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Find count of subarrays having sum in a given range in less than O(n^2).
By code_fille, history, 11 years ago, In English

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.

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11 years ago, hide # |
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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 ith 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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    11 years ago, hide # ^ |
     
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    How could segment tree be used to find the possible values of j?

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    11 years ago, hide # ^ |
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    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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      11 years ago, hide # ^ |
       
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      Could you elaborate how divide and conquer could be used?

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        11 years ago, hide # ^ |
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        B — partial sum, B[0] = 0.

        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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    9 years ago, hide # ^ |
     
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    For segment tree , Is the time complexity is (n*log(sum of all array elements)) ? correct me if i am wrong.

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    8 years ago, hide # ^ |
     
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    Your solution is quite beautiful :)

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