Finding closest sum of numbers to a given number

Question

Say I have a list [1,2,3,4,5,6,7] and I would like to find the closest sum of numbers to a given number. Sorry for the crappy explanation but here\'s an example:

Say I h

Answer 1

This proposal generates all possible combinations, collects them in an object which takes the sum as key and filters then the closest sum to the given value.

function getSum(array, sum) {
    function add(a, b) { return a + b; }

    function c(left, right) {
        var s = right.reduce(add, 0);
        if (s > sum) {
            return;
        }
        if (!result.length || s === result[0].reduce(add, 0)) {
            result.push(right);
        } else if (s > result[0].reduce(add, 0)) {
            result = [right];
        }
        left.forEach(function (a, i) {
            var x = left.slice();
            x.splice(i);
            c(left.slice(0, i), [a].concat(right));
        });
    }

    var result = [];
    c(array, [], 0);
    return result;
}

function print(o) {
    document.write('<pre>' + JSON.stringify(o, 0, 4) + '</pre>');
}

print(getSum([1, 2, 3, 4, 5, 6, 7], 10));
print(getSum([1, 2, 3, 4, 5, 6, 7], 14));
print(getSum([1, 2, 3, 4, 5, 6, 7], 19));

Answer 2

Functions for combine, locationOf, are taken from different answers, written by different authors.

printClosest([0.5,2,4] , 5);
printClosest([1, 2, 3, 4, 5, 6, 7], 28);
printClosest([1, 2, 3, 4, 5, 6, 7], 10.9);
printClosest([1, 2, 3, 4, 5, 6, 7], 10, 2);
printClosest([1, 2, 3, 4, 5, 6, 7], 10, 3);
printClosest([1, 2, 3, 4, 5, 6, 7], 14, 2);

function printClosest(array, value, limit) {
  var checkLength = function(array) {
    return array.length === limit;
  };
  var combinations = combine(array); //get all combinations
  combinations = limit ? combinations.filter(checkLength) : combinations;//limit length if required
  var sum = combinations.map(function(c) { //create an array with sum of combinations
    return c.reduce(function(p, c) {
      return p + c;
    }, 0)
  });
  var sumSorted = sum.slice(0).sort(function(a, b) {//sort sum array
    return a - b;
  });

  index = locationOf(value, sumSorted);//find where the value fits in
  //index = (Math.abs(value - sum[index]) <= Math.abs(value - sum[index + 1])) ? index : index + 1;
  index = index >= sum.length ? sum.length - 1 : index;
  index = sum.indexOf(sumSorted[index]);//get the respective combination

  console.log(sum, combinations, index);

  document.getElementById("result").innerHTML += "value : " + value + " combi: " + combinations[index].toString() + " (limit : " + (limit || "none") + ")<br>";
}


function combine(a) {
  var fn = function(n, src, got, all) {
    if (n == 0) {
      if (got.length > 0) {
        all[all.length] = got;
      }
      return;
    }
    for (var j = 0; j < src.length; j++) {
      fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
    }
    return;
  }
  var all = [];
  for (var i = 0; i < a.length; i++) {
    fn(i, a, [], all);
  }
  all.push(a);
  return all;
}

function locationOf(element, array, start, end) {
  start = start || 0;
  end = end || array.length;
  var pivot = parseInt(start + (end - start) / 2, 10);
  if (end - start <= 1 || array[pivot] === element) return pivot;
  if (array[pivot] < element) {
    return locationOf(element, array, pivot, end);
  } else {
    return locationOf(element, array, start, pivot);
  }
}

<pre id="result"><pre>

Answer 3

From what I understood from your question, I made this snippet. I assumed you did not wanted to have the same digit twice (e.g 14 => 7 + 7).

It is working with your examples.

var arr = [1, 2, 3, 4, 5, 6, 7];

var a = 0, b = 0;
var nb = 14;

for(var i in arr) {
  for(var j in arr) {
    if(i != j) {
      var tmp = arr[i] + arr[j];
      if(tmp <= nb && tmp > a + b) {
        a = arr[i];
        b = arr[j];
      }
    }
  }
}

document.write("Closest to " + nb + " => " + a + " + " + b);

Answer 4

 var data= [1, 2, 3,4,5,6,7];
  var closest = 14;
 for (var x = 0; x < data.length; x++) {
    for (var y = x+1; y < data.length; y++) {
             if(data[x] + data[y] == closet){
                     alert(data[x].toString() + "  " + data[y].toString());
                }
            }
       }

Answer 5

I have a little bit long winded solution to the problem just so it is easier to see what is done.

The main benefits with solution below:

• The second loop will not start from beginning of the array again. What I mean that instead of having loop_boundary for second loop as 0 as you normally would, here it starts from next index. This helps if your numbers array is long. However, if it as short as in example, the impact in performance is minimal. Decreasing first loop's boundary by one will prevent errors from happening.
• Works even when the wanted number is 1 or negative numbers.

Fiddle:

JSFiddle

The code:

var numbers = [1,2,3,4,5,6,7];
var wanted_number = 1;
var closest_range, closest1, closest2 = null;

var loop1_boundary = numbers.length-1;
for(var i=0; i<loop1_boundary; i++) {

    var start_index = i+1;
    var loop2_boundary = numbers.length;

    for(var k=start_index; k<loop2_boundary; k++) {

        var number1 = parseInt(numbers[i]);
        var number2 = parseInt(numbers[k]);

        var sum = number1 + number2;
        var range = wanted_number - sum;

        document.write( number1+' + '+number2 +' < '+closest_range+'<br/>' );

        if(Math.abs(range) < Math.abs(closest_range) || closest_range == null ) {
            closest_range = range;
            closest1 = number1;
            closest2 = number2;
        }

    }

    if(range==0){
        break;
    }
}

document.write( 'closest to given number was '+closest1+' and '+closest2+'. The range from wanted number is '+closest_range );