Find count of subarrays having sum in a given range in less than O(n^2).

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Автор code_fille, история, 11 лет назад, По-английски

Given an array of integers and a range find the count of subarrays whose sum lies in the given range. Do so in less than O(n^2) complexity.

subarray

code_fille
11 лет назад
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RedNextCentury

11 лет назад, скрыть # |

← Rev. 6 →

+10

let us call the range [ l, r ] build the prefix sum array for the given array, let us call it B (B[0]=0 for empty prefix)

now
B[i] - B[j] where 0 ≤ j < i
gives all the sums of subarrays that end in i^th element

So, in order for the sum of subarray [ j + 1...i ] to be between l and r the following should be true:

l ≤ B[i] - B[j] ≤ r
=>
l - B[i] ≤ - B[j] ≤ r - B[i]
=>
B[i] - l ≥ B[j] ≥ B[i] - r

so for each B[i], we should find the number of B[j] s that are in the previous range and j < i, which is done in a segment tree

total complexity: O(nlogn)

NOTE If the values are all positive you can use binary search for O(nlogn)

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code_fille

11 лет назад, скрыть # ^ |

How could segment tree be used to find the possible values of j?

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RedNextCentury

11 лет назад, скрыть # ^ |

← Rev. 3 →

Array C , C[i] is the number of B[j]'s which are equal to i
You need a segment tree to perform 2 types of queries:
1) add 1 to C[x]
2) get the sum of C[x] : s<=x<=e

So for each B[i] you query for the sum of C[x] between B[i]-r and B[i]-l Then you add 1 to C[B[i]]

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Dazzler

8 лет назад, скрыть # ^ |

B[i] can be negative ..how it is possible to store in array?

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Vasiljko

8 лет назад, скрыть # ^ |

Then compress the values

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Dazzler

8 лет назад, скрыть # ^ |

I was trying to solve this question. I tried to compress the values. But it is giving WA. Here is the code. Please help me.

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Vasiljko

8 лет назад, скрыть # ^ |

You will have at most N+2 different values (Including L and R), compress them.

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Dazzler

8 лет назад, скрыть # ^ |

Can you please have a glance at my code?

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Dazzler

8 лет назад, скрыть # ^ |

For each B[i], I have to put B[i]-l and B[i]-r. I have to compress n*2*n values.

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zephyr_23

8 лет назад, скрыть # ^ |

Were you able to solve that question ? If yes, can you please share the code? I am also getting W.A.

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Dazzler

8 лет назад, скрыть # ^ |

Yes, I solved it Here is the code

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SashaMN

11 лет назад, скрыть # ^ |

← Rev. 2 →

If the values are all positive you can solve in O(N) using only 3 pointers.

Note: You can also solve in O(NlogN) without segment tree, just using divide-and-conquer idea.

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code_fille

11 лет назад, скрыть # ^ |

Could you elaborate how divide and conquer could be used?

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SashaMN

11 лет назад, скрыть # ^ |

← Rev. 3 →

B — partial sum, B[0] = 0.

$\text{[math]}$

Now we are going to split our verctor a into 2 parts with size N / 2 each.

2 cases are possible:

1) l, r < N / 2 or l, r > N / 2: We can find the answer recursively for the left part and for the right part.

2) l < N / 2 and r > N / 2: We can find the answer in O(N) using 3 pointers but only if the left part sorted and the right part sorted too. But you can maintain these parts sorted by using merge sort algorithm.

The algorithm is very similar to the algorithm for counting inversions in O(NlogN) time. You can find the explanation on coursera: link

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