Given a sorted array arr[], and a set Q having M queries, where each query has values X and Y, the task is to find the sum of all integers less than X and greater than Y present in the array.
Note: X and Y may or may not be present in the array.
Examples:
Input: arr[] = [3 5 8 12 15], Q = {{5, 12}, {4, 8}}
Output:
18
30
Explanation:
For query 1, X = 5 and Y = 12. Sum = 3( < 5) + 15( > 12) = 18.
For query 2, X = 4 and Y = 8. Sum = 3( < 4) + 15 ( > 8) + 12 ( > 8) = 30.Input: arr[] = [4 7 7 12 15], Q = {{3, 8}, {4, 8}}
Output:
27
27
Explanation:
For query 1, X = 3 and Y = 8. Sum = 12( > 8) + 15 ( > 8) = 27.
For query 2, Sum = 12 + 15 = 27.
Approach:
- Build a prefix sum array where prefix_sum[i] denotes the sum of arr[0] + arr[1] + … arr[i].
- Find the last index i which has a value less than X and extract the prefix sum up to the ith index. The index can be obtained in O(logN) complexity by using bisect_left() or lower_bound() in Python and C++ respectively..
- Similarly, find the first index i in the array with a value greater than Y and calculate the sum prefix_sum[n-1] – prefix_sum[i-1]. Inbuilt functions bisect() and upper_bound() in Python and C++ respectively, perform the required operation in O(logN).
- Sum of the two results calculated in the above two steps is the required answer. Keep repeating these steps for every query.
Below is the implementation of the above approach:
Python3
# Python code for the above program from bisect import bisect, bisect_left def createPrefixSum(ar, n): # Initialize prefix sum # array prefix_sum = [0]*n # Initialize prefix_sum[0] # by ar[0] prefix_sum[0] = ar[0] # Compute prefix sum for # all indices for i in range(1, n): prefix_sum[i] = prefix_sum[i-1]+ar[i] return prefix_sum # Function to return sum of all # elements less than X def sumLessThanX(ar, x, prefix_xum): # Index of the last element # which is less than x pos_x = bisect_left(ar, x) - 1 if pos_x >= 0 : return prefix_sum[pos_x] # If no element is less than x else: return 0 # Function to return sum of all # elements greater than Y def sumGreaterThanY(ar, y, prefix_sum): # Index of the first element # which is greater than y pos_y = bisect(ar, y) pos_y -= 1 if pos_y < len(ar)-1 : return prefix_sum[-1]-prefix_sum[pos_y] # If no element is greater than y else: return 0 def solve(ar, x, y, prefix_sum): ltx = sumLessThanX(ar, x, prefix_sum) gty = sumGreaterThanY(ar, y, prefix_sum) # printing the final sum print(ltx + gty) def print_l(lb, ub): print("sum of integers less than {}".format(lb) + " and greater than {} is ".format(ub), end = '') if __name__ == '__main__': # example 1 ar = [3, 6, 6, 12, 15] n = len(ar) prefix_sum = createPrefixSum(ar, n) # for query 1 q1x = 5 q1y = 12 print_l(q1x, q1y) solve(ar, q1x, q1y, prefix_sum) # for query 2 q2x = 7 q2y = 8 print_l(q2x, q2y) solve(ar, q2x, q2y, prefix_sum) |
sum of integers less than 5 and greater than 12 is 18 sum of integers less than 7 and greater than 8 is 42
Time Complexity: O(N + (M * logN))
Auxiliary Space complexity: O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
