Hackerrank The subarray sums question - time out test cases - algorithm

Question

Here is one question from hackerrank, I have a solution but there is some testcase failed because time limit exceeded. I don't know the better solution for it.
Find Sum of elements in a subarray (if in subarray has 0, sum = sum + number x)
input:
numbers: main array(1-indexed)
queries:
array of query: left index, right index, number x(0-indexed)
output:
array of sum corresponding with queries.
my solution on C++ code:
vector<long> findSum(vector<int>numbers, vector<vector<int>> queries)
{
vector<long>result;
long sum = 0;
int count = 0;
for(auto i : queries)
{
sum = 0;
count = 0;
int l = i[0] - 1;
int r = i[1]-1;
int x = i[2];
for(int j =l; j<r;j++)
{
sum+=numbers[j]==0?x:numbers[j];
}
result.push_back(sum);
}
return result;
}

Answer 1

As suggested by #Annity, you need to maintain two arrays:
Sum of all numbers so far. At any point of the index, it should have the sum of all previous numbers.
Same as above but it should have the total count of all previous Zeros.
You should avoid nested loops to reduce time complexity. Here is a solution in javascript:
function  findSum(numbers,  queries)  {
let  result  =   [];    
let  subArraySum  =   [];    
let  countZero  =  numbers[0]  ==  0  ?  1  :  0;    
let  zeroArr  =   [];    
zeroArr[0] =  countZero;    
subArraySum[0]  =  numbers[0];    
for (let  i = 1;  i <=  numbers.length - 1;  i++)  {               
if (numbers[i]  ==  0) {            
countZero++;            
zeroArr[i]  =  countZero;        
} 
else  {            
zeroArr[i]  =  countZero;         
}         
subArraySum[i]  = subArraySum[i - 1]  +  numbers[i];    
}    
for (let  q  of  queries)  {            
if (q.length  ==  3)  {            
let  i  =  q[0];            
let  j  =  q[1];            
let  x  =  q[2];            
let  sum  =  0;                       
sum  =  subArraySum[j - 1] - ((i - 2  <  0 )  ?  0  :  subArraySum[i - 2]);             
sum  =  sum  +  (zeroArr[j - 1] - ((i - 2  <  0 )  ?  0  :  zeroArr[i - 2])) * x;            
result.push(sum);        
}                  
}    
return  result;
}
console.log(findSum([5, 10, 15], [
[1, 2, 5]
]))
console.log(findSum([0, 10, 15], [
[1, 2, 5]
]))

Answer 2

Here's the solution that occurred to me.
You need to create two new arrays - one for the sums and one for the number of zeroes. Eg. sums[] and zeroes[].
In these two arrays you will store the value for the sum and the number of zeroes from the first element to the current. This is how it goes:
Loop through all of the numbers.
For each number with index i save in sums[i] the cumulative sum of all the elements from the first to the current and in zeroes[i] - the number of zeroes from the first element to the current.
Then loop through all of the queries.
For every query calculate the sum:
You take the sum from the sums array for the element with the right index and deduct the sum for the element before the left index.
(sum = sums[r] - sums[l-1])
If you deduct the count of zeroes in the element before the left index from the number in the element on the right index - you will get the number of zeroes in the interval.
(zero = zeroes[r] - zeroes[l-1])
Then multiply it by the third element in the query and add it to the sum.
And this is how you have the sum for the query.

Answer 3

this is in python
tempRes=0
temp = []
for b in range(len(queries)):
for k in range(queries[b][0]-1,queries[b][1]):
if numbers[k] == 0:
tempRes = tempRes + queries[b][2]
tempRes = tempRes + numbers[k]
temp.append(tempRes)
tempRes = 0
return temp

Answer 4

Your code takes a linear time per query. Invest in a linear time once to precompute an array of partial sums. Then each query will be answered in a constant time.

Answer 5

public static List<Long> findSum(List<Integer> numbers, List<List<Integer>> queries) {
// Write your code here
ArrayList<Long> arr = new ArrayList<>();
for(int i = 0; i < queries.size(); i++) {
int sum = 0;
for(int j = queries.get(i).get(0) - 1; j <= queries.get(i).get(1) - 1; j++) {
if(numbers.get(j) == 0){
sum += queries.get(i).get(2);
} else {
sum += numbers.get(j);
}
}
arr.add((long)sum);
sum = 0;
}
return arr;
}

Answer 6

The logic tries to sum each index with the last index.
But initializes the first index with 0.
The same with counting the zero.
After that, just subtract the left side with the desired index.
Also with the zero value in the index, but multiplid by the queries.
There you have it. I can't post the answer.

