A = []
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
function solution(A) {
 // write your code in JavaScript (Node.js 8.9.4)
 if (A.length > 0) {
 let defArr = [];
 let l = [];
 let all = A.reduce((a, b) => {
 l.push(a)
 return a + b
 })
 l.forEach((item) => {
 defArr.push(Math.abs(all - (item + item)))
 })
 return Math.min(...defArr)
 }
 return 0;
}
console.log(solution(A))

#Problem A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3

We can split this tape in four places:

P = 1, difference = |3 − 10| = 7
P = 2, difference = |4 − 9| = 5
P = 3, difference = |6 − 7| = 1
P = 4, difference = |10 − 3| = 7

Write a function:

function solution(A);

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:
 A[0] = 3
 A[1] = 1
 A[2] = 2
 A[3] = 4
 A[4] = 3
the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [2..100,000]; each element of array A is an integer within the range [−1,000..1,000].

#My solution:

A = []
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
function solution(A) {
 // write your code in JavaScript (Node.js 8.9.4)
 if (A.length > 0) {
 let defArr = [];
 let l = [];
 let all = A.reduce((a, b) => {
 l.push(a)
 return a + b
 })
 l.forEach((item) => {
 defArr.push(Math.abs(all - (item + item)))
 })
 return Math.min(...defArr)
 }
 return 0;
}
console.log(solution(A))

Problem

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3

We can split this tape in four places:

P = 1, difference = |3 − 10| = 7
P = 2, difference = |4 − 9| = 5
P = 3, difference = |6 − 7| = 1
P = 4, difference = |10 − 3| = 7

Write a function:

function solution(A);

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:
 A[0] = 3
 A[1] = 1
 A[2] = 2
 A[3] = 4
 A[4] = 3
the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [2..100,000]; each element of array A is an integer within the range [−1,000..1,000].

My solution:

A = []
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
function solution(A) {
 // write your code in JavaScript (Node.js 8.9.4)
 if (A.length > 0) {
 let defArr = [];
 let l = [];
 let all = A.reduce((a, b) => {
 l.push(a)
 return a + b
 })
 l.forEach((item) => {
 defArr.push(Math.abs(all - (item + item)))
 })
 return Math.min(...defArr)
 }
 return 0;
}
console.log(solution(A))

A = []
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
function solution(A) {
 // write your code in JavaScript (Node.js 8.9.4)
 if (A.length > 0) {
 let defArr = [];
 let l = [];
 let all = A.reduce((a, b) => {
 l.push(a)
 return a + b
 })
 l.forEach((item) => {
 defArr.push(Math.abs(all - (item + item)))
 })
 return Math.min(...defArr)
 }
 return 0;
}
console.log(solution(A))

function solution(A) {
 // write your code in JavaScript (Node.js 8.9.4)
 if(A.length > 0) {
 let defArr = [];
 let l = [];
 let all = A.reduce((a,b) => {
 l.push(a)
 return a + b
 })
 
 l.forEach((item) => {
 defArr.push(Math.abs(all - (item + item)))
 })
 return Math.min(...defArr)
 }
 
 return 0;
 
}