# Program to find kpr sum for all queries for a given list of numbers in Python

Suppose we have a list of numbers nums. We also have a list of queries where queries[i] contains three elements [k, p, r], for each query we shall have to find kpr_sum. The formula for kpr_sum is like below.

$$\mathrm{{𝑘𝑝𝑟}\_{𝑠𝑢𝑚} =\sum_{\substack{𝑖=𝑃}}^{𝑅−1}\sum_{\substack{𝑗=𝑖+1}}^{𝑅}(𝐾 ⊕(𝐴[𝑖]⊕𝐴[𝑗]))}$$

If the sum is too large, then return sum modulo 10^9+7.

So, if the input is like nums = [1,2,3] queries = [[1,1,3],[2,1,3]], then the output will be [5, 4] because for the first element it is (1 XOR (1 XOR 2)) + (1 XOR (1 XOR 3)) + (1 XOR (2 XOR 3)) = 5, similarly for second query, it is 4.

To solve this, we will follow these steps −

• m := 10^9 + 7
• N := size of nums
• q_cnt := size of queries
• C := a new list
• res := a new list
• for i in range 0 to 19, do
• R := an array with single element 0
• t := 0
• for each x in nums, do
• t := t + (x after shifting i times to the right) AND 1
• insert t at the end of R
• insert R at the end of C
• for j in range 0 to q_cnt, do
• (K, P, R) := queries[j]
• d := R - P + 1
• t := 0
• for i in range 0 to 19, do
• n1 := C[i, R] - C[i, P-1]
• n0 := d - n1
• if (K after shifting i times to the right) AND 1 is non-zero, then
• x := quotient of (n1 *(n1 - 1) + n0 *(n0 - 1))/2
• otherwise,
• x := n1 * n0
• t :=(t +(x after shifting i times to the left)) mod m
• insert t at the end of res
• return res

## Example

Let us see the following implementation to get better understanding −

def solve(nums, queries):
m = 10**9 + 7
N = len(nums)
q_cnt = len(queries)
C = []
res = []
for i in range(20):
R = [0]
t = 0
for x in nums:
t += (x >> i) & 1
R.append(t)
C.append(R)
for j in range(q_cnt):
K, P, R = queries[j]
d = R - P + 1
t = 0
for i in range(20):
n1 = C[i][R] - C[i][P-1]
n0 = d - n1
if (K >> i) & 1:
x = (n1 * (n1 - 1) + n0 * (n0 - 1)) >> 1
else:
x = n1 * n0
t = (t + (x << i)) % m
res.append(t)

return res

nums = [1,2,3]
queries = [[1,1,3],[2,1,3]]
print(solve(nums, queries))

## Input

[1,2,3], [[1,1,3],[2,1,3]]

## Output

[5, 4]