I am trying to write an efficient algorithm in JavaScript to solve this task. Please see the next examples of input data and correct results:
Array: [ [-3,-4], [1,2,-3] ] Result: (-4)*(-3) = 12
Array: [ [1,-1], [2,3], [10,-100,20] ] Result: (-1)*3*(-100) = 300
Array: [ [-3,-15], [-3,-7], [-5,1,-2,-7] ] Result: (-15)*(-7)*1 = 105
It can be any number of sub-arrays and any number of elements in each sub-array. What I already found is that I probably should leave only min and max values in the each sub-array, I did it using .map(a => [Math.min(...a), Math.max(...a)])
and sort them using .sort((a, b) => a[0] - b[0])
.
And now I am stuck. Probably there is a way to calculate all possible products but I am sure that it's not an effective way to solve this task.
Please help!
(Generally, for two numbers whose sum is n, the largest product is given by (n/2)2, for three numbers whose sum is n, the largest product is given by (n/3)3... The nearest integer to (n/e) where e = 2.718 is the number of numbers which will give the maximum product.)
Input: nums = [3,4,5,2] Output: 12 Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.
Suppose we have an integer array called nums, we have to find the contiguous subarray within an array (containing at least one number) which has the largest product. So if the array is [2,3,-2,4], the output will be 6, as contiguous subarray [2,3] has max product.
The problem you post can be solved with a simple algorithm. We just need to keep tracking the maximum/minimum when iterating over each sub-array. We can keep finding the next maximum/minimum by multiplying the current maximum/minimum with the max/min value in each sub-array. We pick the maximum when the iterating is over. Its time complexity is O(n)
where n
is total number of elements in an array (i.e. sum of number of elements in each sub-array).
Here's the complete code. find_maximum_product
function keeps tracking the minimum/maximum and returns the maximum eventually, and it also keeps tracking the multipliers and return it:
/**
* arr: array of any number of sub-arrays and
* any number of elements in each sub-array.
* e.g. [[1, -1], [2, 3], [10, -100, 20]]
*/
function find_maximum_product(arr) {
let max = 1;
let min = 1;
let max_multipliers = [];
let min_multipliers = [];
for (let i = 0; i < arr.length; i++) {
const a = Math.max(...arr[i]);
const b = Math.min(...arr[i]);
const candidates = [max * a, max * b, min * a, min * b];
max = Math.max(...candidates);
min = Math.min(...candidates);
let new_max_multipliers;
let new_min_multipliers;
switch (max) {
case candidates[0]:
new_max_multipliers = max_multipliers.concat(a);
break;
case candidates[1]:
new_max_multipliers = max_multipliers.concat(b);
break;
case candidates[2]:
new_max_multipliers = min_multipliers.concat(a);
break;
case candidates[3]:
new_max_multipliers = min_multipliers.concat(b);
break;
}
switch (min) {
case candidates[0]:
new_min_multipliers = max_multipliers.concat(a);
break;
case candidates[1]:
new_min_multipliers = max_multipliers.concat(b);
break;
case candidates[2]:
new_min_multipliers = min_multipliers.concat(a);
break;
case candidates[3]:
new_min_multipliers = min_multipliers.concat(b);
break;
}
max_multipliers = new_max_multipliers;
min_multipliers = new_min_multipliers;
}
if (max >= min) {
return [max, max_multipliers];
}
return [min, min_multipliers];
}
const arrays = [
[
[-3, -4],
[1, 2, -3],
],
[
[1, -1],
[2, 3],
[10, -100, 20],
],
[
[-3, -15],
[-3, -7],
[-5, 1, -2, -7],
],
[
[14, 2],
[0, -16],
[-12, -16],
],
[
[-20, -4, -19, -18],
[0, -15, -10],
[-13, 4],
],
[
[-2, -15, -12, -8, -16],
[-4, -15, -7],
[-10, -5],
],
];
for (let i = 0; i < arrays.length; i++) {
const [max, max_multipliers] = find_maximum_product(arrays[i]);
console.log('Array:', JSON.stringify(arrays[i]));
console.log('Result:', `${max_multipliers.join(' * ')} = ${max}`);
console.log('');
}
UPDATE
Simpler version for just getting the maximum, not getting the multipliers:
/**
* arr: array of any number of sub-arrays and
* any number of elements in each sub-array.
* e.g. [[1, -1], [2, 3], [10, -100, 20]]
*/
function get_maximum_product(arr) {
return arr
.map((a) => [Math.min(...a), Math.max(...a)])
.reduce(
(acc, current) => {
const candidates = [
acc[0] * current[0],
acc[0] * current[1],
acc[1] * current[0],
acc[1] * current[1],
];
return [Math.min(...candidates), Math.max(...candidates)];
},
[1, 1]
)[1];
}
Here is a top-down recurrence that could be adapted to bottom-up (a loop) and utilises O(n)
search space.
Until I can complete it, the reader is encouraged to add a third return value in the tuple, largest_non_positive
for that special case.
// Returns [highest positive, lowest negative]
// Does not address highest non-positive
function f(A, i){
const high = Math.max(...A[i]);
const low = Math.min(...A[i]);
if (i == 0){
if (low < 0 && high >= 0)
return [high, low];
if (low <= 0 && high <= 0)
return [-Infinity, low];
if (low >= 0 && high >= 0)
return [high, Infinity];
}
const [pos, neg] = f(A, i - 1);
function maybeZero(prod){
return isNaN(prod) ? 0 : prod;
}
let hp = maybeZero(high * pos);
let hn = maybeZero(high * neg);
let ln = maybeZero(low * neg);
let lp = maybeZero(low * pos);
if (low < 0 && high >= 0)
return [Math.max(hp, ln), Math.min(hn, lp)];
if (low <= 0 && high <= 0)
return [ln, lp];
if (low >= 0 && high >= 0)
return [hp, hn];
}
var As = [
[[-3,-4], [1,2,-3]],
[[1,-1], [2,3], [10,-100,20]],
[[-3,-15], [-3,-7], [-5,1,-2,-7]],
[[-11,-6], [-20,-20], [18,-4], [-20,1]],
[[-1000,1], [-1,1], [-1,1], [-1,1]],
[[14,2], [0,-16], [-12,-16]],
[[-20, -4, -19, -18], [0, -15, -10],[-13, 4]]
];
for (let A of As){
console.log(JSON.stringify(A));
console.log(f(A, A.length - 1)[0]);
console.log('');
}
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