Answer 7

JS Solution:
There are two methods to solve this problem (Brute force Solution [Nested Loop]) and that's the method that responds with a timeout error throughout the execution of the test, you need to solve it with another time complexity, here is my answer 1st method O(N^2) and then 2nd method Optimized to be O(N) to solving the timeout error.
Answers in JavaScript wish it helps you.
/*
-- Brainstorming --
Have 1-indexed Array [numbers] of length n;
Have numbers of queries (q). [2D Array]
Each Query => [startIdx i, endIdx r, number x]
Find => Sum of numbers between indexes i and r (both included)
if zero occured in range add x instead of it.
*/
// 1st Method: Brute force Solution [Nested Loop]
function findSum(numbers, queries) {
const sum = new Array(queries.length).fill(0);
for (let i = 0; i < queries.length; i++) {
// let j = queries[i][0] - 1 Cause numbers 1-index not 0
for (let j = queries[i][0] - 1; j < queries[i][1]; j++) {
if (numbers[j] === 0) {
sum[i] += queries[i][2];
} else {
sum[i] += numbers[j];
}
}
}
return sum;
}
// 2nd Method: The Optimized Solution Single Loop
function findSum(numbers, queries) {
const sums = [];
const subArraySum = [];
const zerosArr = [];
let zerosCount = 0;
for (let i = 0; i < numbers.length; i++) {
if (numbers[i] === 0) {
zerosCount++;
zerosArr[i] = zerosCount;
} else {
zerosArr[i] = zerosCount;
}
subArraySum[i] = numbers[i] + (subArraySum[i - 1] || 0);
}
for (let q of queries) {
const i = q[0] - 1;
const r = q[1] - 1;
const x = q[2];
let finalSum = subArraySum[r] - (subArraySum[i - 1] || 0) + (zerosArr[r] - (zerosArr[i - 1] || 0)) * x;
sums.push(finalSum);
}
return sums;
}
console.log("1st Test: " + findSum([5,10,10], [[1,2,5]])) // 15
console.log("2nd Test: " + findSum([-5,0], [ [2, 2 ,20] , [1,2,10]])) // 20 , 5
console.log("3rd Test: " + findSum([-1,-1,1,-4,3,-3,-4], [ [1, 4 ,2]])) // -5
console.log("4th Test: " + findSum([1000000000], [[1, 1,100]]))// 1000000000
console.log("5th Test: " + findSum([-1000000000], [[1, 1,100]]))// -1000000000

Answer 8

Try this(in javascript):
function findSum(numbers, queries) {
// Write your code here
let sumArr = [];
for (let i = 0; i < queries.length; i++) {
let sum = 0;
let start = queries[i][0];
let end = queries[i][1] + 1;
let x = queries[i][2];
const newArr = numbers.slice(start, end);
newArr.forEach((item) => {
if (item === 0) {
sum = sum + x;
}
sum = sum + item;
});
sumArr.push(sum);
}
return sumArr;
}

Answer 9

#!/usr/bin/python3
def solution(numbers, queries):
if not isinstance(numbers, list) and not isinstance(queries, list):
return
result = []
total = 0
for query in queries:
if len(query) == 3:
start = query[0]
end = query[1]
additional = query[2]
selection = numbers[start-1:end]
has_zeros = False
for el in selection:
if el == 0:
has_zeros = True
print(f"has_zeros: {has_zeros}")
print(f"selection: {selection}")
total = sum(selection)
if has_zeros:
total += additional
result.append(total)
return result
if "__main__" == __name__:
# 15
_input = [5, 10, 10]
_queries = [[1], [3], [1, 2, 5]]
print(f"input: {_input}")
print(f"queries: {_queries}")
_output = solution(_input, _queries)
print(f"ouput: {_output}")

Answer 10

Solution in python3 :
def findSum(numbers, queries):
a = [0]
b = [0]
for x in numbers:
a.append(a[-1] + x)
b.append(b[-1] + (x == 0)) if in subarray has 0, sum = sum + number x)
return [a[r] - a[l - 1] + x * (b[r] - b[l - 1]) for l, r, x in queries]