I have an array with elements structured like this:
var x = 17;
var arr = [{ value:2, quantity: 4 }, { value:8, quantity: 1 }, { value:3, quantity: 3 }];
How can I identify the indexes of elements that would add up to equal the x number? For example, in this case the desired output would be:
[1, 3, 3, 3]
The goal is to have the lowest length possible for the returned array, such as [0, 0, 0, 1, 2] or [0, 0, 0, 0, 2, 2, 2].
While there are more efficient methods available, the following solution is straightforward and clean. The approach taken treats this problem akin to a linear equation where the combo variable holds the coefficients for each value in the arr array:
// define initial x and arr values
var x = 17;
var arr = [{ value:2, quantity: 4 }, { value:8, quantity: 1 }, { value:3, quantity: 3 }];
// maximums[i] represents the maximum amount of arr[i]'s that can be present in any combination
var maximums = arr.map(function(item){ return Math.floor(x / item.value) });
// an array to store the current coefficients being tested. combo[i]
// corresponds to the coefficient for arr[i]
// all coefficients start at 0 and increment by 1 progressively
var combo = arr.map(function(){ return 0 });
// keep track of the current sum obtained with the coefficients
var sum = 0;
// iterate through the coefficients from left to right, stopping
// when a coefficient exceed its corresponding max value
function next() {
for(var i = 0; i < combo.length; i++) {
combo[i]++;
// adjust the sum due to coefficient increase
sum += arr[i].value;
if(combo[i] <= maximums[i]) {
// stop since coefficient is within range
break;
}else{
// reached maximum for one coeff., reset it and update sum
if(i == combo.length-1) return false;
sum -= arr[i].value * combo[i];
combo[i] = 0;
}
}
return true;
}
// find all combinations until sum matches x
while(next()) {
if(sum == x) break;
}
// if no valid combination is found, halt execution
if(sum != x) {
console.log('not possible');
// otherwise print the successful combination
}else{
for(var i = 0; i < combo.length; i++) {
if(combo[i] == 0) continue;
console.log(combo[i] + ' of (' + JSON.stringify(arr[i]) + ')');
}
}
If aiming for the smallest possible combination always, utilize this alternative function:
function coeffsUsed() {
var count = 0;
for(var i = 0; i < combo.length; i++) {
if(combo[i] > 0) count++;
}
return count;
}
// iterate through all combinations to reach the target sum x
var smallestCombo = {};
var smallest = -1;
while(next()) {
if(sum == x) {
var count = coeffsUsed();
if(smallest == -1 || count < smallest) {
smallest = count;
smallestCombo = combo.slice(0); // clone the array
}
}
}
// if no optimal combination is found, exit
if(smallest == -1) {
console.log('not possible');
// otherwise print the most efficient combination
}else{
for(var i = 0; i < smallestCombo.length; i++) {
if(smallestCombo[i] == 0) continue;
console.log(smallestCombo[i] + ' of (' + JSON.stringify(arr[i]) + ')');
}
}
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