Count of subarrays whose sum is a perfect square

Given an array arr[] with positive and negative elements, the task is to count all subarrays whose sum is a perfect square.

Examples:

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. 

Naive Approach:
A simple solution would be to generate all possible subarrays. While traversing, keep track of the subarray sum. Keep a count of all subarrays whose sum is a perfect square.

Efficient Solution: The idea is to use a prefix sum array to solve the given problem.  

• Create a prefixSum array and store it’s prefix sum.
• Traverse the prefixSum array and identify it’s minimum value i.e (prefixMin).
• Now, create an unordered map that can be used to store the frequency of current prefixSum, while traversing the prefixSum array.
• Initialize the 0th key-index of the map with value 1, as 0 is a perfect square.
• Traverse the prefixSum array with a nested loop.
• For each prefixSum element, the nested loop is going to find the mapKey = (prefixSum[i] – j*j), if available in the map index.
• If (prefixSum[i] – j*j) is already available in the map, we update our counter with the index value of (prefixSum[i] – j*j).
• The idea is to check the current prefixSum value with all the squares (j*j) till the difference reaches prefixMin.
• Now, increment the map with index of the current prefixSum by 1 with every iteration of the outer loop.
• The underlying concept is that we keep searching from (prefixSum[i] – j*j ) because, if one part is the array is (prefixSum[i] – j*j ), then the other part of the array would be (j*j) i.e a perfect square sum.
• You can see in the above diagram that the totalSum is actually the prefixSum, which is used for that purpose.

Below is the implementation of the above approach:

C++

// C++ code for the above approach.
#include <bits/stdc++.h>
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<int, int> 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<Integer,
          Integer> mp = new HashMap<Integer,
                                    Integer>();

  // 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<int,
             int> mp =
             new Dictionary<int,
                            int>();

  // 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

<script>

// 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);

  // Printing the result
  document.write(ans);

// This code is contributed by souravghosh0416.
</script>

Output:

5

Time Complexity: O(N * sqrt(K))
Auxiliary Space: O(N)