Q查询的所有小于X且大于Y的数组元素的总和

给定一个已排序的数组arr []和一个具有M个查询的集合Q ，其中每个查询具有值X和Y ，任务是查找数组中存在的所有小于X且大于Y的整数的总和。
注意： X和Y可能会或可能不会出现在数组中。

例子：

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.

方法：

1. 构建一个前缀和数组，其中prefix_sum [i]表示arr [0] + arr [1] +…arr [i]的和。
2. 找到值小于X的最后一个索引i ，然后提取前缀总和直到第i^个索引。通过分别在Python和C++中使用bisect_left()或lower_bound()，可以以O(logN)复杂度获得索引。
3. 同样，在数组中找到第一个索引i ，其值大于Y，然后计算出总和prefix_sum [n-1] – prefix_sum [i-1] 。分别在Python和C++中的内置函数bisect()和upper_bound()在O(logN)中执行所需的操作。
4. 在以上两个步骤中计算出的两个结果的总和是必需的答案。继续对每个查询重复这些步骤。

下面是上述方法的实现：

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)

输出：

sum of integers less than 5 and greater than 12 is 18
sum of integers less than 7 and greater than 8 is 42

时间复杂度： O(N +(M * logN))
辅助空间复杂度： O(N)