Let A be a non-empty array consisting of N integers.
The abs sum of two for a pair of indices (P, Q) is the absolute value |A[P] + A[Q]|, for 0 ≤ P ≤ Q < N.
For example, the following array A:
A[0] = 1 A[1] = 4 A[2] = -3has pairs of indices (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).
The abs sum of two for the pair (0, 0) is A[0] + A[0] = |1 + 1| = 2.
The abs sum of two for the pair (0, 1) is A[0] + A[1] = |1 + 4| = 5.
The abs sum of two for the pair (0, 2) is A[0] + A[2] = |1 + (−3)| = 2.
The abs sum of two for the pair (1, 1) is A[1] + A[1] = |4 + 4| = 8.
The abs sum of two for the pair (1, 2) is A[1] + A[2] = |4 + (−3)| = 1.
The abs sum of two for the pair (2, 2) is A[2] + A[2] = |(−3) + (−3)| = 6.
Write a function:
function solution(A);
that, given a non-empty array A consisting of N integers, returns the minimal abs sum of two for any pair of indices in this array.
For example, given the following array A:
A[0] = 1 A[1] = 4 A[2] = -3the function should return 1, as explained above.
Given array A:
A[0] = -8 A[1] = 4 A[2] = 5 A[3] =-10 A[4] = 3the function should return |(−8) + 5| = 3.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..100,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].
Invalid result type, integer expected, 'undefined' found Perhaps you are missing a 'return'?stdout:
[ 1, -3, 4 ]
Invalid result type, integer expected, 'undefined' found Perhaps you are missing a 'return'?stdout:
[ 3, 4, 5, -8, -10 ]
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const N = A.length;
A.sort((a, b) => Math.abs(a) - Math.abs(b));
let min = Math.abs(A[0] + A[0]);
for (let i = 0; i < N - 1; i++) {
const current = Math.abs(A[i] + A[i + 1])
}
}
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const N = A.length;
A.sort((a, b) => Math.abs(a) - Math.abs(b));
let min = Math.abs(A[0] + A[0]);
for (let i = 0; i < N - 1; i++) {
const current = Math.abs(A[i] + A[i + 1]);
if (current < min) {
min = current;
}
}
return current;
}
ReferenceError: current is not defined at solution (solution.js:18:5) at solutionWrapper (/tmp/exec.js:402:28) at Promise.resolve.then (/tmp/exec.js:428:24) at <anonymous> at process._tickCallback (internal/process/next_tick.js:188:7) at Function.Module.runMain (module.js:686:11) at startup (bootstrap_node.js:187:16) at bootstrap_node.js:608:3
ReferenceError: current is not defined at solution (solution.js:18:5) at solutionWrapper (/tmp/exec.js:402:28) at Promise.resolve.then (/tmp/exec.js:428:24) at <anonymous> at process._tickCallback (internal/process/next_tick.js:188:7) at Function.Module.runMain (module.js:686:11) at startup (bootstrap_node.js:187:16) at bootstrap_node.js:608:3
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const N = A.length;
A.sort((a, b) => Math.abs(a) - Math.abs(b));
let min = Math.abs(A[0] + A[0]);
for (let i = 0; i < N - 1; i++) {
const current = Math.abs(A[i] + A[i + 1]);
if (current < min) {
min = current;
}
}
return min;
}
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const N = A.length;
A.sort((a, b) => Math.abs(a) - Math.abs(b));
let min = Math.abs(A[0] + A[0]);
for (let i = 0; i < N - 1; i++) {
const current = Math.abs(A[i] + A[i + 1]);
if (current < min) {
min = current;
}
}
return min;
}
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const N = A.length;
A.sort((a, b) => Math.abs(a) - Math.abs(b));
let min = Math.abs(A[0] + A[0]);
for (let i = 0; i < N - 1; i++) {
const current = Math.abs(A[i] + A[i + 1]);
if (current < min) {
min = current;
}
}
return min;
}
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');
function solution(A) {
const N = A.length;
A.sort((a, b) => Math.abs(a) - Math.abs(b));
let min = Math.abs(A[0] + A[0]);
for (let i = 0; i < N - 1; i++) {
const current = Math.abs(A[i] + A[i + 1]);
if (current < min) {
min = current;
}
}
return min;
}
The solution obtained perfect score.