Last Updated: 11 Nov, 2020
Sum Of Infinite Array
Moderate
Asked in company
Given an array “A” of N integers and you have also defined the new array “B” as a concatenation of array “A” for an infinite number of times.
For example, if the given array “A” is [1,2,3] then, infinite array “B” is [1,2,3,1,2,3,1,2,3,.......].
Now you are given Q queries, each query consists of two integers “L“ and “R”(1-based indexing). Your task is to find the sum of the subarray from index “L” to “R” (both inclusive) in the infinite array “B” for each query.
Note :
The value of the sum can be very large, return the answer as modulus 10^9+7.
Input format :
The first line of input contains a single integer T, representing the number of test cases or queries to be run.
Then the T test cases follow.
The first line of each test case contains a single integer N, denoting the size of the array “A”.
The second line of each test case contains N single space-separated integers, elements of the array “A”.
The third line of each test case contains a single integer Q, denoting the number of queries.
Then each of the Q lines of each test case contains two single space-separated integers L, and R denoting the left and the right index of the infinite array “B” whose sum is to be returned.
Output format :
For each test case, print Q space-separated integers that denote the answers of the given Q queries.
Print the answer to each test case in a separate line.
Note :
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints :
1 <= T <= 100
1 <= N <= 10^4
1 <= A[i] <= 10^9
1 <= Q <= 10^4
1 <= L <= R <= 10^18
Time Limit: 1sec
Approaches
- Instead of creating a new infinite array B which has a repeated array A elements in the form [A1, A2,... AN, A1, A2,... AN, A1, A2,... AN…....]. We will traverse array A, again and again, to find the sum as array A is only repeating in infinite array B.
- So the brute force approach is, for each query,
- we run a loop from L to R, and for each index i, add the value at index (i%N) of the array A i.e A[i%N] to sum. So this way we can find the sum of the required subarray from index L to R in an infinite array B.
The better idea is to first create the sum array in which sumArray[i] stores the sum from (A[0]+....+A[i]). Now instead of iterating from L to R like in the previous approach, we find the sum using our sumArray[].
Let’s take an example, where array A = [1,2,3] and we have to find the sum of the subarray of the infinite array(as shown in below fig) from index 3 to 10 (0-based indexing).
- Now instead of iterating from 3 to 10 and then calculate the sum. We can observe one thing that we are going to traverse the array A again and again so instead of doing this, we can first find the sum from index 0 to index 10 say a, and then find the sum from index 0 to 2 say b, then subtract b from a as a-b, which is the sum of subarray from index 3 to 10 in an infinite array B.
- Now to find the sum, from index 0 to any index X, we first find how many number of times the given array A can comes completely upto index X. which can be simply found by X / N say count , and sum will be count * sumArray[N] where N is the length of array A. Now for the remaining part of the subarray sum can be found by sumArray[ (X % N)].
- Consider array A = {1, 2, 3} and we have to find the sum between L= 1 and R = 5. Then Till index 5 the array A repeats itself one time i.e {1, 2, 3, 1, 2} which can be calculated as R/N i.e 1 and the remaining elements till index 5 are {1, 2} which can be calculated as R%N. So the sum till index 5 is R/N * sumArray(N) + sumArray(R%N).
So this way we can find the sum without iterating from L to R.