Finding subarrays in an array where length equals P * (sum of elements)

Question

How would we go about testing all combinations of subarrays in a array where length of each subarray is equal to P times the sum of subarray elements.

A brief example: Edit:

 A = [2,-1,3,0,1,2,1] , P =2

Desired result:

1. Length = 2,P * Sum of elements = 1 . Subarrays are [2,-1] , [0,1]

Edit Constraint :

N represents number of elements in an array
1 <= N <= 1e5
-1000 <= P <= 1000
|A[i]| <= 1e6

What kind of problem set(for eg:NP-hard?) does these kind of questions fall into ? Language : C#

Answer 1

This problem falls in P. Here's an O(n) solution.

Let's do some algebra with prefix sums:

j - i = p * (prefix_sum_j - prefix_sum_i)
j - i = p * prefix_sum_j - p * prefix_sum_i
j - p * prefix_sum_j = i - p * prefix_sum_i
p * prefix_sum_j - j = p * prefix_sum_i - i

JavaScript code with testing against brute force.

const f = (nums, p) =>
  nums.reduce(([map, sum, answer], val, i) => {    
    const newSum = sum + val;
    answer += p * newSum == i + 1;
    answer += map[p * newSum - i] || 0;
    map[p * newSum - i] = (map[p * newSum - i] || 0) + 1;
    return [map, newSum, answer];
  }, [{}, 0, 0])[2];


console.log('Input: [2,-1,3,0,1,2,1], 2')
console.log('Output: ' + f([2,-1,3,0,1,2,1], 2));


function bruteForce(A, p){
  let result = 0;

  for (let windowSize=1; windowSize<=A.length; windowSize++){
    for (let start=0; start<A.length-windowSize+1; start++){
      let sum = 0;

      for (let end=start; end<start+windowSize; end++)
        sum += A[end];
      
      if (p * sum == windowSize)
        result += 1;
    }
  }

  return result;
}


var numTests = 500;
var n = 20;
var m = 20;
var pRange = 10;

console.log('\nTesting against brute force...')

for (let i=0; i<numTests; i++){
  const A = new Array(n);
  for (let j=0; j<n; j++)
    A[j] = Math.floor(Math.random() * m) * [1, -1][Math.floor(Math.random()*2)];
  
  const p = Math.floor(Math.random() * pRange) * [1, -1][Math.floor(Math.random()*2)];

  _f = f(A, p);
  _brute = bruteForce(A, p);

  //console.log(String([_f, _brute, p, JSON.stringify(A)]));

  if (_f != _brute){
    console.log('MISMATCH!');
    console.log(p, JSON.stringify(A));
    console.log(_f, _brute);
    break;
  }
}

console.log('Done testing.')

In case it helps readers, the function, f, as a for loop could look like:

function f(A, p){
  const seen = {}; // Map data structure
  let sum = 0;
  let result = 0;
  
  for (let i=0; i<A.length; i++){
    sum += A[i];
    result += p * sum == i + 1; // Boolean evaluates to 1 or 0
    result += seen[p * sum - i] || 0;
    seen[p * sum - i] = (seen[p * sum - i] || 0) + 1;
  }
  
  return result;
}

My (first) attempt at C# code:

using System;
using System.Collections.Generic;

public class Solution{
  static int f(int[] A, int p){
    var seen = new Dictionary<int, int>();
    int sum = 0;
    int result = 0;
  
    for (int i=0; i<A.Length; i++){
      sum += A[i];
      result += Convert.ToInt32( p * sum == i + 1 );

      int key = p * sum - i;

      if (seen.ContainsKey(key)){
        result += seen[key];
        seen[key] += 1;

      } else {
        seen[key] = 1;
      }
    }
  
    return result;
  }


  public static void Main(){
    int[] A = new int[7]{2, -1, 3, 0, 1, 2, 1};
    int p = 2;

    Console.WriteLine(f(A, p));
  }
}

Answer 2

I tried to solve this problem using dynamic programming. In my solution I have used 2 nested for loops for making the dp matrix so it should have time complexity of O(n^2) (not counting the 3 nested for loops used to print the solution). Since this problem can be solved using brute force method as well as using dynamic programming in polynomial time it has P complexity.

using System;

public class Program
{
    public static void Main()
    {
        int n = 7;
        int p = 2;
        int[, ] arr = new int[n + 1, n + 1];
        int[] nums = new int[]{2, -1, 3, 0, 1, 2, 1};
        for (int i = 1; i <= n; i++)
        {
            for (int j = i; j <= n; j++)
            {
                arr[i, j] = arr[i - 1, j - 1] + nums[j - 1];
            }
        }

        for (int j = 0; j <= n; j++)
        {
            for (int k = 0; k <= n; k++)
            {
                Console.Write(string.Format("{0} ", arr[j, k]));
            }

            Console.Write(Environment.NewLine);
        }

        Console.Write(Environment.NewLine + Environment.NewLine);
        for (int i = 1; i <= n; i++)
        {
            Console.WriteLine(string.Format("For length {0}:  ", i));
            for (int j = i; j <= n; j++)
            {
                if (p * arr[i, j] == i)
                {
                    Console.Write(string.Format("{0} {1}: ", (j - i + 1), j));
                    for (int k = j - i + 1; k <= j; k++)
                    {
                        Console.Write(string.Format("{0},", nums[k - 1]));
                    }

                    Console.Write(Environment.NewLine);
                }
            }

            Console.Write(Environment.NewLine);
        }

        Console.Write(Environment.NewLine);
    }
}

you can test this code on dotnetfiddle (this is the first c# code I have ever written so maybe some more optimizations could be done in code).