further
    if(mp.ContainsKey(prefixSum[i]))
    {
      mp[prefixSum[i]]++;   
    }
    else
    {
      mp.Add(prefixSum[i], 1);
    }
  }
 
  return countSubs;
}
 
// Driver code
public static void Main(String[] args)
{
  int []arr = {2, 3, -5,
               6, -7, 4};
  int n = arr.Length;
  long ans = countSubarrays(arr, n);
 
  // Printing the result
  Console.Write(ans);
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript code for
// the above approach.
 
// Function to find count of
// subarrays whose sum is
// a perfect square.
function countSubarrays(arr, n)
{
  // To search for index with
  // (current prefix sum - j*j)
  let mp = new Map();
  
  // Storing the prefix sum
  let prefixSum = Array.from({length: n}, (_, i) => 0);
  
  // Used to track the minimum
  // value in prefixSum
  let prefixMin = 0;
  
  prefixSum[0] = arr[0];
  prefixMin = Math.min(prefixMin,
                       prefixSum[0]);
  
  // Calculating the prefixSum
  // and tracking the prefixMin
  for (let i = 1; i < n; i++)
  {
    prefixSum[i] = prefixSum[i - 1] + arr[i];
  
    // Below statement is used if
    // array contains
    // negative numbers
    prefixMin = Math.min(prefixMin,
                         prefixSum[i]);
  }
  
  // Counts the no of subarrays
  // with perfect square sum
  let countSubs = 0;
  
  // As 0 is a perfect square,
  // so we initialize 0th
  // index-key with value 1
  mp.set(0, 1);
  
  // Here we count the perfect
  // square subarray sum by
  // searching if there is a
  // prefix with
  // sum = (current prefixSum - (sq*sq))
  for (let i = 0; i < n; i++)
  {
    for (let j = 0;
             prefixSum[i] - j *
             j >= prefixMin; j++)
    {
      if (mp.has(prefixSum[i] - j * j))
  
        // Increasing our subarray count
        countSubs += mp.get(prefixSum[i] -
                            j * j);
    }
  
    // Increasing the current prefixSum
    // index value in map by 1 to count
    // the other perfect squares while
    // traversing further
    if(mp.has(prefixSum[i]))
    {
      mp.set(prefixSum[i],
      mp.get(prefixSum[i]) + 1);
    }
    else
    {
      mp.set(prefixSum[i], 1);
    }
  }
  
  return countSubs;
}
 
// Driver code
     
      let arr = [2, 3, -5,
               6, -7, 4];
  let n = arr.length;
  let ans = countSubarrays(arr, n);
  
  // Prleting the result
  document.write(ans);
 
// This code is contributed by souravghosh0416.

输出：

5

时间复杂度： O(N * sqrt(K))
辅助空间： O(N)

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