c# - 在数组中查找长度等于 P *(元素总和)的子数组

Question

我们将如何测试数组中子数组的所有组合，其中
每个子数组的长度等于子数组元素总和的 P 倍。
一个简单的例子:
编辑:

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

想要的结果:

长度 = 2, 电话 * 元素总和 = 1 。子数组是 [2,-1] , [0,1]

编辑
约束:

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

这些问题属于什么样的问题集(例如:NP-hard？)？语言:C#

Answer 1

这个问题属于P .这是一个 O(n)解决方案。
让我们用前缀和做一些代数:

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

带有蛮力测试的 JavaScript 代码。

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.')

如果对读者有帮助，该函数f ，作为 for循环看起来像:

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;
}

我的(第一次)尝试 C# 代码:

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));
  }
}