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Range Sum using Segment Tree

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Given an integer array "nums" and two integers "lower" and "upper", return the number of range sums that lie in "[lower, upper]" inclusive.

A range sum "S(i, j)" is defined as the sum of the elements in "nums" between indices "I" and "j" inclusive: "nums[i] + nums[i+1] + ... + nums[j]".

Example 1
    
  
  • Input: nums = [-2, 5, -1], lower = -2, upper = 2
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  • Output: 3
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  • Explanation: 
    We need to count the number of range sums that lie in [-2, 2]. 
    The range sums are: S(0, 0) = -2, which is in the range [-2, 2]; 
    S(0, 1) = -2 + 5 = 3, which is NOT in the range [-2, 2]; 
    S(0, 2) = -2 + 5 + (-1) = 2, which is in the range [-2, 2]; 
    S(1, 1) = 5, which is NOT in the range [-2, 2]; 
    S(1, 2) = 5 + (-1) = 4, which is NOT in the range [-2, 2]; S(2, 2) = -1, which is in the range [-2, 2]. 
    There are 3 range sums that lie in [-2, 2].
    •
Example 2
    
  
  • Input: nums = [0, -3, 2, -3, 4], lower = 3, upper = 5
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  • Output: 2
    • 
  
  • Explanation: 
    We need to count the number of range sums that lie in [3, 5]. 
    The range sums that lie in [3, 5] are: S(2, 4) = 2 + (-3) + 4 = 3, and S(4, 4) = 4. 
    There are 2 range sums that lie in [3, 5].
    •
Constraints
    
  
  • 1 <= nums.length <= 10^4
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  • -2^31 <= nums[i] <= 2^31 - 1
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  • -10^5 <= lower <= upper <= 10^5
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  • The sum of elements in any range [i, j] will not exceed 2^31 - 1
    • 
  
  • Time Complexity: O(n log n), where n is the length of the array
    • 
  
  • Space Complexity: O(n)
    •
ArraysTreeAccentureGoldman Sachs
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Hint
Solution

Solution Hints

    
  
  • Calculate the prefix sums of the array to make range sum queries faster
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  • Use a segment tree to efficiently query and update range sums
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  • For each prefix sum, count how many previous prefix sums fall within the range [prefixSum - upper, prefixSum - lower]
    • 
  
  • The difference between two prefix sums gives the sum of elements between those indices
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  • We need to handle potential overflow due to large integers
    •

Steps to solve by this approach:

 Step 1: Calculate the prefix sums of the array to make range sum queries faster.

 Step 2: Sort all the prefix sums and create a mapping from values to indices (with deduplication).

 Step 3: Implement a segment tree data structure to efficiently count elements within a range.

 Step 4: For each prefix sum, find how many previous prefix sums fall within the range [prefixSum - upper, prefixSum - lower].

 Step 5: Update the segment tree with the current prefix sum after counting.

 Step 6: Use TreeMap to map prefix sum values to indices efficiently.

 Step 7: Be careful with potential integer overflow by using long type for prefix sums.

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